With the most recent US Presidential election in the books, let me examine a few numbers closely. The two most populous states, (California and New York) gave Hillary Clinton the advantage in the popular vote. Donald Trump had more votes in the other 48 states overall. The four states with the most population make up 1/3 of the total US population, while the other 46 states account for only 67% of the population. This is why the electoral college is in place, so a few states can't determine the election when the vast majority of the other states show support for the other candidate. Each state has a proportional amount of electoral votes depending on the population in the state. Each state has its fair stake in the outcome of the election. It's a process that has worked and will continue to work, even if sometimes (on a rare occasion) the popular vote is for the candidate that ultimately loses.

For the hypothesis test, we need to find the test statistics

t = (x-bar - Mu)/standard error

standard error = sample standard deviation/square root of n

From here we compare the test statistic to the critical value t or you can compute the p-value and compare to alpha level of the test at .05

If the p-value < alpha, then reject Ho. Notice in the case of your problem, there is a rejection of the null hypothesis in both cases.

For the confidence interval it's mean +/- tcritical*standard error.

To calculate the Q1 (25th percentile) take n times .25 and that value is the data value that is Q1. If n(.25) does not come out even, round up to the next integer. Remember the data values must be sorted in order from lowest to highest for this. Q3 is found by taking n(.75) and round up if necessary and that is the value in order of Q3, IQR is interquartile range and is Q3-Q1. We use that to find out if there are any outliers. Any data value less than Q1 - 1.5(IQR) or greater than Q3 + 1.5(IQR) is an outlie

Mean (x + y) = Expected value (X + Y) = E(x) + E(y) which means the mean is meanx + meany, so is meanx = 75 and meany =70, then the mean (x +y) =  75 + 70 = 145

The Var(x + y) = Var(x) + Var(y) if x and y are independent.
Var(x) = 6^2 = 36
Var(y) = 8^2 + 64
Var(x+y) = 100. The standard deviation is the square root of the variance, so standard deviation = 10

The mean of the difference is the difference of the means E(X - Y) = mean (x - y) = meanx - meany

So the difference of the means is 75-70 = 5

The standard deviation of the difference is the square root of the Var(x-y)

Var(x - y) = Var(x) + Var(y) if x and y are independent, same as for Var(x +y) . Therefore the answer is 10

Just noticed a few sales on the books I wrote a few years ago. I had forgotten about them since sales were minimal. I was just pleased to put out my work on algebra. Now looking at completing a long standing project involving math for kids K through 5.  Check out the link below for the books.

http://www.lulu.com/spotlight/KKauffman1969

As the school year is under way in many areas, don't forget my self help algebra books on lulu.com. Working on a kids book that has been stagnant lately, but getting back to it. Thanks for checking!

http://www.lulu.com/spotlight/KKauffman1969

Remember that the t distribution is more wide than the standard normal distribution, but at n gets larger and larger, the t distribution become approximately normal in nature.

The t-test is used when population standard deviation is not known. When we know that population is normally distributed, then we can use Z and also when population standard deviation is known, we can use Z

Note that if you want to figure out if a function has a slant asymptote, you have to realize that the equation of a slant asymptote is linear, so the exponent of the leading coefficient of the numerator of the function must be one greater than that of the denominator.

When doing long division (denominator into numerator), it might not divide evenly. Do not concern yourself with the remainder. That slant or oblique asymptote is just the linear portion.

Suppose you roll two 6 sides dice and we want to see the outcomes and probability distribution for the difference between the two dice. 3 on the first and 2 on the second would be 3-2=1 and 2 on the first and 3 on the second would be 2-3 = -1

To get a 0 when subtracting the numbers on the dice, the numbers must be the same. This happens 6 ways.

(1,1), (2,2), (3,3), (4,4), (5,5), (6,6).. since there are 36 possibilities, we have 6/36 = 1/6

For problem 6, to subtract numbers to get -2, the second number must be two larger than the first number.

(1,3), (2,4), (3,5), (4,6) . So we have 4 out of 36 outcomes that will subtract to -2. The probability is 4/36 = 1/9

For problem 7, It's basically the same as problem 6, except now the first number is two larger than the second number.

