Happy MMXIV !! For those not good with Roman Numerals,
M = 1,000
X = 10
I = 1
V = 5
Not used in 2014 are some commonly used Roman Numerals
C = 100
L = 50
Remember IV = 4, IX = 10, CM = 900
Monday, December 30, 2013
Monday, December 23, 2013
Thursday, December 19, 2013
There are a few ways to remember how to factor the sum and difference of cubes. First, remember all the perfect cubes from 1 to 10. Those cubes are as follows
1, 8, 27, 64, 125, 216, 343, 512, 729, 1000
The formulas for factoring sum and difference of cubes are as follows:
a^3+ b^3 = (a+b)(a^2 - ab + b^2)
a^3 - b^3 = (a-b)(a^2 + ab + b^2)
You can remember the terms by thinking of a and b in the first set of parentheses then squaring a, multiplying a and b and squaring b. The signs between terms for sum of cubes is same, negative, positive and for difference of cubes it's negative, positive, positive.
You can remember the signs another way.. Same, Opposite, Always Positive. If you remember the word SOAP it will help you get the correct signs when factoring.
1, 8, 27, 64, 125, 216, 343, 512, 729, 1000
The formulas for factoring sum and difference of cubes are as follows:
a^3+ b^3 = (a+b)(a^2 - ab + b^2)
a^3 - b^3 = (a-b)(a^2 + ab + b^2)
You can remember the terms by thinking of a and b in the first set of parentheses then squaring a, multiplying a and b and squaring b. The signs between terms for sum of cubes is same, negative, positive and for difference of cubes it's negative, positive, positive.
You can remember the signs another way.. Same, Opposite, Always Positive. If you remember the word SOAP it will help you get the correct signs when factoring.
Thursday, December 12, 2013
When verifying trigonometric identities, it's generally a good idea to get everything in terms of sine and cosine. That's not always the case, but many times it's helpful.
Remember that sin^2(x) + cos^2(x) = 1
1 + tan^2(x) = sec^2(x)
1 + cot^2(x) = csc^2(x)
csc(x) = 1/sin(x)
sec(x)= 1/cos(x)
tan(x) = sin(x)/cos(x)
cot(x) = 1/tan(x) = cos(x)/sin(x)
Because all the trig functions are based off of sine and cosine, you can clearly see why this would be a benefit when verifying.
It's also a benefit when solving trigonometric equations.
Remember that sin^2(x) + cos^2(x) = 1
1 + tan^2(x) = sec^2(x)
1 + cot^2(x) = csc^2(x)
csc(x) = 1/sin(x)
sec(x)= 1/cos(x)
tan(x) = sin(x)/cos(x)
cot(x) = 1/tan(x) = cos(x)/sin(x)
Because all the trig functions are based off of sine and cosine, you can clearly see why this would be a benefit when verifying.
It's also a benefit when solving trigonometric equations.
Saturday, December 7, 2013
Suppose you flip a coin 6 times and want to know the probability of obtaining 4 or more heads in the 6 tosses. This is a binomial distribution problem with n = 6, x = 4 and p = .5.
Suppose you flip a coin and want to know what the probability that the first head you obtain is on the 3 third roll. This is a geometric probability distribution problem.
Whenever you want to know the probability of some event occurring for the first time on the xth trial, you have a geometric probability distribution problem.
Suppose you flip a coin and want to know what the probability that the first head you obtain is on the 3 third roll. This is a geometric probability distribution problem.
Whenever you want to know the probability of some event occurring for the first time on the xth trial, you have a geometric probability distribution problem.
Monday, December 2, 2013
Remember with conditional probability, such as probability of event A given event B is defined as follows:
P(B/A) = P(A and B)/P(A)
P(A and B) = P(A)P(B) if events A and B are independent, therefore if A and B are independent, P(B/A) = P(B) and P(A/B) = P(A)
If A and B are not independent then P(A and B) = P(A/B)P(B)
Saturday, November 30, 2013
Tips for taking the SAT math
Are you a struggling math student preparing for the SAT's? Are you a
parent who has a child who has struggled with the SAT's? I have been
helping students prepare for the math portion of the SAT for several
years and have tips to help students conquer SAT math.
First, I suggest students purchase "The Official SAT Study Guide" , which has practice exams exactly like the real test, as well as exam topics to review. The math foundation needed for the SAT is a solid knowledge of concepts in algebra, geometry, statistics, graph reading and word problem skills.
This test is broken down into two sections, multiple choice and free response. The multiple choice section is scored so that any question answered incorrectly deducts 1/4 of a point, therefore guessing is not wise unless the possible choices can be narrowed to two or three. There is no penalty for incorrect answers on the free response questions, so guessing when not knowing an answer on these problems is encouraged.
When encountering a question about factoring, if you do not know how to factor, you can look at the answers, multiply them and the one gives the original problem is the correct answer. Some questions are all in terms of variables. In these situations, substituting numbers for the variable is a good option. Work the problem with these substituted value and check your answer with the answers with your numbers substituted. Choose numbers a few times to make sure you get the same result each time to confirm your correct answer.
Another tip when dealing with word problems write an algebraic equation for the problem while reading it. For example, if the question is "What number is four more than three times that number?", the equation is x - 4 = 3x and is found as follows: What number (unknown x) is four more than (subtract 4 and set equal to) three times that number (3 times x).
Usually the mathematical operations involved in solving problems on the SAT is not difficult. The problem often lies in understanding what the question is asking. Brush up on algebra concepts, such as factoring, distributing, exponents, functions, graphs of equations. Geometric concepts to review include areas of polygons, three dimensional figures, volume and Pythagorean theorem. Statistical concepts to review include simple probability, mean, median and mode. It's important to utilize the formulas given in the beginning of the section.
The exam is timed, so complete the questions that you know and come back to those that are difficult, if time permits. Answer all the easy questions correctly to get a decent score. Get some of the medium difficulty questions correct and do not guess on the hard questions. The questions are arranged from easiest to hardest.
If you took the test previously, get your exam results which shows your answers, correct answers, the type of question (algebra, geometry, statistics, etc) and the difficulty of the question. It will give you something to refer to so you know what areas to study for the next time around. Contact your school to find SAT tutors in your area or search for a tutor on one of many tutoring websites. Finally, do not get overwhelmed with the material. A positive attitude is another key to success. Don't have yourself defeated before you take the exam. Good luck!
First, I suggest students purchase "The Official SAT Study Guide" , which has practice exams exactly like the real test, as well as exam topics to review. The math foundation needed for the SAT is a solid knowledge of concepts in algebra, geometry, statistics, graph reading and word problem skills.
This test is broken down into two sections, multiple choice and free response. The multiple choice section is scored so that any question answered incorrectly deducts 1/4 of a point, therefore guessing is not wise unless the possible choices can be narrowed to two or three. There is no penalty for incorrect answers on the free response questions, so guessing when not knowing an answer on these problems is encouraged.
When encountering a question about factoring, if you do not know how to factor, you can look at the answers, multiply them and the one gives the original problem is the correct answer. Some questions are all in terms of variables. In these situations, substituting numbers for the variable is a good option. Work the problem with these substituted value and check your answer with the answers with your numbers substituted. Choose numbers a few times to make sure you get the same result each time to confirm your correct answer.
Another tip when dealing with word problems write an algebraic equation for the problem while reading it. For example, if the question is "What number is four more than three times that number?", the equation is x - 4 = 3x and is found as follows: What number (unknown x) is four more than (subtract 4 and set equal to) three times that number (3 times x).
Usually the mathematical operations involved in solving problems on the SAT is not difficult. The problem often lies in understanding what the question is asking. Brush up on algebra concepts, such as factoring, distributing, exponents, functions, graphs of equations. Geometric concepts to review include areas of polygons, three dimensional figures, volume and Pythagorean theorem. Statistical concepts to review include simple probability, mean, median and mode. It's important to utilize the formulas given in the beginning of the section.
The exam is timed, so complete the questions that you know and come back to those that are difficult, if time permits. Answer all the easy questions correctly to get a decent score. Get some of the medium difficulty questions correct and do not guess on the hard questions. The questions are arranged from easiest to hardest.
If you took the test previously, get your exam results which shows your answers, correct answers, the type of question (algebra, geometry, statistics, etc) and the difficulty of the question. It will give you something to refer to so you know what areas to study for the next time around. Contact your school to find SAT tutors in your area or search for a tutor on one of many tutoring websites. Finally, do not get overwhelmed with the material. A positive attitude is another key to success. Don't have yourself defeated before you take the exam. Good luck!
Saturday, November 23, 2013
When solving a system of linear equations, you can use substitution, elimination to solve. But you also set up the system with matrices and solve using Cramer's rule.
For example, the system 2x + 3y = 10
-3x - 6y = 22
Can be solved as follows:
matrix A = 2 3
-3 -6
Get the determinant of A = (2)(-6) - (-3)(3) = -3
Now to solve for x, set up another matrix, substituting 10 for 2 in the matrix
22 -3
the new matrix looks like 10 3
22 -3
Get the determinant (10)(-3) - (22)(3) = -30 - 66 = -86
x is the determinant x divided by determinant A = -86/-3 = 86/3
Now do the same thing to get y
the new matrix is 2 10
-3 22
Determinant is (2)(22) - (-3)(10) = 44 + 30 = 74
y is 74/-3 = -74/3
For example, the system 2x + 3y = 10
-3x - 6y = 22
Can be solved as follows:
matrix A = 2 3
-3 -6
Get the determinant of A = (2)(-6) - (-3)(3) = -3
Now to solve for x, set up another matrix, substituting 10 for 2 in the matrix
22 -3
the new matrix looks like 10 3
22 -3
Get the determinant (10)(-3) - (22)(3) = -30 - 66 = -86
x is the determinant x divided by determinant A = -86/-3 = 86/3
Now do the same thing to get y
the new matrix is 2 10
-3 22
Determinant is (2)(22) - (-3)(10) = 44 + 30 = 74
y is 74/-3 = -74/3
Tuesday, November 19, 2013
It's important to understand the definitions of conic sections. Here's the definition of a hyperbola and a parabola.
The conic section defined by all the points where the difference of the distance from any point drawn to the foci is constant is a hyperbola.
The conic section comprised of all the points that are equidistant from a focus point and a directrix line is a parabola.
The conic section defined by all the points where the difference of the distance from any point drawn to the foci is constant is a hyperbola.
The conic section comprised of all the points that are equidistant from a focus point and a directrix line is a parabola.
Saturday, November 9, 2013
Notice the relationship between the product of the 9's tables from to 10 and the sum of the digits of each product. The sums of the digits of each product equals 9. Does anyone know why this is the case?
1 X 9 = 9 0 + 9 = 9
2 X 9 = 18 1 + 8 = 9
3 X 9 = 27 2 + 7 = 9
4 X 9 = 36 3 + 6 = 9
5 X 9 = 45 4 + 5 = 9
6 X 9 = 54 5 + 4 = 9
7 X 9 = 63 6 + 3 = 9
8 X 9 = 72 7 + 2 = 9
9 X 9 = 81 8 + 1 = 9
10 X 9 = 90 9 + 0 = 9
1 X 9 = 9 0 + 9 = 9
2 X 9 = 18 1 + 8 = 9
3 X 9 = 27 2 + 7 = 9
4 X 9 = 36 3 + 6 = 9
5 X 9 = 45 4 + 5 = 9
6 X 9 = 54 5 + 4 = 9
7 X 9 = 63 6 + 3 = 9
8 X 9 = 72 7 + 2 = 9
9 X 9 = 81 8 + 1 = 9
10 X 9 = 90 9 + 0 = 9
Sunday, November 3, 2013
This may be a repeat post from a long time ago, but it's a topic of importance, knowing how to simplify square roots.
In a first year algebra class, students will encounter problems involving simplifying square roots. But many times the teacher simply doesn't explain the process by which this is accomplished in a manner that students can understand. I will explain a way that will make simplifying square roots easier for students of any ability level.
The first method I use when teaching students how to calculate square root is to look to see if the number is a perfect square first. I suggest students memorize the perfect squares from 1 to 25 as follows: 1, 4, 9, 16, 25, 49, 64, 81, 100, 121, 144, 169, 196, 225, 256, 289, 324, 361, 400, 441, 484, 529, 576, 625. So if you are asked to calculate the square root of 441, you automatically know the answer is 21 or -21, since a negative number times another negative number yields a positive number.
For non-perfect squares or larger numbers that you are unsure of, if the number is a perfect square, I suggest using a factor tree. For example, suppose you want to simplify a square root of 48. It is not a perfect square since 6 times 6 equals 36 and 7 times 7 equals 49. No whole number times itself equals 48. So break it down into factors. I always suggest trying to find a perfect square as one of the factors and in this case, 16 times 3 equals 48 and 16 is a perfect square. Remember: Since you are dealing with square root, the factors are also square root. So the square root of 48 equals the square root of 16 times the square root of 3. Three is a prime number so you cannot break down 3 any farther using a factor tree. We know that the square root of 16 is 4, so the answer is 4 times the square root of 3.
Another more difficult example---say we need to find the square root of 2025. With such a large number, most people won't know if this is a perfect square, so use the factor tree. Know that any number ending in 5 is divisible by 5. So 5 times 405 is 2025. But 405 can be broken down into factors, using the same rule, therefore 5 times 81 is 405. Now we have the square root of 5 times the square root of 5 times the square root of 81. Notice then that the square root of 5 times the square root of 5 equals the square root of 25. Now this problem becomes simple because you notice we have two perfect squares here, 25 and 81. The square root of 25 is 5 and square root of 81 is 9. So the answer is 5 times 9, which is 45 and -45, since -45 times -45 equals 2025. This problem is actually a perfect square, but if you do not recognize it as such, you can use the factor tree method I just described.
When dealing with the square root of a negative number, imaginary numbers come into play. The square root of -1 equals an imaginary number denoted as "i". So in the above problems if we had the square root of -48, you have square root of -1 times root of 3 times square root of 16. The answer is 4i times the square root of 3, and -4i times square root of 3. If we have the square root of -2025, the answer is simply 45i and -45i.
Hope my method helps you when trying to figure out the square root of both positive and negative numbers.
In a first year algebra class, students will encounter problems involving simplifying square roots. But many times the teacher simply doesn't explain the process by which this is accomplished in a manner that students can understand. I will explain a way that will make simplifying square roots easier for students of any ability level.
The first method I use when teaching students how to calculate square root is to look to see if the number is a perfect square first. I suggest students memorize the perfect squares from 1 to 25 as follows: 1, 4, 9, 16, 25, 49, 64, 81, 100, 121, 144, 169, 196, 225, 256, 289, 324, 361, 400, 441, 484, 529, 576, 625. So if you are asked to calculate the square root of 441, you automatically know the answer is 21 or -21, since a negative number times another negative number yields a positive number.
For non-perfect squares or larger numbers that you are unsure of, if the number is a perfect square, I suggest using a factor tree. For example, suppose you want to simplify a square root of 48. It is not a perfect square since 6 times 6 equals 36 and 7 times 7 equals 49. No whole number times itself equals 48. So break it down into factors. I always suggest trying to find a perfect square as one of the factors and in this case, 16 times 3 equals 48 and 16 is a perfect square. Remember: Since you are dealing with square root, the factors are also square root. So the square root of 48 equals the square root of 16 times the square root of 3. Three is a prime number so you cannot break down 3 any farther using a factor tree. We know that the square root of 16 is 4, so the answer is 4 times the square root of 3.
Another more difficult example---say we need to find the square root of 2025. With such a large number, most people won't know if this is a perfect square, so use the factor tree. Know that any number ending in 5 is divisible by 5. So 5 times 405 is 2025. But 405 can be broken down into factors, using the same rule, therefore 5 times 81 is 405. Now we have the square root of 5 times the square root of 5 times the square root of 81. Notice then that the square root of 5 times the square root of 5 equals the square root of 25. Now this problem becomes simple because you notice we have two perfect squares here, 25 and 81. The square root of 25 is 5 and square root of 81 is 9. So the answer is 5 times 9, which is 45 and -45, since -45 times -45 equals 2025. This problem is actually a perfect square, but if you do not recognize it as such, you can use the factor tree method I just described.
When dealing with the square root of a negative number, imaginary numbers come into play. The square root of -1 equals an imaginary number denoted as "i". So in the above problems if we had the square root of -48, you have square root of -1 times root of 3 times square root of 16. The answer is 4i times the square root of 3, and -4i times square root of 3. If we have the square root of -2025, the answer is simply 45i and -45i.
Hope my method helps you when trying to figure out the square root of both positive and negative numbers.
Saturday, October 26, 2013
When trying to find possible roots of an equation, we can start by using the rational root theorem to find all the possible real roots. You can also use Descartes rule of signs to possibly eliminate positive or negative roots.
If you have a cubic equation you can use synthetic division to find a root, which will leave you ith coefficients to a quadratic equation.
From here you can use quadratic formula, complete the square or other methods of factoring to get the remaining roots.
If you have a cubic equation you can use synthetic division to find a root, which will leave you ith coefficients to a quadratic equation.
From here you can use quadratic formula, complete the square or other methods of factoring to get the remaining roots.
Tuesday, October 22, 2013
Remember the important angles on the unit circle, which is a circle with radius 1. The angles are found by using the pythagorean theorem to get the sides of right triangles inscribed inside the unit circle and the trig functions sine, cosine and tangent.
0, 30, 45, 60, 90, 120, 135, 150, 180, 210, 225, 240, 270, 300, 315, 330, 360
These angle measures can be converted to radians by multiplying each by Pi/180.
0, 30, 45, 60, 90, 120, 135, 150, 180, 210, 225, 240, 270, 300, 315, 330, 360
These angle measures can be converted to radians by multiplying each by Pi/180.
Thursday, October 17, 2013
When graphing the any absolute value function, it's important to know the graph of the parent function f(x) = |x|. This looks like a v with the vertex at the origin and slope of -1 from -infinity to 0, and slope of 1 from 0 to infinity.
From here we can graph any of them in this form a|x +/- h| +/- k
If a is positive, the shape is a v, if it's negative it's an upside down v and the value of a determines the slope.
If it's x + h, the graph shifts h units to the left and if it's x - h, it shifts h units to the right.
If it's + k, it shifts k units up and if it's -k, it shifts k units down. The point we are shifting is the vertex.
For example,
f(x) = 4|x + 3| - 2, the vertex moves 3 to the left and 2 down to (3, -2) and the slope is 4 from 0 to infinity and -4 from -infinity to 0.
From here we can graph any of them in this form a|x +/- h| +/- k
If a is positive, the shape is a v, if it's negative it's an upside down v and the value of a determines the slope.
If it's x + h, the graph shifts h units to the left and if it's x - h, it shifts h units to the right.
If it's + k, it shifts k units up and if it's -k, it shifts k units down. The point we are shifting is the vertex.
For example,
f(x) = 4|x + 3| - 2, the vertex moves 3 to the left and 2 down to (3, -2) and the slope is 4 from 0 to infinity and -4 from -infinity to 0.
Thursday, October 10, 2013
You can tell from a residual plot whether or not a model is linear. If the residuals are uniform, then yes. If not, then the linear model does not fit.
For example, if you have 6 residuals and all of them are negative, then that is not uniform. If all are positive, it is not uniform. If half are positive and half are negative and they are about the same distance from the center line on the graph, then the linear model is the correct model.
For example, if you have 6 residuals and all of them are negative, then that is not uniform. If all are positive, it is not uniform. If half are positive and half are negative and they are about the same distance from the center line on the graph, then the linear model is the correct model.
Sunday, October 6, 2013
Some of the most difficult concepts to understand in statistics are the
various types of sampling. Specifically it can be confusing to
distinguish between stratified sampling and cluster sampling. I have
used the following explanations of these sampling techniques during my
13 years experience as a math tutor.
A simple random sample is the most common type of sampling technique used in statistics. However, designs that are used to sample populations across large areas, are more complex than the simple random sample. In some instances, populations are divided into homogeneous groups, called strata. Then a simple random sample is selected from each strata. This kind of sampling is known as stratified sampling.
The question that often arises is, "why would we want to make things more complicated by using a stratified sample?" Suppose we want to learn about fundraising for a high school baseball team. The school is 55% boys and 45% girls, and we expect that boys and girls have different ideas on the fundraising. If a simple random sample is used to choose 200 students, we could possibly get 130 boys and 70 girls or 45 boys and 155 girls. Because of this, the amount of variability could be large. So to reduce the variability, you can sample 55% boys and 45% girls. This kind of "forced representative balance" will ensure that the percentage of boys and girls in the sample is identical to that in the population. This is a better method that the simple random sample because it should give a more accurate representation of the opinion of all the students in the school.
Now suppose we want to find out what the high school freshmen think about the food served in the cafeteria. We could use the simple random sample or stratified sampling but it's too time-consuming to find every student that was selected in the sample. But the freshmen homerooms are all in one of ten rooms on the ground floor of the school. So, we could sample two or three homerooms and sample every student in those homerooms. The population was divided into representative clusters and a few clusters were sampled in their entirety. This type of sampling is called cluster sampling.
What is the difference between stratified and cluster sampling? Clusters are heterogeneous and resemble the population in its entirety. Stratified sampling is done to make sure the sample represents different groups in the population, and samples are taken randomly within each strata. Clusters are chosen to make sampling more affordable or practical for the given situation.
An example which will more clearly display the differences in the two types of sampling is examining a pizza. Suppose you have a professional taster whose job is to check each pizza for quality. Samples need to be eaten from selected pizzas, with the crust, sauce, cheese and toppings tested.
You could taste a slice of pizza as a customer would eat a slice. When doing so, you'll learn about the pizza as a whole. The slice would be a cluster sample since it contains all the ingredients of the pizza.
If you select some tastes of the crust at random, of the cheese at random, of the sauce at random, and of the toppings at random, you will still get a pretty good judgment of the overall quality of the pizza. This kind of sampling would be stratified.
Cluster samples slice across the layers to obtain clusters, while stratified sampling represent the population by selecting some from each layer, which reduces the amount of variability.
This guide should help students better understand the differences between stratified sampling and cluster sampling.
A simple random sample is the most common type of sampling technique used in statistics. However, designs that are used to sample populations across large areas, are more complex than the simple random sample. In some instances, populations are divided into homogeneous groups, called strata. Then a simple random sample is selected from each strata. This kind of sampling is known as stratified sampling.
The question that often arises is, "why would we want to make things more complicated by using a stratified sample?" Suppose we want to learn about fundraising for a high school baseball team. The school is 55% boys and 45% girls, and we expect that boys and girls have different ideas on the fundraising. If a simple random sample is used to choose 200 students, we could possibly get 130 boys and 70 girls or 45 boys and 155 girls. Because of this, the amount of variability could be large. So to reduce the variability, you can sample 55% boys and 45% girls. This kind of "forced representative balance" will ensure that the percentage of boys and girls in the sample is identical to that in the population. This is a better method that the simple random sample because it should give a more accurate representation of the opinion of all the students in the school.
Now suppose we want to find out what the high school freshmen think about the food served in the cafeteria. We could use the simple random sample or stratified sampling but it's too time-consuming to find every student that was selected in the sample. But the freshmen homerooms are all in one of ten rooms on the ground floor of the school. So, we could sample two or three homerooms and sample every student in those homerooms. The population was divided into representative clusters and a few clusters were sampled in their entirety. This type of sampling is called cluster sampling.
What is the difference between stratified and cluster sampling? Clusters are heterogeneous and resemble the population in its entirety. Stratified sampling is done to make sure the sample represents different groups in the population, and samples are taken randomly within each strata. Clusters are chosen to make sampling more affordable or practical for the given situation.
An example which will more clearly display the differences in the two types of sampling is examining a pizza. Suppose you have a professional taster whose job is to check each pizza for quality. Samples need to be eaten from selected pizzas, with the crust, sauce, cheese and toppings tested.
You could taste a slice of pizza as a customer would eat a slice. When doing so, you'll learn about the pizza as a whole. The slice would be a cluster sample since it contains all the ingredients of the pizza.
If you select some tastes of the crust at random, of the cheese at random, of the sauce at random, and of the toppings at random, you will still get a pretty good judgment of the overall quality of the pizza. This kind of sampling would be stratified.
Cluster samples slice across the layers to obtain clusters, while stratified sampling represent the population by selecting some from each layer, which reduces the amount of variability.
This guide should help students better understand the differences between stratified sampling and cluster sampling.
Saturday, October 5, 2013
A common mistake students make when dealing with correlation is assuming
that correlation equals causation. Although putting a regression line
through points on a scatter plot may tempt one to say that the
x-variable causes the y-variable, this is not the case. Other factor or
factors may be driving both variables being observed. This third
variable is known as a lurking variable.
When data is observed, as opposed to data obtained from a designed experiment, there is no way to be certain that a lurking variable is not the cause of an apparent association between the variables.
For example, suppose a scatter plot shows the average life expectancy of men for 20 different countries is plotted against the number of doctors per person in the country. Note that we must check the conditions of correlation before interpreting the correlation. The conditions are that both variables must be quantitative, and life expectancy and number of doctors are both quantitative variables. Second, the pattern of the scatter plot must be quite straight. Since there is no scatter plot shown, for the sake of argument, we will assume this condition is met. Finally, no outliers can be present. Once again, we will assume this condition is met.
Suppose the data shows that there is a strong positive association, r2 = 0.81, between the variables. The seems to confirm what we'd expect that the more doctors per person that a country has, the longer the life expectancy. So, we may think that the countries will lower life expectancies need more doctors. But could there be other factors that increase life expectancy? Yes, of course. One cannot say that only an increase in doctors will cause life expectancy to increase. There are certainly lurking variables involved here.
Suppose more data was taken from these countries and this time we plot average life expectancy against the number of tv's per household in the country. The scatter plot shows a very strong association of r2 = 0.88. Fitting the linear model, we may use the number of tv's as a predictor of life expectancy. But this is an absurd way of thinking. If we just use the r2 value to determine causation, we would think we need to send more tv's to the countries with lower life expectancies than doctors.
What is most likely the cause for higher life expectancy is much more than just doctors per person and tv's per household. Higher living standards may be more of a reason that life expectancy is hire, increase the number of tv's and increase the number of doctors.