This gives us (3,1), (4,2), (5,3), (6,4). Again this is 4/36, simplified to 1/9

For the last problem, we want the possible outcomes with corresponding probabilities

We can get 0 if the numbers are the same, which I showed in problem 5, that is probability 1/6

We can get 1 when subtracting the numbers if we have (2,1), (3,2), (4,3), (5,4), (6,5) that is 5/36

We can get -1 when subtracting if we have (1,2), (2,3), (3,4), (4,5), (5,6), again with probability 5/36

We can get 2 as shown in problem 7 with probability of 1/9
We can get -2 as shown in problem 6 with probability of 1/9
We can get 3 with rolls of (4,1), (5,2), (6,3) with probability of 3/36 = 1/12
We can get -3 with rolls (1,4), (2,5), (3,6) with probability of 3/36 = 1/12
We can get 4 with rolls (5,1), (6,2) with probability of 2/36 = 1/18
We can get -4 with rolls (1,5), (2,6) with probability of 2/36 = 1/18
We can get 5 with a roll of (6,1) with probability 1/36
We can get -5 with a roll of (1,6) with probability of 1/36

Putting it all together we get this model

outcome Probability
5 1/36
4 1/18
3 1/12
2 1/9
1 5/36
0 1/6
-1 5/36
-2 1/9
-3 1/12
-4 1/18
-5 1/36

Notice that the probabilities will add to 1. That must always be the case for a legitimate probability model and the probabilities for each event must be between 0 and 1 inclusive

Suppose you have the following distribution

x = 0, 2, 5

p(x) = 1/4, 1/4, 1/2

find the mean and variance.

To get the mean you take the sum of x(P(x))

so for x = 0, 1,5 with P(x) = 1/4, 1/4, and 1/2

You get 0(1/4) + 1(1/4) + 5(1/2) = 0 + 1/4 + 5/2 = 2.75

To calculate sigma squared (variance)

It's the [sum (x- mean)^2P(x)]/n

So we have (0 - 2.75)^2 + (2-2.75)^2 + (5-2.75)^2

The equals 7.5625 + 0.5625 + 5.0625 = 13.1875
now take 13.1875/3 = 4.396

For a sample size of two you can have these possibilities (0,0), (0,2), (0,5), (2,0), (2,2), (2,5), (5,0), (5,2), (5,5)

The means are the two numbers added and divided by two. That gives us
(0 + 0)/2 = 0
(0 + 2)/2 = 1
(0 + 5)/2 = 2.5
(2 + 0)/2 = 1
(2 + 2)/2 = 2
(2 + 5)/2 = 3.5
(5 + 0)/2 = 2.5
(5 +2)/2 = 3.5
(5 + 5)/2 + 5

So you can have mean of 0 with (0, 0) with probability (1/4)(1/4) = 1/16
mean of 1 with (0, 2) and (2,0) with probability 2(1/4)(1/4) = 1/8
mean of 2 with (2, 2) with probability of (1/4)(1/4) = 1/16
mean of 2.5 with (0,5) and (5,0) with probability of 2(1/4)(1/2) = 1/4
mean of 3.5 with (2,5) and (5,2) with probability of 2(1/4)(1/2) = 1/4
mean of 5 with (5,5) with probability of (1/2)(1/2) = 1/4

When conducting an ANOVA, it's important to realize that the test in and of itself will not tell which of the means in the null hypothesis are different from each other. The only thing we will know is whether or not all the means are equal or if one or more of the means are different. If you want to know which mean or means are different from the others you must perform one of the post hoc tests. You can find more information about these with a Google search on the topic.

Suppose you need to take a log transformation on set of data that is non-linear in nature.  You can tell this from a scatterplot.  Take ln (natural logarithm) of each x value and y value to complete the transformation. In doing so, you will get some errors in x for ln(0) and negative values for ln of values between 0 and 1. So to compensate for that add 1 to each x value before taking natural log. Then you will eliminate such problems and can get the regression equation and scatterplot accordingly

If we are now testing to see if two variances are equal, as compared to where we wanted to see if the variance equaled a certain value, we use an F test instead of a Chi-square test.  So the test statistic is an F statistic  equal to s1^2/s2^2 and we compare to an F critical value found in any F chart. The decision rule is reject if F falls in the rejection region and do not reject if F is not in the rejection region.  The hypotheses for this test would be Ho: variance1 = variance2,  Ha: variance1 does not equal variance2

Remember when solving a system of equations, you can either work using substitution or elimination. When dealing with a 3 variable, 3 equation system, generally you want to eliminate a variable from two equations, then eliminate the same variable from another set of two equations. From there solve for one of the two variables in the newly formed system of 2 equations with two variables. From here you can do back substitution to find the values of the other variables. Be sure to check all answers with each equation to make sure they satisfy each.