It's very easy to use regression and assuming causation from it. But beware that lurking variables may be a cause for an apparent association and that regression never can used to show that one variable causes another.
When data is observed, as opposed to data obtained from a designed experiment, there is no way to be certain that a lurking variable is not the cause of an apparent association between the variables.
For example, suppose a scatter plot shows the average life expectancy of men for 20 different countries is plotted against the number of doctors per person in the country. Note that we must check the conditions of correlation before interpreting the correlation. The conditions are that both variables must be quantitative, and life expectancy and number of doctors are both quantitative variables. Second, the pattern of the scatter plot must be quite straight. Since there is no scatter plot shown, for the sake of argument, we will assume this condition is met. Finally, no outliers can be present. Once again, we will assume this condition is met.
Suppose the data shows that there is a strong positive association, r2 = 0.81, between the variables. The seems to confirm what we'd expect that the more doctors per person that a country has, the longer the life expectancy. So, we may think that the countries will lower life expectancies need more doctors. But could there be other factors that increase life expectancy? Yes, of course. One cannot say that only an increase in doctors will cause life expectancy to increase. There are certainly lurking variables involved here.
Suppose more data was taken from these countries and this time we plot average life expectancy against the number of tv's per household in the country. The scatter plot shows a very strong association of r2 = 0.88. Fitting the linear model, we may use the number of tv's as a predictor of life expectancy. But this is an absurd way of thinking. If we just use the r2 value to determine causation, we would think we need to send more tv's to the countries with lower life expectancies than doctors.
What is most likely the cause for higher life expectancy is much more than just doctors per person and tv's per household. Higher living standards may be more of a reason that life expectancy is hire, increase the number of tv's and increase the number of doctors.
It's very easy to use regression and assuming causation from it. But beware that lurking variables may be a cause for an apparent association and that regression never can used to show that one variable causes another.
Tuesday, October 1, 2013
Linear regression is one of the many courses I studied while
earning my BS in Statistics from Lehigh University. I have been tutoring
statistics for the past 13 years and there is often confusion among
students as to when the linear regression model can be used. Sometimes
the model, although tempting to use, doesn't apply. Certain conditions
and assumptions must be checked. The following paragraphs I explain each
condition and assumption as I would to any of my students over the
years.
The linear regression model has two easily estimated parameters, a significant measure of how well the model fits the data, and it has the ability to predict new values. The first condition is the quantitative variables condition. When a measured variable with units answers questions about the amount or quantity of what is being measured, it is a quantitative variable. Examples of quantitative variables are cost, scores, temperature, height, and weight.
A linear regression model makes many assumptions. First, the relationship between the variables must be linear. This assumption cannot be checked, per say, but can be checked by viewing the scatter plot. The scatter plot will help check the straight enough condition, which means the data in the scatter plot should be straight enough to make sense. For example, if the data shows more of a curved relationship between the variables and you try to use a linear model, stop. You cannot use it, the model won't mean a thing.
To summarize the scatter of the data in the plot by using the standard deviation, all of the residuals should have the same spread, or variance. Therefore, we need the equal variance assumption. Check for changing spread of the scatter plot. If you notice that the spread thickens at any part of the plot, then the does the plot thicken? condition does not hold and the linear model does not apply.
Finally, we have to check for outliers, which are points which are drastically far above or below the rest of the data points. These points can dramatically change a regression model, such as the slope, which can mislead us about the relationship between the variables in the model. Therefore, be certain that the outlier condition is also met.
Although the linear regression model is widely used and a very powerful tool in statistics to predict values, several assumptions and conditions must be check to make sure the model is appropriate. If the model is inappropriate, do not use it. The results will be misleading.
The linear regression model has two easily estimated parameters, a significant measure of how well the model fits the data, and it has the ability to predict new values. The first condition is the quantitative variables condition. When a measured variable with units answers questions about the amount or quantity of what is being measured, it is a quantitative variable. Examples of quantitative variables are cost, scores, temperature, height, and weight.
A linear regression model makes many assumptions. First, the relationship between the variables must be linear. This assumption cannot be checked, per say, but can be checked by viewing the scatter plot. The scatter plot will help check the straight enough condition, which means the data in the scatter plot should be straight enough to make sense. For example, if the data shows more of a curved relationship between the variables and you try to use a linear model, stop. You cannot use it, the model won't mean a thing.
To summarize the scatter of the data in the plot by using the standard deviation, all of the residuals should have the same spread, or variance. Therefore, we need the equal variance assumption. Check for changing spread of the scatter plot. If you notice that the spread thickens at any part of the plot, then the does the plot thicken? condition does not hold and the linear model does not apply.
Finally, we have to check for outliers, which are points which are drastically far above or below the rest of the data points. These points can dramatically change a regression model, such as the slope, which can mislead us about the relationship between the variables in the model. Therefore, be certain that the outlier condition is also met.
Although the linear regression model is widely used and a very powerful tool in statistics to predict values, several assumptions and conditions must be check to make sure the model is appropriate. If the model is inappropriate, do not use it. The results will be misleading.
Wednesday, September 25, 2013
An easy way to graph logarithmic functions, exponential functions and square root functions is to know the basic graph of the parent functions, then shift accordingly.
f(x) = sqrt(x), f(x) - sqrt(x) flips across the x axis, f(x) = sqrt(-x) flips across the y-axis
f(x) = sqrt(x - h) shifts h units to the right, f(x) = sqrt(x +h) shifts h units to the left, f(x) = sqrt(x + h) + k shifts h units left and k units up, f(x) = sqrt(x) - k, shifts k units down.
The shifts and flips are very similar with f(x) = log(x), f(x) = e^x, etc.
f(x) = sqrt(x), f(x) - sqrt(x) flips across the x axis, f(x) = sqrt(-x) flips across the y-axis
f(x) = sqrt(x - h) shifts h units to the right, f(x) = sqrt(x +h) shifts h units to the left, f(x) = sqrt(x + h) + k shifts h units left and k units up, f(x) = sqrt(x) - k, shifts k units down.
The shifts and flips are very similar with f(x) = log(x), f(x) = e^x, etc.
Friday, September 20, 2013
Wednesday, September 18, 2013
There is an easy way to determine the domain of a function. If the denominator is 1 and there are no radicals in the numerator, the domain is all real numbers.
If there is a variable in the denominator and no radical in the denominator, the domain is all real numbers except the value of x that will give a zero in the denominator.
If there is a radical with an even root, square root, fourth root, etc. You have to solve for x such that the value under the radical is greater than or equal to zero. The domain is all real numbers except for that value.
If there is a radical with an odd root, then the domain is all real numbers.
If there is a variable in the denominator and no radical in the denominator, the domain is all real numbers except the value of x that will give a zero in the denominator.
If there is a radical with an even root, square root, fourth root, etc. You have to solve for x such that the value under the radical is greater than or equal to zero. The domain is all real numbers except for that value.
If there is a radical with an odd root, then the domain is all real numbers.
Thursday, September 12, 2013
Remember there are several ways to go about graphing a line.
From the equation you can plot the y-intercept and use the slope to find another point on the line.
For example, for the equation y = 2x + 4, the y-intercept if 4 and the slope is 2. Plot a point at (0,4) and use the slope to move up 2 and to the right 1 and plot another point at (1,6). Now draw a line through the points.
You can also find both intercepts by putting 0 in for x and then 0 in for y. Using the equation above, you get 0 = 2x + 4, therefore x = -2. The x-intercept is -2, so plot a point at (-2,0). Putting a 0 in for x we get y = 2(0) + 4, so the y-intercept is (0,4). Plot that point and draw a line through the points.
You can also pick any 2 values for x and find the corresponding y values. For y = 2x +4, suppose we pick 1 for x, therefore y = 6. Now pick x =2 and you get y = 8. Plot points at (1,6) and (2,8) and draw a line through the points.
From the equation you can plot the y-intercept and use the slope to find another point on the line.
For example, for the equation y = 2x + 4, the y-intercept if 4 and the slope is 2. Plot a point at (0,4) and use the slope to move up 2 and to the right 1 and plot another point at (1,6). Now draw a line through the points.
You can also find both intercepts by putting 0 in for x and then 0 in for y. Using the equation above, you get 0 = 2x + 4, therefore x = -2. The x-intercept is -2, so plot a point at (-2,0). Putting a 0 in for x we get y = 2(0) + 4, so the y-intercept is (0,4). Plot that point and draw a line through the points.
You can also pick any 2 values for x and find the corresponding y values. For y = 2x +4, suppose we pick 1 for x, therefore y = 6. Now pick x =2 and you get y = 8. Plot points at (1,6) and (2,8) and draw a line through the points.
Thursday, September 5, 2013
Here's some good math jokes.
Q: Why do they never serve beer at a math party?
A: Because you can't drink and derive...
Q: Why didn't the quarter roll down the hill with the nickel?
A: Because it had more cents.
Q: Did you hear about the constipated mathematician?
A: He worked it out with a pencil.
Q: How many molecules in a bowl of guacamole?
A: Avacado's Number
Q: What happened to the plant in math class?
A: It grew square roots.
Q: Why did the chicken cross the mobius strip?
A: To get to the same side.
Q: How do you make seven an even number?
A: Take the s out!
Q: Why should the number 288 never be mentioned?
A: It's two gross.
Q: Why couldn't the moebius strip enroll at the school?
A: They required an orientation.
Q: What does a mathematician do about constipation?
A: He works it out with a pencil.
Q: Why is a math book always unhappy?
A: Because it always has lots of problems.
Q: Why don't you do arithmetic in the jungle?
A: Because if you add 4+4 you get ate!
Q: Why did I divide sin by tan?
A: Just cos.
Q: Where do math teachers go on vacation?
A: To Times Square.
Q: Why is 6 afraid of 7?
A: Because 7 8 9
Q: What does the zero say to the the eight?
A: Nice belt!
Q: Why did the mutually exclusive events break up?
A: They had nothing in common.
Q: How is an artificial christmas tree like the fourth root of -68?
A: Neither has real roots.
source: http://www.jokes4us.com/miscellaneousjokes/mathjokes/
source: http://www.jokes4us.com/miscellaneousjokes/mathjokes/
Tuesday, September 3, 2013
Just a question to ponder. In most high school math sequences, algebra is followed by geometry, second year algebra, pre-calculus, (which involves trigonometry), then calculus, or statistics.
Why is not algebra II immediately following algebra I? Geometry and trigonometry work hand in hand, so why not bunch those two back to back following the algebra courses?
Why is not algebra II immediately following algebra I? Geometry and trigonometry work hand in hand, so why not bunch those two back to back following the algebra courses?
Thursday, August 29, 2013
I got a question today about the difference between nominal, ordinal, ratio and interval data.
Nominal
applies to data that consist of names, labels or categories.. Ex:
Atlantic, Pacific, India, Arctic....... names of oceans
Ordinal
applies to data that can be arranged in order.. example: in a class of
320 students, John ranked 25th, Joe ranked 35th, Amy ranked 10th, Julie
ranked 4th
Interval
applies to data that also can be arrange in order but the differences
in order are important, they are not important in ordinal ex: temperature.. you can order temperatures and observe meaningful difference
Ratio:
data is arranged in order and there is a ratio between the values of
the data... ex: length of fish in a river.. 18 inch fish is 3 times the
length of a 6 inch fish.. 6 inch fish is 3 times the length of a 2 inch
fish, etc
Saturday, August 24, 2013
For those kids that like math and wish to boost their skills while having fun solving puzzles, here's a good site. It covers math from 1st grade through 6th grade as well as algebra.
http://www.mathsisfun.com/puzzles/
The specific category of puzzles are
starter, measuring, logic, shape, number, algebra, assorted math puzzles and quizzes, card, tricky, Einstein, Sam Loyd, symmetry jigsaw, puzzle games
Give them a shot and have fun!
http://www.mathsisfun.com/puzzles/
The specific category of puzzles are
starter, measuring, logic, shape, number, algebra, assorted math puzzles and quizzes, card, tricky, Einstein, Sam Loyd, symmetry jigsaw, puzzle games
Give them a shot and have fun!
Tuesday, August 20, 2013
Sunday, August 18, 2013
Enjoy some of these math quotes and jokes!
Mathematics is made of 50 percent formulas, 50 percent proofs, and 50 percent imagination.
"A mathematician is a device for turning coffee into theorems" (P. Erdos)
Addendum: American coffee is good for lemmas.
An engineer thinks that his equations are an approximation to reality. A physicist thinks reality is an approximation to his equations. A mathematician doesn't care.
Old mathematicians never die; they just lose some of their functions.
Mathematicians are like Frenchmen: whatever you say to them, they translate it into their own language, and forthwith it means something entirely different. -- Goethe
Mathematics is the art of giving the same name to different things. -- J. H. Poincare
What is a rigorous definition of rigor?
There is no logical foundation of mathematics, and Gödel has proved it!
Mathematics is made of 50 percent formulas, 50 percent proofs, and 50 percent imagination.
"A mathematician is a device for turning coffee into theorems" (P. Erdos)
Addendum: American coffee is good for lemmas.
An engineer thinks that his equations are an approximation to reality. A physicist thinks reality is an approximation to his equations. A mathematician doesn't care.
Old mathematicians never die; they just lose some of their functions.