Suppose one travels from point (2,3) to point (11,8)  at a constant speed for one hour. How far did they travel in 35 minutes and where is their location in terms of x and y?

The key is to use the distance formula and slope. For the distance formula, recall it is the square root[(x2-x1)^2 + (y2-y1)^ You can apply that to get the total distance traveled at approximately 10.3, Now I use the slope to determine how much rise and run there is in 35 minutes. Take the slope of see that there is 5 rise and 9 run over the course of 60 minutes. Now take 5(35/60) and 9(35/60) to see how far the x and y coordinates move in the 35 minutes.

we have to know that to find concavity we need the second derivative and since we are looking for concave up, it's where y'' > 0. Given that y is an integral, y' is the integral evaluated at x, so it's y' = 6/(1+2x+ x^2), then for the second derivative I used the quotient rule. I set the second derivative > 0 and see that x > -1 gives us concave up

USA Today reported that the state with the longest mean life span is Hawaii, where the population mean life span is 77 years. A random sample of 20 obituary notices in the Honoululu Advertiser gave the following information about the life span in years of Honolulu residents.
72,68,81,93,56,19,78,94,83,84,77,69,85,97,75,71,86,47,66,27
We are testing to see if the information indicates that the mean life span for Honolulu residents is less than 77.
Ho: Mu = 77
H1: Mu < 77
since population standard deviation is not known, we use t test with 19 df and we are testing at alpha = .05. So the critical value t for a left -tailed test is -1.729 (value obtain on any critical value chart for t distribution)
The results do not have any real limitations since the sample is simple random, no bias involved in sampling.
test statistic t = (x-bar - 77)/(standard deviation/square root(n))
x-bar = 71.4
s= 20.65
t = (71.4 - 77)/(20.65/sqrt(20))
t= -1.213
Since -1.213 > -1.729, we do not reject Ho and conclude there is not sufficient evidence to support the claim that the population mean life span for residence in Hononlulu is less than 77 years.

Suppose you want to test to see if exam scores between men and women in a chemistry class are different. In that case the test would be two-tailed because we are not specifying a test that men score higher than women or women score higher than men. We are just looking for a difference, which could go in either direction, thus 2-tailed.
A one tailed test would be testing to see if men score higher than women, which would then have a null hypothesis of Ho: Mu1 = Mu2 and Ha: Mu1 > Mu2 where Mu1 is the mens mean and Mu2 is the women's mean. Likewise you could test to see if women score higher than men's score, so Ho would be the same and Ha would be Mu1 < Mu2.
If the significance level of the test is .05, then for a two tailed test, there would be .025 on the left tail and .025 on the right tail, and if it is a Z test, the critical values would be -1.96 and 1.96.
For a right tailed test, still with significance of .05, the critical value for a Z test would be 1.645 (notice it is lower because ALL of the .05 is i one-tail, making easier to reject). Likewise the critical value if this was a left-tailed test would be -1.645.

Confidence infrerval for proportion example

 "Many teens understand the risks of texting behind the wheel," said Amanda
Lenhart, co-author of the Pew report, "but the desire to stay connected is so strong for teens and
their parents that safety sometimes takes a back seat to staying in touch with friends and family."
Overall, 81 percent of U.S. residents use cell-phone while driving, according to the National
Highway Traffic Safety Administration. Consider a random sample of 100 U.S. residents from
which 89 admitted using cell-phones while driving.
Perform now an appropriate test hypothesis to check if the proportion of Americans who used
their cellphone while driving is significantly higher than 0.81 at 5% level.


For this problem, you need to recognize that this will be a test for proportion, so the parameter is population proportion p, then from the problem we know the hypotheses must be p = .81 and p > .81. The level of significance is given in the problem at .05. Remember that the null hypothesis always contains an equals sign. The conditions we have to see if np and n(1-p) are greater than 5. Some texts say they must be greater than 10, not sure what your book says, but either way they are both greater than 10 as well. Now we need the test statistic Z, which is p^-p divided by square root(pq/n), where q = 1-p.  Now once we have that test statistic, compare it to the critical value which we can find using the standard normal distribution chart and Z at .05 significance for right-tailed test is 1.645.