Mathematicians are like Frenchmen: whatever you say to them, they translate it into their own language, and forthwith it means something entirely different. -- Goethe
Mathematics is the art of giving the same name to different things. -- J. H. Poincare
What is a rigorous definition of rigor?
There is no logical foundation of mathematics, and Gödel has proved it!
Wednesday, August 14, 2013
the equation for a hyperbola is (x-h)^2/a^2 - (y-k)^2/b^2 = 1 if the transverse axis is horizontal and (y-k)^2/a^2 - (x-h)^2/b^2 = 1 if the transverse axis is the vertical axis.
The transverse axis it eht line segment connecting the vertices. The foci are located c units from the center, where c^2 = a^2 + b^2. The center, vertices and foci all lie along the transverse axis. The hyperbola opens along the transverse axis as well.
The transverse axis it eht line segment connecting the vertices. The foci are located c units from the center, where c^2 = a^2 + b^2. The center, vertices and foci all lie along the transverse axis. The hyperbola opens along the transverse axis as well.
Friday, August 9, 2013
Tips for graphing an ellipse.
The equation of an ellipse is (x-h)^2/a^2 + (y-k)^2/b^2 = 1 for an ellipse that is elongated along the x-axis.
The longer axis of the ellipse is the major axis and is of length 2a. The shorter axis of the ellipse is the minor axis and is of length 2b.
Plot the coordinate of the center (h,k) and move "a" units in each direction from the center along the major axis. Move "b" units from the center in each direction along the minor axis. These 4 points are the vertices. Then draw a smooth curve through the 4 points.
If you want to plot the foci, you need a value for c, which is found by c^2 = a^2 - b^2. Move c units from the center along the major axis in both directions and plot the point.
The equation of an ellipse is (x-h)^2/a^2 + (y-k)^2/b^2 = 1 for an ellipse that is elongated along the x-axis.
The longer axis of the ellipse is the major axis and is of length 2a. The shorter axis of the ellipse is the minor axis and is of length 2b.
Plot the coordinate of the center (h,k) and move "a" units in each direction from the center along the major axis. Move "b" units from the center in each direction along the minor axis. These 4 points are the vertices. Then draw a smooth curve through the 4 points.
If you want to plot the foci, you need a value for c, which is found by c^2 = a^2 - b^2. Move c units from the center along the major axis in both directions and plot the point.
Monday, August 5, 2013
When graphing a parabola from an equation in standard form, use these steps:
1. plot the vertez
2. plot the focus
3. draw the directrix line
4. find the latus rectum
5. draw the parabola
For example, graph the parabola with the equation (x - 2)^2 = 12(y + 1)
The form of a parabola that opens up or down is in the form (x - h)^2 = 4p(y -k), the vertex is (h,k), the focus is p units from the vertex in the direction the parabola opens, the directix is the line p units from the vertex in the opposite direction. The latus rectum is 4p. Plot a point 2p from the focus in both directions to determine the width of the parabola.
1. the vertex is (2, -1)
2. the focus is 3 units from the vertex.. Since p is positive the parabola opens up and the focus is at (2, 2)
3. the directrix is the horizontal line y = -1
4. the latus rectum is 12, so move 6 units to the left and right of the focus and plots those points.. The points are (8,2) and (-4,2)
5. now draw the parabola
1. plot the vertez
2. plot the focus
3. draw the directrix line
4. find the latus rectum
5. draw the parabola
For example, graph the parabola with the equation (x - 2)^2 = 12(y + 1)
The form of a parabola that opens up or down is in the form (x - h)^2 = 4p(y -k), the vertex is (h,k), the focus is p units from the vertex in the direction the parabola opens, the directix is the line p units from the vertex in the opposite direction. The latus rectum is 4p. Plot a point 2p from the focus in both directions to determine the width of the parabola.
1. the vertex is (2, -1)
2. the focus is 3 units from the vertex.. Since p is positive the parabola opens up and the focus is at (2, 2)
3. the directrix is the horizontal line y = -1
4. the latus rectum is 12, so move 6 units to the left and right of the focus and plots those points.. The points are (8,2) and (-4,2)
5. now draw the parabola
Thursday, August 1, 2013
How are the conic sections formed? Use paper and tape to create a double-napped cone. Then, using scissors, you can but the cone in 4 different ways.
1. A cut that is parallel to the base of the cone. The cross section on the surface of the cone is a circle.
2. A cute that is parallel to one of the sides of the cone. The cut will be through the other side of the cone and the base. This cross-section is a parabola.
3. A cut through both sides of the cone that is not parallel to the base. The cross-section is an ellipse.
4. A cut vertical through both cones. The cross-section is a hyperbola.
1. A cut that is parallel to the base of the cone. The cross section on the surface of the cone is a circle.
2. A cute that is parallel to one of the sides of the cone. The cut will be through the other side of the cone and the base. This cross-section is a parabola.
3. A cut through both sides of the cone that is not parallel to the base. The cross-section is an ellipse.
4. A cut vertical through both cones. The cross-section is a hyperbola.
Sunday, July 28, 2013
How to Work with Normal Distributions
used and important continuous probability distributions seen in statistics is the normal distribution. The normal distribution is characterized by a bell shaped curve with the highest point above the mean. The distribution is symmetrical and approaches the horizontal axis but never touches it. But how do we work with normal distributions? At the end of this article, you should be able to calculate probabilities for a random variable that has a normal distribution.Suppose a random variable x has a normal distribution with a mean m and standard deviation s. To find areas and probabilities for such a random variable, convert x values to z using the formula z = (x - m)/s. Then use a table for the area of a standard normal distribution, found in most statistics textbooks, to find the corresponding areas and probabilities.
Consider the following example:
Let x have a normal distribution with m = 15 and s = 4. Find the probability that an x value chosen at random from this distribution is between 8 and 19.
In symbols, we write this problem as P(8 ≤ x ≤ 19).
Since the probabilities correspond with areas under the normal distribution curve, we can find the area under the curve from x = 8 to x = 19. To do this, we convert the x values to z values.
For x = 8, z = (8 - 15)/4, therefore z = -1.75. For x = 19, z = (19 - 15)/4, therefore z = 1. Writing this in symbols, we get P(-1.75 ≤ z ≤ 1), This is the same as thinking of the area to the left of z = 1 minus the area to the left of z = -1.75. These values can be found using a table for the area of a standard normal distribution. Using the table, we get 0.8413 - 0.0401 = 0.8012.
Suppose we want to know the probability that a value chosen at random is greater than a certain value? Using the same value for mean and standard deviation, suppose we want the probability that x > 12?
This is the same as one minus the probability of x less than or equal to 12. In symbols, this is 1 - P(x ≤ 12). Converting the x value to z value, we get z - (12 - 15)/4 = -0.75. The z value corresponding to -0.75 is 0.2266. Therefore, P(z > -0.75) = 1- .2266 = .7734.
This guide should give a student a basic understanding of how to calculate probabilities for a random variable that has a normal distribution.
Wednesday, July 24, 2013
Normal Approximation to the Binomial Distribution]
Sppose that the probability a new vaccine will protect adults from a certain disease is 0.8. The vaccine is given to 200 adults. What is the probability that more than 170 of those adults will be protected by the vaccine?
This question is in the category of a binomial experiment with n trials equal to 200 and the probability of success p equal to 0.8, and the number of successes r equal to 170. There is a formula for the binomial distribution which can be used to calculate the probability. But this would be very time consuming and easy to make some errors in calculation. But there is a simpler way to solve this problem, and that is by using the normal distribution to approximate the binomial distribution. However, certain conditions must be present in order to use this method.First, we must consider the binomial distribution with n = number of trials, p = probability of success on a single trial, q = 1 - p = probability of failure on a single trial, and r = number of successes. Then if np > 5 and nq > 5, then r has a binomial distribution which you can approximate with the normal distribution. The mean is approximated by np and the standard deviation is approximated by the square root of npq. This approximation becomes more accurate as the sample size n increases.
Now we can put this all together with an example. Suppose the owner of a hotel needs to install new air conditioner units to 25 of the rooms. From past experience with a noted brand, he knows that the air conditioner unit is guaranteed for 5 years, and the probability that it will last 10 years is 0.35. What is the probability that 12 or more of the units will last more than 10 years?
In this problem, n = 25, p = 0.35, q = 0.65. We want the probability that r is greater than 10. We can use the normal approximation to the binomial since np = 8.75, which is greater than 5. Also, nq = 16.25, which is greater than 5. The mean is 8.75 and the standard deviation is square root of npq, which is 2.38. Using the normal distribution, we calculate the z to be (9.5 - 8.75)/2.38 = 0.32. Note that we use 9.5 instead of 10 because of a continuity correction, which converts r to a continuous normal random variable x by subtracting 0.5, since r in this case is the left-point of the interval. Now we find the probability that z > 0.32, which is 0.3745. This value is found by using a table for the areas of a standard normal distribution.
This guide should help assist students having difficulties understanding the normal approximation to the binomial.
Saturday, July 20, 2013
Estimating Population Mean Using Sample Data
Suppose we have a situation where we want to estimate the
population mean when the population standard deviation is known. There
are techniques that can be developed to accomplish this, but first there
are some basic assumptions about the random variable from which we
obtain a sample.
First, we have a simple random sample of size n which is drawn from a population containing x values. The value of the population standard deviation sigma, is known. The methods will work for any simple size n as long as x is normally distributed. When the distribution of x is unknown, we need a sample size of greater than or equal to 30 to use the normal approximation. But if the data is severely skewed, a larger sample size of 50 or greater is more appropriate.
A point estimate is an estimate of a population parameter given by a single number. Therefore, the sample mean is used as a point estimate for the population mean. But take note, that even with a large sample size, the sample mean x-bar is not exactly equal to the population mean mu. There is a margin of error, found by taking the absolute value of x-bar minus mu.
But when mu is not known, we cannot determine the margin of error. This is usually the case since the population mean is rarely known. If it was, there would be no need to use x-bar as a point estimate for mu.
The reliability of an estimate is determined by a confidence level. But what is a confidence level? For a confidence level x, there critical value Zc is the number such that the area under the standard normal curve between -Zc and Zc equals c. Remember there is typically a table with areas of a standard normal distribution at the back of a statistics textbook.
Using the table, if we want a level of confidence to be .7, the critical value is 1.04. For a .75 level of confidence, the critical value is 1.15. Typically we will want a confidence level to be .9, .95 or .99. The critical values for these levels of confidence are 1.645, 1.96 and 2.58, respectively.
Knowing how to find critical values for certain confidence levels will help determine confidence intervals for mu. A confidence interval for mu is an interval calculated from sample data so that c is the probability of generating an interval containing the actual value of mu.
For example, suppose John and Steve go to the track to jog. They like to jog 3 miles and from a random sample of 50 jogging sessions their mean time to complete the 3 miles is x-bar = 23.25 minutes and the standard deviation sigma = 2.5 minutes. Find a 99% confidence interval for mu, which is the mean jogging time for the entire distribution of their jogging sessions.
The critical value Zc = 2.58. From the Central Limit Theorem, we know that x-bar is approximately normally distributed. The confidence interval is x-bar +/- E, where E is the margin of error found by taking Zc(sigma/square root of n). So E in this problem is 0.912. Therefore the 99% confidence interval is 23.25 +/- 0.912. The lower limit of the interval is 22.338 minutes and the upper level of the interval is 24.162 minutes. The conclusion is that we are 99% certain that the interval contains the population mean time mu.
First, we have a simple random sample of size n which is drawn from a population containing x values. The value of the population standard deviation sigma, is known. The methods will work for any simple size n as long as x is normally distributed. When the distribution of x is unknown, we need a sample size of greater than or equal to 30 to use the normal approximation. But if the data is severely skewed, a larger sample size of 50 or greater is more appropriate.
A point estimate is an estimate of a population parameter given by a single number. Therefore, the sample mean is used as a point estimate for the population mean. But take note, that even with a large sample size, the sample mean x-bar is not exactly equal to the population mean mu. There is a margin of error, found by taking the absolute value of x-bar minus mu.
But when mu is not known, we cannot determine the margin of error. This is usually the case since the population mean is rarely known. If it was, there would be no need to use x-bar as a point estimate for mu.
The reliability of an estimate is determined by a confidence level. But what is a confidence level? For a confidence level x, there critical value Zc is the number such that the area under the standard normal curve between -Zc and Zc equals c. Remember there is typically a table with areas of a standard normal distribution at the back of a statistics textbook.
Using the table, if we want a level of confidence to be .7, the critical value is 1.04. For a .75 level of confidence, the critical value is 1.15. Typically we will want a confidence level to be .9, .95 or .99. The critical values for these levels of confidence are 1.645, 1.96 and 2.58, respectively.
Knowing how to find critical values for certain confidence levels will help determine confidence intervals for mu. A confidence interval for mu is an interval calculated from sample data so that c is the probability of generating an interval containing the actual value of mu.
For example, suppose John and Steve go to the track to jog. They like to jog 3 miles and from a random sample of 50 jogging sessions their mean time to complete the 3 miles is x-bar = 23.25 minutes and the standard deviation sigma = 2.5 minutes. Find a 99% confidence interval for mu, which is the mean jogging time for the entire distribution of their jogging sessions.