Why is the t-distribution more variable and flatter than the normal distribution?

For z scores and t scores, both have x-bar - mu in the numerator but the denominators are different. For z, the bottom does not vary provided all samples are the same size and chosen from same population, basically the standard error is the same for every sample. But with t distribution the standard deviation and thus the standard error change from sample to sample, causing more variability which makes flatter and more spread out. With larger sample size this variability decreases.

Hypothesis test example

Among private universities in the United States, the mean ratio of students to professors is 35.2 (i.e., 35.2 students for each professor) with a standard deviation of 8.8. a. What is the probability that in a random sample of 25 private universities that the mean student-to-professor ratio exceeds 38?

Suppose a random sample of 25 universities is selected and the observed mean student-to- professor ratio is 38. Is there evidence that the reported mean ratio actually exceeds 35.2?

 For this we need P(X-bar > 38) so we get a Z score, where Z = (x-bar - mean)/(standard deviation/square root(n))
Z= (38-35.2)/(8.8/5)
Z = 3.2/1.76
Z = 1.82
Z(1.82) obtained from a standard normal distribution chart is .9656. Since we want P(X > 38) we take 1- .9656 = .0344
For part b
Ho: Mu = 35.2
Ha: Mu > 35.2
Test statistic : (x-bar - Mu)/(standard deviation/square root(n))
The population standard deviation is known, so use Z
Z = (38-35.2)/(8.8/5)
Z = 3.2/1.76 = 1.81
Decision rule: Critical value for .05 significance it 1.645, so we reject Ho if test statistic > 1.645 and do not reject otherwise
Decision is to reject Ho
Conclusion is that the student to professor ratio exceeds 35.2

In statistics, if a problem asks for the sensitivity, specificity or accuracy or a test, you just need the following formulas:

Sensitivity = (true positives)/(true pos. + false neg)
Specificity = (true negatives)/(true neg. + false pos)
Accuracy = (true negative + true positive)/total

These values are generally easy to find given a 2 x 2 matrix with the data for the problem

The weights of professional baseball players follow a normal distribution
with a mean of 200 pounds and a standard deviation of 25 pounds.
Find the probability that a randomly selected professional baseball
player has a weight less than 225 pounds, P(x < 225) - ?

Z value is (x-mean)/standard deviation
x = 225
mean= 200
standard deviation = 25
Z = (225 - 200)/25 = 1
Now look up Z(1) on the chart and you will see it is .8413. That is the answer since they are asking for P(x < 225) if it was P(x > 225) then would have to take 1- .8413

Here's an example of solving equations using trig functions.


2sin*sin2x - cosx =0

(know that sin2x = 2sinxcosx, therefore we have

4sin^xcosx - cosx =0

cosx(4sin^2(x) - 1) = 0

cosx = 1,  4sin^2(x) = 1

x = 0,       sin^2(x) = 1/4, sinx = +/- sqrt(2)/2

x= pi/4, 3pi/4, 5pi/4 and 7pi/4

Here's an example of solving for dy/dx using implicit differentiation


y^2 + 2xy^3 - 3x^2 = 14

2y(dy/dx) + 2x(3y^2)(dy/dx) + 2y^3 - 6x = 0

2y(dy/dx) + 6xy^2(dy/dx) = 6x - 2y^3

dy/dx(2y +6xy^2) = 6x - 2y^3

dy/dx = (6x-2y^3)/(6xy^2 + 2y)

dy/dx = (3x-y^3)/(3xy^2 + y)

For Normal distribution with mean μ = 400
and standard deviation σ = 100 find probability
that x is between 340 and 714: P( 340 < x < 714)

0.3427

0.5462

0.7249

0.9140

We have mean = 400 and standard deviation of 100, now we want P(X > 619), Z = (619 - 400)/100 = 2.19

Z(2.19) = .9857, so we take 1- .9857 since the probability is greater than, so we get .0143
For the next question, mean is 400 and standard deviation is 100, want: P( 340 < x < 714)
Z = (340 - 400)/100 = -0.6 and Z = (714 - 400)/100 = 3.14

So now we find Z(3.14) and Z(-0.6) and subtract them to get the probability in between. .9992 - .2743 = .7249

A quick note:

For the test just comparing two means when population standard deviation is not known and sample size is small, you can do a t-test . The formula for the test statistic is t = (x1-bar - x2-bar)/sqrt(s1^2/n1 +s2^2/n2), then compare to the critical value t , found using any t-distribution chart.