The critical value Zc = 2.58. From the Central Limit Theorem, we know that x-bar is approximately normally distributed. The confidence interval is x-bar +/- E, where E is the margin of error found by taking Zc(sigma/square root of n). So E in this problem is 0.912. Therefore the 99% confidence interval is 23.25 +/- 0.912. The lower limit of the interval is 22.338 minutes and the upper level of the interval is 24.162 minutes. The conclusion is that we are 99% certain that the interval contains the population mean time mu.
Monday, July 15, 2013
In statistics, suppose x is a random variable with normal
distribution with mean mu and standard deviation sigma. Now let x-bar be
the sample mean from a sample of size n from the x distribution. The
following assumptions can now be made:
1. The x-bar distribution is normally distributed, just like the x distribution.
2. The mean of the x-bar distribution is mu.
3. The standard deviation of the x-bar distribution is sigma divided by the square root of n.
This theorem states that the x-bar will be normally distributed if the x distribution is normal, no matter what the sample size is. The mean will always be the same as the mean of the x distribution and the standard deviation is always sigma divided by the square root of n.
What happens if we don't know the shape of the x distribution? The Central Limit Theorem for any probability distribution states that if x has any distribution with mean mu and standard deviation sigma, then x-bar with sample size n will have a distribution that approximates the normal distribution as n gets larger. That means that as a sample size gets larger, the distribution of x-bar will always approach the normal distribution. But how large must the sample size get? Generally speaking, a sample size of 30 or greater will give a reasonable approximation to the normal distribution.
Here's an example using the Central Limit Theorem.
Suppose x has a normal distribution with mean mu = 15 and standard deviation sigma = 4. If you draw random samples of size 5 from the x distribution and x-bar is the sample mean, what can be said about the x-bar distribution?
Even though the sample size is small, much less than 30, you could say the x-bar distribution is approximately normal, since the x distribution is normal. The means mean of x-bar is 15 and standard deviation is 15 divided by square root of 4 which is 7.5.
For another example, suppose the x distribution has mean mu = 100 and standard deviation sigma = 20. But there is no information about the shape of the x distribution. If samples are drawn of size 35 from the x distribution, what can be said about the x-bar distribution? Since the sample size is greater than or equal to 30, the x-bar distribution will be approximately normally distributed with mean = 100 and standard deviation equal to 20 divided by the square root of 35, which is 3.4.
If the sample size were 10 but we did not know the shape of the x distribution, you could not say that x-bar distribution is approximately normal because the sample size is too low.
This guide should help students who are having difficulty understanding the Central Limit Theorem.
1. The x-bar distribution is normally distributed, just like the x distribution.
2. The mean of the x-bar distribution is mu.
3. The standard deviation of the x-bar distribution is sigma divided by the square root of n.
This theorem states that the x-bar will be normally distributed if the x distribution is normal, no matter what the sample size is. The mean will always be the same as the mean of the x distribution and the standard deviation is always sigma divided by the square root of n.
What happens if we don't know the shape of the x distribution? The Central Limit Theorem for any probability distribution states that if x has any distribution with mean mu and standard deviation sigma, then x-bar with sample size n will have a distribution that approximates the normal distribution as n gets larger. That means that as a sample size gets larger, the distribution of x-bar will always approach the normal distribution. But how large must the sample size get? Generally speaking, a sample size of 30 or greater will give a reasonable approximation to the normal distribution.
Here's an example using the Central Limit Theorem.
Suppose x has a normal distribution with mean mu = 15 and standard deviation sigma = 4. If you draw random samples of size 5 from the x distribution and x-bar is the sample mean, what can be said about the x-bar distribution?
Even though the sample size is small, much less than 30, you could say the x-bar distribution is approximately normal, since the x distribution is normal. The means mean of x-bar is 15 and standard deviation is 15 divided by square root of 4 which is 7.5.
For another example, suppose the x distribution has mean mu = 100 and standard deviation sigma = 20. But there is no information about the shape of the x distribution. If samples are drawn of size 35 from the x distribution, what can be said about the x-bar distribution? Since the sample size is greater than or equal to 30, the x-bar distribution will be approximately normally distributed with mean = 100 and standard deviation equal to 20 divided by the square root of 35, which is 3.4.
If the sample size were 10 but we did not know the shape of the x distribution, you could not say that x-bar distribution is approximately normal because the sample size is too low.
This guide should help students who are having difficulty understanding the Central Limit Theorem.
Thursday, July 11, 2013
Asymptotes
When dealing with graphing rational functions you have my asymptotes. The asymptotes may be vertical, horizontal or slant.
If the highest exponent of all terms in the numerator is less than the highest exponent of all terms in the denominator, the horizontal asymptote is y = 0
for example f(x) = x/(x^3 + 3)
If the highest exponent of all terms in the numerator is one greater than the highest exponent of all terms in the denominator, there is a vertical asymptote and slant asymptote, which is found by doing long division
for example f(x) = (x^2 + 2x + 3)/(x + 1)
Doing long division you get x +1 with a remainder of 2.. the slant asymptote is y = x + 1, ignore the remainder.
If the highest exponent of all terms in the numerator equals the highest exponent of all terms in the denominator, the horizontal asymptote is the coefficients of the highest terms divided.
for example, f(x) = 2x/(x + 2), the horizontal asymptote is y= 2x/x, y = 2
Veritcal asymptotes, if they exist are found by setting the denominator to zero and solving for x.
If the highest exponent of all terms in the numerator is less than the highest exponent of all terms in the denominator, the horizontal asymptote is y = 0
for example f(x) = x/(x^3 + 3)
If the highest exponent of all terms in the numerator is one greater than the highest exponent of all terms in the denominator, there is a vertical asymptote and slant asymptote, which is found by doing long division
for example f(x) = (x^2 + 2x + 3)/(x + 1)
Doing long division you get x +1 with a remainder of 2.. the slant asymptote is y = x + 1, ignore the remainder.
If the highest exponent of all terms in the numerator equals the highest exponent of all terms in the denominator, the horizontal asymptote is the coefficients of the highest terms divided.
for example, f(x) = 2x/(x + 2), the horizontal asymptote is y= 2x/x, y = 2
Veritcal asymptotes, if they exist are found by setting the denominator to zero and solving for x.
Monday, July 8, 2013
Definition of the day
A function is continuous, if and only if an infinitely small increment of the independent variable x produces always an infinitely small increment change of f(x). Therefore, functions cannot have gaps or jumps.
Friday, July 5, 2013
End behavior
The end behavior is behavior of a graph as x tends to -infinity and x tends to +infinity.
Positive leading coefficient : f(x) = x^2, as x tends to -infinity, f(x) tends to infinity
as x tends to +infinity, f(x) tends to infinity
f(x) = x^3, as x tends to -infinity, f(x) tends to -infinity, as x tends to infinity, f(x) tends to infinity
Negative leading coefficient: f(x) = -x^2, as x tends to -infinity, f(x) tends to -infinity.
as x tends to +infinity, f(x) tends to infinity
f(x) = -x^3, as x tends to -infinity, f(x) tends to infinity, as x tends to infinity, f(x) tends to -infinity
Positive leading coefficient : f(x) = x^2, as x tends to -infinity, f(x) tends to infinity
as x tends to +infinity, f(x) tends to infinity
f(x) = x^3, as x tends to -infinity, f(x) tends to -infinity, as x tends to infinity, f(x) tends to infinity
Negative leading coefficient: f(x) = -x^2, as x tends to -infinity, f(x) tends to -infinity.
as x tends to +infinity, f(x) tends to infinity
f(x) = -x^3, as x tends to -infinity, f(x) tends to infinity, as x tends to infinity, f(x) tends to -infinity
Tuesday, July 2, 2013
Domain of f(x)/g(x)
Suppose you have two functions f(x) and g(x). Now you take f(x)/g(x)
and want the domain of this new function formed. What you need to do
is get the domain of the numerator and the domain of the denominator and
find the intersection of the two domains.
For example:
f(x) = x^2 - 4
g(x) = sqrt(x^2 - 9)
f(x)/g(x) = (x^2 - 4)/sqrt(x^2 - 9)
domain of numerator is all real numbers.
domain of denominator is (-inf, -3) U (-3, inf)
The intersection is (-inf, -3) U (-3, inf)
For example:
f(x) = x^2 - 4
g(x) = sqrt(x^2 - 9)
f(x)/g(x) = (x^2 - 4)/sqrt(x^2 - 9)
domain of numerator is all real numbers.
domain of denominator is (-inf, -3) U (-3, inf)
The intersection is (-inf, -3) U (-3, inf)
Thursday, June 27, 2013
How to Work with Normal Distributions
One of the most widely used and important continuous probability
distributions seen in statistics is the normal distribution. The normal
distribution is characterized by a bell shaped curve with the highest
point above the mean. The distribution is symmetrical and approaches the
horizontal axis but never touches it. But how do we work with normal
distributions? At the end of this article, you should be able to
calculate probabilities for a random variable that has a normal
distribution.
Suppose a random variable x has a normal distribution with a mean m and standard deviation s. To find areas and probabilities for such a random variable, convert x values to z using the formula z = (x - m)/s. Then use a table for the area of a standard normal distribution, found in most statistics textbooks, to find the corresponding areas and probabilities.
Consider the following example:
Let x have a normal distribution with m = 15 and s = 4. Find the probability that an x value chosen at random from this distribution is between 8 and 19.
In symbols, we write this problem as P(8 ≤ x ≤ 19).
Since the probabilities correspond with areas under the normal distribution curve, we can find the area under the curve from x = 8 to x = 19. To do this, we convert the x values to z values.
For x = 8, z = (8 - 15)/4, therefore z = -1.75. For x = 19, z = (19 - 15)/4, therefore z = 1. Writing this in symbols, we get P(-1.75 ≤ z ≤ 1), This is the same as thinking of the area to the left of z = 1 minus the area to the left of z = -1.75. These values can be found using a table for the area of a standard normal distribution. Using the table, we get 0.8413 - 0.0401 = 0.8012.
Suppose we want to know the probability that a value chosen at random is greater than a certain value? Using the same value for mean and standard deviation, suppose we want the probability that x > 12?
This is the same as one minus the probability of x less than or equal to 12. In symbols, this is 1 - P(x ≤ 12). Converting the x value to z value, we get z - (12 - 15)/4 = -0.75. The z value corresponding to -0.75 is 0.2266. Therefore, P(z > -0.75) = 1- .2266 = .7734.
This guide should give a student a basic understanding of how to calculate probabilities for a random variable that has a normal distribution.
Suppose a random variable x has a normal distribution with a mean m and standard deviation s. To find areas and probabilities for such a random variable, convert x values to z using the formula z = (x - m)/s. Then use a table for the area of a standard normal distribution, found in most statistics textbooks, to find the corresponding areas and probabilities.
Consider the following example:
Let x have a normal distribution with m = 15 and s = 4. Find the probability that an x value chosen at random from this distribution is between 8 and 19.
In symbols, we write this problem as P(8 ≤ x ≤ 19).
Since the probabilities correspond with areas under the normal distribution curve, we can find the area under the curve from x = 8 to x = 19. To do this, we convert the x values to z values.
For x = 8, z = (8 - 15)/4, therefore z = -1.75. For x = 19, z = (19 - 15)/4, therefore z = 1. Writing this in symbols, we get P(-1.75 ≤ z ≤ 1), This is the same as thinking of the area to the left of z = 1 minus the area to the left of z = -1.75. These values can be found using a table for the area of a standard normal distribution. Using the table, we get 0.8413 - 0.0401 = 0.8012.
Suppose we want to know the probability that a value chosen at random is greater than a certain value? Using the same value for mean and standard deviation, suppose we want the probability that x > 12?
This is the same as one minus the probability of x less than or equal to 12. In symbols, this is 1 - P(x ≤ 12). Converting the x value to z value, we get z - (12 - 15)/4 = -0.75. The z value corresponding to -0.75 is 0.2266. Therefore, P(z > -0.75) = 1- .2266 = .7734.
This guide should give a student a basic understanding of how to calculate probabilities for a random variable that has a normal distribution.
Monday, June 24, 2013
The normal distribution is a distribution that is symmetrical
about a vertical line through the mean. The curve of the distribution is
bell-shaped with the highest point over the mean. The curve will never
cross or touch the horizontal axis. The differences in normal
distributions makes calculating the area under the curve in a specific
interval difficult. Therefore, there is a method to make this easier by
changing raw scores into standardized scores known as "z scores".
The standardization of scores must be done in such a way that we can use one table for all of the normal distributions. This is done by considering how many standard deviations a data value lies from the mean. This makes it to compare a value from one normal distribution to a value from another normal distribution.
For example, suppose Frank and Tom are taking a history course. Frank is in the 8 o'clock class and Tom is in the 10 o'clock class. Since each class has a large number of students, the scores on the final exam each follow a normal distribution. Frank's section had a class average of 75 and his score was a 83. In Tom's section, the average was a 68 and Tom scored a 76. When comparing scores they both felt good that they scored 8 points higher than the average. But which one did better with respect to the rest of the students in the class?
Suppose a graph of the data shows that Frank scored higher than most of the other students in his class, but Tom's score places him more in the middle of the upper half. Then you know that Frank's score is clearly much better with respect to the students in his class than Tom's score is. But how can you tell if the distributions are not graphically represented?
You have to change the raw scores to z scores to see how many standard deviations from the mean each score lies. Suppose the standard deviation of scores in Frank's section is 5 and the standard deviation of scores in Tom's section is 6. The z score is the sample value minus the mean value divided by the standard deviation. Therefore, the z score for Frank is 8/5 = 1.6, and the z score for Tom is 8/6 = 1.33. It's clear that Frank had a better score relative to the class since his score fell 1.6 standard deviations about the mean and Tom's score was only 1.33 standard deviations about the mean.
Note that when standardizing the scores, the mean of the original distribution is always zero, which is logical since the mean lies zero standard deviations from itself. Any value above the mean has a positive z score, and any value lower than the mean has a negative z score.
This guide should help anyone understand how to change a raw score to a z score, which makes comparing scores from different normal distributions easy.
The standardization of scores must be done in such a way that we can use one table for all of the normal distributions. This is done by considering how many standard deviations a data value lies from the mean. This makes it to compare a value from one normal distribution to a value from another normal distribution.
For example, suppose Frank and Tom are taking a history course. Frank is in the 8 o'clock class and Tom is in the 10 o'clock class. Since each class has a large number of students, the scores on the final exam each follow a normal distribution. Frank's section had a class average of 75 and his score was a 83. In Tom's section, the average was a 68 and Tom scored a 76. When comparing scores they both felt good that they scored 8 points higher than the average. But which one did better with respect to the rest of the students in the class?
Suppose a graph of the data shows that Frank scored higher than most of the other students in his class, but Tom's score places him more in the middle of the upper half. Then you know that Frank's score is clearly much better with respect to the students in his class than Tom's score is. But how can you tell if the distributions are not graphically represented?
You have to change the raw scores to z scores to see how many standard deviations from the mean each score lies. Suppose the standard deviation of scores in Frank's section is 5 and the standard deviation of scores in Tom's section is 6. The z score is the sample value minus the mean value divided by the standard deviation. Therefore, the z score for Frank is 8/5 = 1.6, and the z score for Tom is 8/6 = 1.33. It's clear that Frank had a better score relative to the class since his score fell 1.6 standard deviations about the mean and Tom's score was only 1.33 standard deviations about the mean.
Note that when standardizing the scores, the mean of the original distribution is always zero, which is logical since the mean lies zero standard deviations from itself. Any value above the mean has a positive z score, and any value lower than the mean has a negative z score.
This guide should help anyone understand how to change a raw score to a z score, which makes comparing scores from different normal distributions easy.
Friday, June 21, 2013
Remember in a two column geometric proof, the statements go on the left side and the reasons on the right side. The reasons should be postulates and axioms. The given statement is almost always first in this kind of proof.
Indirect proofs are done in paragraph form. You first assume that the statement you are to prove is false. Then you use deductive reasoning and come up with two statements that cannot both be true. There is the contradiction that shows the earlier assumption is false, therefore the statement that is to be proven is true.
Indirect proofs are done in paragraph form. You first assume that the statement you are to prove is false. Then you use deductive reasoning and come up with two statements that cannot both be true. There is the contradiction that shows the earlier assumption is false, therefore the statement that is to be proven is true.
Sunday, June 16, 2013
Thursday, June 13, 2013
Math tidbit of the day.
When calculating square roots and cube roots, it's easier if you know the perfect squares from 1 to 25 and the perfect cubes from 1 to 10
perfect squares : 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225, 256, 289, 324, 361, 400, 441, 484, 529, 576, 625
perfect cubes: 1, 8, 27, 64, 125, 216, 343, 512, 729, 1000
When calculating square roots and cube roots, it's easier if you know the perfect squares from 1 to 25 and the perfect cubes from 1 to 10
perfect squares : 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225, 256, 289, 324, 361, 400, 441, 484, 529, 576, 625
perfect cubes: 1, 8, 27, 64, 125, 216, 343, 512, 729, 1000
Tuesday, June 11, 2013
Saturday, June 8, 2013
When graphing functions, it's important to know the basic parent functions. From there you can shift the graph up, down, left and right accordingly. Theses shifts are also known as translations.
f(x) = x^2, f(x) = sqrt(x), f(x) = x^3, f(x)= |x| are some examples of parent functions
f(x) = (x - 2)^2 + 3 has the same graph as f(x) = x^2 except it is translated 2 units left and 3 units up.
f(x) = sqrt(x + 1) - 3 is the same as the graph of f(x) = sqrt(x) except it is translated 1 unit right and 3 units down.
f(x) = (x + 4)^3 - 1 is the same as the graph of f(x) = x^3 except it is translated 4 units right and 1 unit down.
f(x)= |x - 2| + 1 is the same as the graph of f(x)= |x| except it is translated 2 units right and 1 unit up.
f(x) = x^2, f(x) = sqrt(x), f(x) = x^3, f(x)= |x| are some examples of parent functions
f(x) = (x - 2)^2 + 3 has the same graph as f(x) = x^2 except it is translated 2 units left and 3 units up.
f(x) = sqrt(x + 1) - 3 is the same as the graph of f(x) = sqrt(x) except it is translated 1 unit right and 3 units down.
f(x) = (x + 4)^3 - 1 is the same as the graph of f(x) = x^3 except it is translated 4 units right and 1 unit down.
f(x)= |x - 2| + 1 is the same as the graph of f(x)= |x| except it is translated 2 units right and 1 unit up.
Wednesday, June 5, 2013
When trying to identify angles on the unit circle with their corresponding sine and cosine value, there is an easy way to do this.
Remember that the angles start at 0, then 30, 45, 60, 90 in the first quadrant. So they angle increases by 30, 15, 15 and 30. Likewise they increase by the same amount in the 2nd , 3rd and 4th quadrants.
So the remaining angles on the unit circle are 120, 135, 150, 180, 210, 225, 240, 270, 300, 315, 330, 360.
The sine and cosine values in the first quadrant are 1/2, sqrt(2)/2, sqrt(3), 2 - Sine
sqrt(3)/2, sqrt(2)/2, 1/2 - Cosine
These values remain the same in all quadrants with the exception of + or -.
Just remember the signs are follows
Quadrant, Sine, Cosine
1 + +
II + -
III - -
IV - +
Remember that the angles start at 0, then 30, 45, 60, 90 in the first quadrant. So they angle increases by 30, 15, 15 and 30. Likewise they increase by the same amount in the 2nd , 3rd and 4th quadrants.
So the remaining angles on the unit circle are 120, 135, 150, 180, 210, 225, 240, 270, 300, 315, 330, 360.
The sine and cosine values in the first quadrant are 1/2, sqrt(2)/2, sqrt(3), 2 - Sine
sqrt(3)/2, sqrt(2)/2, 1/2 - Cosine
These values remain the same in all quadrants with the exception of + or -.
Just remember the signs are follows
Quadrant, Sine, Cosine
1 + +
II + -
III - -
IV - +
Thursday, May 30, 2013
The Easy Way to Simplify Square Roots
In a first year algebra class, students will encounter problems involving simplifying square roots. But many times the teacher simply doesn't explain the process by which this is accomplished in a manner that students can understand. I will explain a way that will make simplifying square roots easier for students of any ability level.The first method I use when teaching students how to calculate square root is to look to see if the number is a perfect square first. I suggest students memorize the perfect squares from 1 to 25 as follows: 1, 4, 9, 16, 25, 49, 64, 81, 100, 121, 144, 169, 196, 225, 256, 289, 324, 361, 400, 441, 484, 529, 576, 625. So if you are asked to calculate the square root of 441, you automatically know the answer is 21 or -21, since a negative number times another negative number yields a positive number.
For non-perfect squares or larger numbers that you are unsure of, if the number is a perfect square, I suggest using a factor tree. For example, suppose you want to simplify a square root of 48. It is not a perfect square since 6 times 6 equals 36 and 7 times 7 equals 49. No whole number times itself equals 48. So break it down into factors. I always suggest trying to find a perfect square as one of the factors and in this case, 16 times 3 equals 48 and 16 is a perfect square. Remember: Since you are dealing with square root, the factors are also square root. So the square root of 48 equals the square root of 16 times the square root of 3. Three is a prime number so you cannot break down 3 any farther using a factor tree. We know that the square root of 16 is 4, so the answer is 4 times the square root of 3.
Another more difficult example---say we need to find the square root of 2025. With such a large number, most people won't know if this is a perfect square, so use the factor tree. Know that any number ending in 5 is divisible by 5. So 5 times 405 is 2025. But 405 can be broken down into factors, using the same rule, therefore 5 times 81 is 405. Now we have the square root of 5 times the square root of 5 times the square root of 81. Notice then that the square root of 5 times the square root of 5 equals the square root of 25. Now this problem becomes simple because you notice we have two perfect squares here, 25 and 81. The square root of 25 is 5 and square root of 81 is 9. So the answer is 5 times 9, which is 45 and -45, since -45 times -45 equals 2025. This problem is actually a perfect square, but if you do not recognize it as such, you can use the factor tree method I just described.
When dealing with the square root of a negative number, imaginary numbers come into play. The square root of -1 equals an imaginary number denoted as "i". So in the above problems if we had the square root of -48, you have square root of -1 times square root of 3 times square root of 16. The answer is 4i times the square root of 3, and -4i times square root of 3. If we have the square root of -2025, the answer is simply 45i and -45i.
Monday, May 27, 2013
It's easy to remember the formula for volume of a cylinder if you know the area of a circle.. A cylinder is basically a tube, which has two bases, which are circles and a height.
So the area of a circle is Pi(r^2)
Think of the cylinder as a bunch of CD's stacked up top of each other until you reach the total height of the cylinder. So it's obvious to multiply the area of the circle by the height to get the volume of the cylinder.
So the area of a circle is Pi(r^2)
Think of the cylinder as a bunch of CD's stacked up top of each other until you reach the total height of the cylinder. So it's obvious to multiply the area of the circle by the height to get the volume of the cylinder.
Friday, May 24, 2013
Remember the change of base formula when solving a logarithm that is not base 10.
Log(base 3) 11 = Log 11/Log 3
It's easy to see how this comes about when rewriting the log in exponential form and then solving for y.
y = log(base 3) 11
3^y = 11
Take the log of both sides to get
log(3^y) = log 11
ylog 3 = log 11
y = log11/log3
Instead of having to rewrite as an exponential equation and solving, simply use the change of base formula. In general
Log(base b)a = Log a/Log b
Log(base 3) 11 = Log 11/Log 3
It's easy to see how this comes about when rewriting the log in exponential form and then solving for y.
y = log(base 3) 11
3^y = 11
Take the log of both sides to get
log(3^y) = log 11
ylog 3 = log 11
y = log11/log3
Instead of having to rewrite as an exponential equation and solving, simply use the change of base formula. In general
Log(base b)a = Log a/Log b
Friday, May 17, 2013
When calculating limits, try to factor the numerator and denominator first. Then simplify before evaluating the limit. If there is a radical in the denominator, rationalize it first.
Make sure when doing sided limits, when taking the limit from the left (-) and from the right (+), they must be the same or the limit does not exist.
Make sure when doing sided limits, when taking the limit from the left (-) and from the right (+), they must be the same or the limit does not exist.
Tuesday, May 14, 2013
Students taking a course in basic statistics will learn about many
types of probability distributions. The most widely used and most
important is the normal probability distribution. What are the
characteristics of this distribution and how is it graphed?
The normal distribution is a continuous distribution with numerous applications. We need to learn about some of the properties of this distribution.
The normal distribution is defined in terms of its mean and standard deviation. The graph will give one some idea of the main features of this normal distribution. The graph of the normal distribution is called the normal curve. It's also called the bell-shaped curve since it very much resembles a bell.
The curve is symmetrical about the vertical line that extends up from the mean. The highest point on this graph is about the mean. The standard deviation controls the amount of spread in the curve. It is very close to the x-axis at mean + 3 times standard deviation and mean - 3 times standard deviation. This implies that when the standard deviation is small, the curve is less spread out and more spread out when the standard deviation is large.
Summarizing the important properties:
1. The curve is bell-shaped with the highest point at the mean.
2. The curve never touches the x-axis.
3. The curve is symmetrical about a vertical line through the mean.
4. The points between the curve cupping upward and downward occur at the mean plus or minus the standard deviation.
The empirical rule is a rule used for the normal distribution, and all other symmetrical, bell-shaped distributions. It states the approximately 68% of the data values fall within one standard deviation of the mean. Approximately 95% of the data values fall within two standard deviations of the mean. Approximately 99.7% of the data values fall within three standard deviations of the mean.
This guide should help assist students having difficulty understand the basics of the normal probability distribution and its graph.
The normal distribution is a continuous distribution with numerous applications. We need to learn about some of the properties of this distribution.
The normal distribution is defined in terms of its mean and standard deviation. The graph will give one some idea of the main features of this normal distribution. The graph of the normal distribution is called the normal curve. It's also called the bell-shaped curve since it very much resembles a bell.
The curve is symmetrical about the vertical line that extends up from the mean. The highest point on this graph is about the mean. The standard deviation controls the amount of spread in the curve. It is very close to the x-axis at mean + 3 times standard deviation and mean - 3 times standard deviation. This implies that when the standard deviation is small, the curve is less spread out and more spread out when the standard deviation is large.