we calculate Chi-square by getting the sum(Observed - Expected)^2/(Expected)



You get the expected in each cell by taking row total times column total divided by overall total
Expected Low Drunk Driving and Low Under Age is (22)(20)/50 = 8.8
Expected High Drunk Driving and Low Under Age is (28)(20)/50 = 11.2
Expected Low Drunk Driving and High Under Age is (22)(30)/50 = 13.2
Expected High Drunk Driving and High Under Age is (28)(30)/50 = 16.8

so Chi-square statistic is (8-8.8)^2/8.8 + (14-13.2)^2/13.2 + (12 -11.2)^2/11.2 + (16- 16.8)^2/16.8 = 0.216

Now we get the p-value, with df of (row-1)(column - 1). There are 2 rows and 2 columns in the contingency table, so df = 1. I use this site.... http://www.socscistatistics.com/pvalues/chidistribution.aspx
The p-value is 0.642105. Since the p-value > .05

Suppose you want lim x-> a   [f(x) - f(a)](x-a) where a = 2


get [f(x) - f(a)](x-a) with a = 2, then take the limit as x -> 2
f(x) - f(a)/(x-a) =( 4.5x^2 -3x + 2 - (4.5(2)^2 - 3(2) + 2)]/(x- 2)

= (4.5x^2 - 3x + 2 - 14)/(x-2)
= (4.5x^2 -3x -12)/(x-2)
= (4.5x + 6)(x-2)/(x-2)
= 4.5x + 6

Now we take lim x-> 2 (4.5x + 6) = 4.5(2) + 6 = 15

Suppose you have an angle and you want to know where its terminal side lies. How do you go about figuring this out?  It's simply seeing where the angle ends. For example, if you have the 4 quadrants, first ending at 90 degrees, 2nd at 180, 3rd at 270 and 4th at 360, then a 225 degree angle would have its terminal side in the third quadrant since 225 falls between 180 and 270.

If the angle is negative in value, simply add 360 each time until you get a number between 0 and 360 and follow the same guidelines as above.

For example, an angle of -556 degrees would have it's terminal side in quadrant 2. Add 360 to -556 and you get -196, then add  360 again to get 164, which is a second quadrant angle

Here's an easy way to simplify higher powers of i.

First know the first 3 powers of i

i^1 = sqrt(-1)
i^2 = -1
i^3 = -i


To get any power of i greater than 4, simply divide by 4 and the integer value remainder gives the answer. It will be one of the above.

Example:

i^45....   45 divided by 4 is 11 with a remainder of 1, so the answer is sqrt(-1)

i^202..... 202 divided by 4 is 50 with a remainder of 2, so the answer is -1

i^79.... 79 divide by 4 is 19 with a remainder of 3, so the answer is -i

Chi square test

Suppose  the frequency tablef or the counties and votes for a particular candidate are as follows
Miami Dade 6
Broward 8
Palm Beach 34
Pinellas 10
Hillsbourough 8
Total is 66

Relative frequencies are each frequency divided by the total
Miami Dade = 6/66 = .09
Broward = 8/66 = .12
Palm Beach = 34/66 = .52
Pinellas = 10/56 = .18
Hillsborough = 8/66 = .12

b. If the votes were equal across each county, you'd expect 66/5 = 13.2 for each. So we get a X^2 statistic
(6 - 13.2)^2/13.2 + (8-13.2)^2/13.2 + (34-13.2)^2/13.2 + (10 - 13.2)^2/13.2 + (8-13.2)^2/13.2 = 41.59
The critical value for the test for X^2, 4df at alpha = .05 = 9.488
Since the 41.59 > 9.488 critical value, reject Ho

c. The conclusion is that there is significant evidence say that the proportion of voters for Buchanon is not the same across all the counties.

remember key trigonometric identities that can help you in intergration

sin^2 + cos^2 = 1
1 + tan^2 = sec^2
1 + cot^2 + csc^2

Also you need to know the derivatives of trig functions

sinx ..... derivative    cosx
cosx.....  derivative  -sinx
tanx ......derivative   sec^2 x
cotx ....  derivative  -csc^2 x
secx....   derivative   tanxsecx
cscx ....  derivative   -cotxcscx

the trig identities come in handy to do substitutions to obtain antiderivatives.