Summarizing the important properties:
1. The curve is bell-shaped with the highest point at the mean.
2. The curve never touches the x-axis.
3. The curve is symmetrical about a vertical line through the mean.
4. The points between the curve cupping upward and downward occur at the mean plus or minus the standard deviation.
The empirical rule is a rule used for the normal distribution, and all other symmetrical, bell-shaped distributions. It states the approximately 68% of the data values fall within one standard deviation of the mean. Approximately 95% of the data values fall within two standard deviations of the mean. Approximately 99.7% of the data values fall within three standard deviations of the mean.
This guide should help assist students having difficulty understand the basics of the normal probability distribution and its graph.
Thursday, May 9, 2013
Some Properties of the Binomial Distribution
The
main focus of a binomial experiment is to find the probability of r
successes in n trials. How are these probabilities calculated? How is
mean and standard deviation of a binomial distribution calculated? How
is a binomial distribution displayed graphically?
Suppose you wish to flip a fair coin three times. What is the probability of obtaining two heads in the three tosses? This is an example of a binomial experiment, with probability of success 0.5. We could list the possibilities as {HHH, HHT, HTH, THH, HTT, TTH, THT, TTT}. You notice from the sample space that the probability is 3 out of 8, or 3/8. But most times it's much too time consuming to list all possibilities. So there is a formula that we can use:
P(r) = n!/[r!(n - r)!]* prq(n-r)
where n = number of trials
p = probability of success on a trial
q = probability of failure on a trial
r = random variable representing the number of successes out of n trials
Using the coin tossing example above, P(0) = 1/8, P(1) = 3/8, P(2) = 3/8, P(3) = 1/8.
To graph a binomial distribution, place the r values on the horizontal axis. Next, place the P(r) values on the vertical axis. Then construct a bar over each r value. Extend these bars to r - 0.5 to r + 0.5. The height of each bar must be P(r).
In the above example the bar widths would be -0.5 to 0.5, 0.5 to 1.5, 1.5 to 2.5, and 2.5 to 3.5. You cannot have a 0.5 of a success but to have a bar, there needs to be a width, so you add and subtract 0.5 from each value of r.
To help describe the graph of the binomial distribution, the mean and standard deviation are very helpful. For this distribution, the mean is np and the standard deviation is the square root of npq.
In our example with the coins, the mean is 3(0.5) = 1.5, and the standard deviation is square root of 3(0.5)(0.5) = 0.87.
This simple guide should help students understand some properties of the binomial distribution including calculating probabilities, graphing, and computer the mean and standard deviation.
Suppose you wish to flip a fair coin three times. What is the probability of obtaining two heads in the three tosses? This is an example of a binomial experiment, with probability of success 0.5. We could list the possibilities as {HHH, HHT, HTH, THH, HTT, TTH, THT, TTT}. You notice from the sample space that the probability is 3 out of 8, or 3/8. But most times it's much too time consuming to list all possibilities. So there is a formula that we can use:
P(r) = n!/[r!(n - r)!]* prq(n-r)
where n = number of trials
p = probability of success on a trial
q = probability of failure on a trial
r = random variable representing the number of successes out of n trials
Using the coin tossing example above, P(0) = 1/8, P(1) = 3/8, P(2) = 3/8, P(3) = 1/8.
To graph a binomial distribution, place the r values on the horizontal axis. Next, place the P(r) values on the vertical axis. Then construct a bar over each r value. Extend these bars to r - 0.5 to r + 0.5. The height of each bar must be P(r).
In the above example the bar widths would be -0.5 to 0.5, 0.5 to 1.5, 1.5 to 2.5, and 2.5 to 3.5. You cannot have a 0.5 of a success but to have a bar, there needs to be a width, so you add and subtract 0.5 from each value of r.
To help describe the graph of the binomial distribution, the mean and standard deviation are very helpful. For this distribution, the mean is np and the standard deviation is the square root of npq.
In our example with the coins, the mean is 3(0.5) = 1.5, and the standard deviation is square root of 3(0.5)(0.5) = 0.87.
This simple guide should help students understand some properties of the binomial distribution including calculating probabilities, graphing, and computer the mean and standard deviation.
Sunday, May 5, 2013
Computing Quartiles and Box-and-Whisker Plots
Recall that the median of a set of data is the middle value when ordered in ascending or descending order. Since half the data fall above the median and half the data fall below the median, the median is the 50th percentile. In general, the Pth percentile of a distribution is the value where P% of the data falls at or below it and the rest of the data falls above it.
We can expand on the topic of percentile by introducing special percentiles known as quartiles, which are percentiles that divide the data into fourths. The first quartile, known as Q1 is the 25th percentile, the second quartile, known as Q2 is the median, and the third quartile, known as Q3 is the 75th percentile. It is important to know how to find Q1, Q2, and Q3 in order to draw a box-and-whisker plot.
To compute quartiles, first order the data from smallest to largest. Next, find the median. The first quartile is the median of the lower half of the data. The third quartile is the median of the upper half of the data. The interquartile range, Q3 - Q1, gives the spread of the middle half of the data.
Example: Find Q1, Q2, Q3, and the interquartile range of the following set of data.
5, 10, 8, 9, 10, 2, 5, 6, 11, 15, 8
First, we order the data from smallest to largest: 2, 5, 5, 6, 8, 8, 9, 10, 10, 11, 15. The median is 8 since half 2, 5, 5, 6, 8 falls at or below it and 9, 10, 10, 11, 15 falls at or above it. The median of the lower half of the data is 5, which is Q1. The median of the upper half of the data is 10, which is Q3. The interquartile range is Q3 - Q1, which is 5.
The quartiles are used with the maximum and minimum values of a data set to create a box-and-whisker plot. These are very useful to describe a data set. To make a box-and-whisker plot, draw a vertical scale that includes the highest and lowest data values. Then mark Q1, the median and Q3. Draw a box around Q1 and Q3. Draw a line through the box where the median is. The whiskers are then drawn, which are vertical lines from Q3 to the maximum value and from Q1 to the lowest value.
In the data set above, the box would be around 5 and 10. The median line is through 8 in the box. The whiskers are drawn from Q1 to the lowest data value of 2, and from Q3 to the largest data value of 15.
This guide should help students learn the basics about percentiles, quartiles and constructing box-and-whisker plots.
Friday, May 3, 2013
Scatter Diagrams and Linear Correlation
After a scatter diagram is made for a set of data, a line is drawn through the points. This line is known as the "line of best fit". But how do we determine which is the "best" line through a set of points? It is the line that comes closest to each point in the scatter diagram. The "least squares line", which can be computed by hand using the data values, or more easily by a computer, will be the best fit line. This line will contain the mean of the x value and the mean of the y value. In fact, the coordinate is (x mean, y mean).
Sometimes the data is dispersed in a way that there is no "best" line. If the points are a poor fit to any line, it makes no sense to try to find a line of best fit. When the points are scattered in a way that there is not a "good" fit, then there is no linear correlation between the x and y values. Picture many randomly scattered points, almost as if looking into the sky at a bunch of stars. There will be very little if any linear correlation between the points. If the points are scattered in a way that you can visually see where a line would go or the points almost form a line, then there will be linear correlation and strong linear correlation the more the points form a straight line.
The measurement that determines the strength of linear association between variables is known as the "sample correlation coefficient r". Also known as Pearson's correlation coefficient, named after statistician Karl Pearson.
The correlation coefficient is a measurement between -1 and 1. A correlation coefficient of -1 means there is perfect negative linear correlation between x and y. On the scatter diagram the points would form a perfect line with negative slope. A correlation coefficient of 1 means there is perfect positive linear correlation between x and y. On a scatter diagram the points would form a perfect line with positive slope. The closer r is 1, the stronger the positive correlation, and the closer r is to -1, the stronger the negative correlation. Basically this means that the closer r is to 1 and -1, the more the line describes the relationship between the variables.
In linear correlation, the explanatory variable is x and the response variable is y. These are also known as the independent and dependent variables, respectively. The value for r can be calculated by hand from the data pairs using a tedious formula or can be easily calculated using a computer.
This guide should help assist students learning the basics about scatter diagrams and linear correlation.
Wednesday, May 1, 2013
when working with polar coordinates, remember that there are several ways to represent the same point. For example, the point (2, Pi) is the same as the point (-2, 0), (-2, -Pi), (2, 0).
You can check this by converting the polar coordinates to rectangular coordinates.
If the polar coordinate is represented as (r, A) where r is the radius and A is the angle, then
x = rcos(A), y = rsin(A)
Using the above example,
x= 2cos(Pi)
y= 2sin(Pi) converts to (-2,0) in rectangular coordinates
Likewise using the same formula for the polar coordinates (-2, 0), (-2, -Pi), (2, 0) also convert to (-2, 0) in rectangular coordinates.
You can check this by converting the polar coordinates to rectangular coordinates.
If the polar coordinate is represented as (r, A) where r is the radius and A is the angle, then
x = rcos(A), y = rsin(A)
Using the above example,
x= 2cos(Pi)
y= 2sin(Pi) converts to (-2,0) in rectangular coordinates
Likewise using the same formula for the polar coordinates (-2, 0), (-2, -Pi), (2, 0) also convert to (-2, 0) in rectangular coordinates.
Saturday, April 27, 2013
Suppose you have an region bounded by the curves y = x^3 an y = x^2. Now you rotate the region about the x axis to for a three dimension figure. What is the volume of this newly formed figure?
To calculate the volume, you need to know the equation of the curve that forms the outer edge of the shape minus the equation of the curve that forms the inner edge of the shape. The bounds of integration are the points of intersection of the curves.
Set the 2 equations equal to each other and solve for x.
x^3 = x^2
x^3 - x^2 = 0
x^2(x - 1) = 0
x = 0 and x = 1. This can also be seen by a graph of the two equations.
Now that we have the points of intersection, we take pi times the integral of the outer radius squared minus the inner radius squared
integrate from 0 to 1 of Pi[(x^2)^2 - (x^3)^2]
integral from 0 to 1 of Pi[x^4 - x^6]
Pi[x^5/5 - x^7/7] evaluated from 0 to 1
Pi(1/5 - 1/7) = Pi(7/35 - 5/35) = Pi(2/35)
To calculate the volume, you need to know the equation of the curve that forms the outer edge of the shape minus the equation of the curve that forms the inner edge of the shape. The bounds of integration are the points of intersection of the curves.
Set the 2 equations equal to each other and solve for x.
x^3 = x^2
x^3 - x^2 = 0
x^2(x - 1) = 0
x = 0 and x = 1. This can also be seen by a graph of the two equations.
Now that we have the points of intersection, we take pi times the integral of the outer radius squared minus the inner radius squared
integrate from 0 to 1 of Pi[(x^2)^2 - (x^3)^2]
integral from 0 to 1 of Pi[x^4 - x^6]
Pi[x^5/5 - x^7/7] evaluated from 0 to 1
Pi(1/5 - 1/7) = Pi(7/35 - 5/35) = Pi(2/35)
Thursday, April 25, 2013
Tuesday, April 23, 2013
A way to measure the spread of data about the mean is by using the
standard deviation. If the spread is small, the standard deviation is
small. If the spread is large, the standard deviation is large. The
proportion of the data that falls within a certain number of standard
deviations from the mean is very definite and precise with a bell shaped
curve. But what can be said about the proportion of data spread about
the mean for other distributions such as skewed, symmetric, or other
shapes? Chebyshev's Theorem will solve this problem.
The basis of Chebyshev's Theorem is that no matter how large or small a data set is from a population or a sample, the proportion of data that lies within k standard deviations is at least 1 - 1/k2. For k = 2, the proportion is 1 - 1/4 = .75. For k = 3, the proportion is 1 - 1/9 = .889. For k = 4, the proportion is 1 - 1/16 = .938. These results mean that at least 75% of the data must fall within 2 standard deviations from the mean, 88.9% must fall within 3 standard deviations from the mean, and at least 93.8% must fall within 4 standard deviations from the mean.
Many distributions will have much greater percentages of the data falling within specified intervals. For example, in the well known normal distribution, which is bell shaped, 95% of the data falls within 2 standard deviations, 99.7% falls within 3 standard deviations from the mean, and virtually 100% falls within 4 standard deviations from the mean.
Here's an example using Chebyshev's Theorem.
Suppose students at a local college volunteer to work on community projects, such as cleaning parks, renovating playgrounds, and planting trees. A professor in charge of the program kept track of the time in hours that each student worked. Suppose a random sample of x students in the program were picked and the mean hours the students worked was 24.5 hours, and the standard deviation was 1.4 hours. From this information we can find an interval which at least 75%, 88.9% and 93.8% of the students worked.
Interval which at least 75% worked is 24.5 +/- 2(1.4) = 21.7 to 27.3 hours.
Interval which at least 88.9% worked is 24.5 +/- 3(1.4) = 20.3 to 28.7 hours.
Interval which at least 93.8% worked is 24.5 +/- 4(1.4) = 18.9 to 30.1 hours.
This guide should help students better understand Chebyshev's Theorem and how it can be applied.
The basis of Chebyshev's Theorem is that no matter how large or small a data set is from a population or a sample, the proportion of data that lies within k standard deviations is at least 1 - 1/k2. For k = 2, the proportion is 1 - 1/4 = .75. For k = 3, the proportion is 1 - 1/9 = .889. For k = 4, the proportion is 1 - 1/16 = .938. These results mean that at least 75% of the data must fall within 2 standard deviations from the mean, 88.9% must fall within 3 standard deviations from the mean, and at least 93.8% must fall within 4 standard deviations from the mean.
Many distributions will have much greater percentages of the data falling within specified intervals. For example, in the well known normal distribution, which is bell shaped, 95% of the data falls within 2 standard deviations, 99.7% falls within 3 standard deviations from the mean, and virtually 100% falls within 4 standard deviations from the mean.
Here's an example using Chebyshev's Theorem.
Suppose students at a local college volunteer to work on community projects, such as cleaning parks, renovating playgrounds, and planting trees. A professor in charge of the program kept track of the time in hours that each student worked. Suppose a random sample of x students in the program were picked and the mean hours the students worked was 24.5 hours, and the standard deviation was 1.4 hours. From this information we can find an interval which at least 75%, 88.9% and 93.8% of the students worked.
Interval which at least 75% worked is 24.5 +/- 2(1.4) = 21.7 to 27.3 hours.
Interval which at least 88.9% worked is 24.5 +/- 3(1.4) = 20.3 to 28.7 hours.
Interval which at least 93.8% worked is 24.5 +/- 4(1.4) = 18.9 to 30.1 hours.
This guide should help students better understand Chebyshev's Theorem and how it can be applied.
Saturday, April 20, 2013
Knowing the graphs of "parent functions", such as f(x) = x^2 and f(x) = |x| makes it simple to graph variations of such functions.
For example,
f(x) = |x| + 3 has the same graph as f(x) = |x| except it's translated 3 units up. Similarly f(x) = |x| - 3 would be the same as f(x) = |x| except it is translated 3 units down.
For f(x) = (x - 3)^2, the graph is the same as for f(x) = x^2 except it is translated 3 units to the right. Similarly, f(x) = (x + 3)^2 has the same graph as f(x) = x^2 except it is translated 3 units to the left.
You can also have translations horizontally and vertically in the same function
f(x) = (x + 4)^2 - 3 is translated 4 units left and 3 units down from the graph of f(x) = x^2
For example,
f(x) = |x| + 3 has the same graph as f(x) = |x| except it's translated 3 units up. Similarly f(x) = |x| - 3 would be the same as f(x) = |x| except it is translated 3 units down.
For f(x) = (x - 3)^2, the graph is the same as for f(x) = x^2 except it is translated 3 units to the right. Similarly, f(x) = (x + 3)^2 has the same graph as f(x) = x^2 except it is translated 3 units to the left.
You can also have translations horizontally and vertically in the same function
f(x) = (x + 4)^2 - 3 is translated 4 units left and 3 units down from the graph of f(x) = x^2
Thursday, April 18, 2013
Graphs Used to Display Qualitative Data
Bar graph
A bar graph can be horizontal or vertical that are uniformly spaced and have uniform width. The length of each bar show the frequency or percentage of occurrence, depending on what is to be displayed in the data. The lengths of the bars show the variable and the values of the variable being displayed. The graph must be titled with labels for each bar and a scale or precise value for the length of each bar.
For example: Suppose you wish to display the population of Chicago for years 1982, 1992, 2002, and 2012. The population is on the vertical axis and the years are on the horizontal axis. The length of the bars represent the population for each year. The title of the graph might be "Population For Chicago".
Suppose you wish to break down the population into male and female. This type of graph is known as a clustered bar graph because there are two bars for each year the population is measured. One bar will be for male population, and one bar for female population.
A type of bar graph in which the height of each bar represents frequency is known as a Pareto chart. A distinguishing feature of this bar graph is that the bars are arranged from left to right in decreasing order of frequency. Therefore, the graph will have the highest bar at the far left and lowest bar at the far right.
Circle graph
In a circle graph, also known as a "pie chart", wedges of a circle represent a percent of a population which have a common trait.
For example, suppose you want to know how much time Americans aged 25 to 50 watch television after 7 pm on a weeknight. Suppose in a sample of 200, 25 people watch up to 1 hour of television, 50 people watch between 1 and 2 hours of television, 100 watch between 2 and 3 hours of television, and 25 people watch over 3 hours of television.
That means there are 4 wedges of the pie, 12.5% represent up to 1 hour of television, 12.5% represent 3 or more hours, 25% represent 1 to 2 hours, and 50 percent represent 2 to 3 hours.
Draw the circle and make appropriate sized wedges to show the various percentages, and notice that the total percentages is 100. Label each piece and mark each piece of the chart with the designated numbers of degrees.
Time-series graph
When you want to track a change over time, the best type of graph to use is a time-series graph. For example, suppose you start exercising and you ride a bike for 45 minutes. You want to monitor your progress over a month period of time to see how much farther you can bike at the end of the month compared to the beginning of the month.
The time-series graph shows data measurements in chronological order. Time is placed on the horizontal axis, and the variable of interest is placed on the vertical axis. The basic time-series graph is made by connecting the data points by lines.
This article should give students a solid foundation for understanding graphs that can used to display qualitative data, as well as quantitative data.
Monday, April 15, 2013
When quantitative data is used in statistics, a histogram is a great way
to provide a useful visual display. But when the data is qualitative,
other types of graphs must be used. These graphs may also be used to
display quantitative data. The graphs are the bar graph, circle graph,
and time-series graph.
Bar graph
A bar graph can be horizontal or vertical that are uniformly spaced and have uniform width. The length of each bar show the frequency or percentage of occurrence, depending on what is to be displayed in the data. The lengths of the bars show the variable and the values of the variable being displayed. The graph must be titled with labels for each bar and a scale or precise value for the length of each bar.
For example: Suppose you wish to display the population of Chicago for years 1982, 1992, 2002, and 2012. The population is on the vertical axis and the years are on the horizontal axis. The length of the bars represent the population for each year. The title of the graph might be "Population For Chicago".
Suppose you wish to break down the population into male and female. This type of graph is known as a clustered bar graph because there are two bars for each year the population is measured. One bar will be for male population, and one bar for female population.
A type of bar graph in which the height of each bar represents frequency is known as a Pareto chart. A distinguishing feature of this bar graph is that the bars are arranged from left to right in decreasing order of frequency. Therefore, the graph will have the highest bar at the far left and lowest bar at the far right.
Circle graph
In a circle graph, also known as a "pie chart", wedges of a circle represent a percent of a population which have a common trait.
For example, suppose you want to know how much time Americans aged 25 to 50 watch television after 7 pm on a weeknight. Suppose in a sample of 200, 25 people watch up to 1 hour of television, 50 people watch between 1 and 2 hours of television, 100 watch between 2 and 3 hours of television, and 25 people watch over 3 hours of television.
That means there are 4 wedges of the pie, 12.5% represent up to 1 hour of television, 12.5% represent 3 or more hours, 25% represent 1 to 2 hours, and 50 percent represent 2 to 3 hours.
Draw the circle and make appropriate sized wedges to show the various percentages, and notice that the total percentages is 100. Label each piece and mark each piece of the chart with the designated numbers of degrees.
Time-series graph
When you want to track a change over time, the best type of graph to use is a time-series graph. For example, suppose you start exercising and you ride a bike for 45 minutes. You want to monitor your progress over a month period of time to see how much farther you can bike at the end of the month compared to the beginning of the month.
The time-series graph shows data measurements in chronological order. Time is placed on the horizontal axis, and the variable of interest is placed on the vertical axis. The basic time-series graph is made by connecting the data points by lines.
This article should give students a solid foundation for understanding graphs that can used to display qualitative data, as well as quantitative data.
Bar graph
A bar graph can be horizontal or vertical that are uniformly spaced and have uniform width. The length of each bar show the frequency or percentage of occurrence, depending on what is to be displayed in the data. The lengths of the bars show the variable and the values of the variable being displayed. The graph must be titled with labels for each bar and a scale or precise value for the length of each bar.
For example: Suppose you wish to display the population of Chicago for years 1982, 1992, 2002, and 2012. The population is on the vertical axis and the years are on the horizontal axis. The length of the bars represent the population for each year. The title of the graph might be "Population For Chicago".
Suppose you wish to break down the population into male and female. This type of graph is known as a clustered bar graph because there are two bars for each year the population is measured. One bar will be for male population, and one bar for female population.
A type of bar graph in which the height of each bar represents frequency is known as a Pareto chart. A distinguishing feature of this bar graph is that the bars are arranged from left to right in decreasing order of frequency. Therefore, the graph will have the highest bar at the far left and lowest bar at the far right.
Circle graph
In a circle graph, also known as a "pie chart", wedges of a circle represent a percent of a population which have a common trait.
For example, suppose you want to know how much time Americans aged 25 to 50 watch television after 7 pm on a weeknight. Suppose in a sample of 200, 25 people watch up to 1 hour of television, 50 people watch between 1 and 2 hours of television, 100 watch between 2 and 3 hours of television, and 25 people watch over 3 hours of television.
That means there are 4 wedges of the pie, 12.5% represent up to 1 hour of television, 12.5% represent 3 or more hours, 25% represent 1 to 2 hours, and 50 percent represent 2 to 3 hours.
Draw the circle and make appropriate sized wedges to show the various percentages, and notice that the total percentages is 100. Label each piece and mark each piece of the chart with the designated numbers of degrees.
Time-series graph
When you want to track a change over time, the best type of graph to use is a time-series graph. For example, suppose you start exercising and you ride a bike for 45 minutes. You want to monitor your progress over a month period of time to see how much farther you can bike at the end of the month compared to the beginning of the month.
The time-series graph shows data measurements in chronological order. Time is placed on the horizontal axis, and the variable of interest is placed on the vertical axis. The basic time-series graph is made by connecting the data points by lines.
This article should give students a solid foundation for understanding graphs that can used to display qualitative data, as well as quantitative data.
Saturday, April 13, 2013
When dealing with the unit circle, sometimes it's easier to remember angles and convert them to radians.
The angles along the unit circle in degrees are
0, 30, 45, 60, 90, 120, 135, 150, 180, 210, 225, 240, 270, 300, 315, 330
To convert angles from degrees to radians, multiply by Pi/180. Therefore the above angles in radians are
0, Pi/6, Pi/4, Pi/3, Pi/2, 2Pi/3 3Pi/4, 5Pi/6, Pi, 7Pi/6, 5Pi/4, 4Pi/3, 3Pi/2, 5Pi/3, 7Pi/4, 11Pi/6
The values for sin and cos along the unit circle will be +/- 1/2, +/- sqrt(2)/2, +/- sqrt(3)/2. The + or - depends on what quadrant the angle falls.
The angles along the unit circle in degrees are
0, 30, 45, 60, 90, 120, 135, 150, 180, 210, 225, 240, 270, 300, 315, 330
To convert angles from degrees to radians, multiply by Pi/180. Therefore the above angles in radians are
0, Pi/6, Pi/4, Pi/3, Pi/2, 2Pi/3 3Pi/4, 5Pi/6, Pi, 7Pi/6, 5Pi/4, 4Pi/3, 3Pi/2, 5Pi/3, 7Pi/4, 11Pi/6
The values for sin and cos along the unit circle will be +/- 1/2, +/- sqrt(2)/2, +/- sqrt(3)/2. The + or - depends on what quadrant the angle falls.
Wednesday, April 10, 2013
Understanding Sampling Techniques
There are five different types of sampling techniques that are commonly used. Those techniques are as follows:
Simple Random
In a simple random sample, "n" objects are selected from a population in such a way that each sample of size "n" is equally liked to be chosen from the population. Also, every member of the population is equally likely to be chosen.
Stratified
in a stratified sample, the population is divided into subgroups called strata. All members of each strata fit a specified trait, such as height, gender, income level, education, and so on. Then random samples are drawn from each strata.
Systematic
In a systematic sample, each member of the population is numbered in order. Then from a specified point, select every so many members to be included in the sample. For example, you can choose to have every 4th, 10th, 15th member, etc, to be part of the sample.
Cluster
In cluster sampling, the population is divided into specific pre-existing groups. Many times these groups are geographic regions. Then, randomly select a group of clusters, and every member from the cluster is part of the sample.
Convenience
In convenience sampling, the sample is chosen by using data from members of the population that are most convenient to use or easiest to obtain.
Here's some examples of the different types of sampling techniques.
Simple random: Assign each teacher in the Wilson School District a number, then select the teachers to be included in the sample using a random number generator.
Stratified: Group sports franchises according to sport: baseball, football, basketball, hockey. Then select a random sample of 12 teams from each sport.
Systematic: Use the Wilson School District faculty directory. Number all of the teachers. Select a starting point in the list of teachers and select every 25th to be included in the sample. Continue selecting members of the sample in this manner until 50 teachers are selected.
Cluster: Divide a state into regions by using the counties. Pick a random sample of 10 counties and include all the businesses in each selected county.
Convenience: Choose 10 newspaper reporters from a local newspaper. Have each reporter select a neighborhood in the area and interview a business owner from any business found. The reporter is finished after interviewing 15 business owners.
This guide should help assist any student having difficulty understanding and distinguishing between the different types of sampling techniques.
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