## Saturday, October 31, 2015

Steps for hypothesis test

For the first part we want to prove the claim that mean life of time of battery exceeds 400 hours, so Ho would be that the mean is 400 and Ha is that the mean is greater than 400
step 1: Ho: Mu = 400
Ha: Mu > 400
for the second step, recall that since sample size is small and standard deviation is not known, have to use the t-distribution, so test statistic is t
t = (sample mean - population mean)/(sample standard deviation/square root (n))
x-bar = 473.46
s = 210.77 as done on the calculator
n = 13
t = (473.46 - 400)/(210.77/square root(13))
t = 1.26
For the next step,we know using the p-value approach that we reject if p-value is less than alpha level of the test. So we reject Ho if p-value < .025
For the next step, I obtained the p-value from this site..
http://www.socscistatistics.com/pvalues/tdistribution.aspx
Putting in the values for t-statistic and one-tailed test, p-value is .11581
Last step, since .115811 > .025, we do not reject Ho since there is not significant evidence to conclude mean battery life is more than 400 hours.

## Friday, October 23, 2015

A random sample of size 49 is taken from a population with mean 31and standard deviation of 12.
What are the expected value and the standard deviation of the sample mean ?
Describe the probability distribution of .
What is the probability that the sample mean is greater than 32?
What is the probability that the sample mean falls between 27 and 29?
What is the probability that the sample mean will be within ±3 of the population mean?

The distribution is the normal distribution which is symmetrical about the mean. Symmetric means it looks the same on both sides of the mean.
P(sample mean > 32)
We need to find a Z-score, which shows how many standard deviations away from the mean a value is.
Z = (sample mean - population mean)/(standard deviation/square root(n))
Z = (32 - 31)/(12/square root (49))
Z = 1/(12/7)
Z = 0.58

Now find Z(0.58) on the standard normal distribution chart. Since the chart shows probabilities less than, we need to take 1- Z(0.58)
Here is the chat I used for all these. http://www.regentsprep.org/regents/math/algtrig/ats7/zchart.htm
Z(0.58) = .7190
1- .7190 = .2010
The probability it falls between 27 and 29 ... P(27 < X < 29)
We need 2 Z-scores now, one for 27 and one for 29
Z = (27 -31)/(12/sqrt(49))
Z = -4/(12/7)
Z = -2.33
and
Z = (29 - 31)/(12/sqrt(49))
Z = -2/(12/7)
Z = -1.17
Now we get Z(-1.17) and Z(-2.33) and subtract them
.1210 - .0099 = .1111
Probability sample mean is within +/- 3standard deviations we simply find Z(3) and Z(-3) and subtract
So we get .9987 - .0013 = .9974

## Sunday, October 18, 2015

he 2003 Statistical Abstract of the United States reported the percentage of people 18 years of age and older who smoke. Suppose that a study designed to collect new data on smokers and nonsmokers uses a preliminary estimate of the proportion who smoke of .26.
a. How large a sample should be taken to estimate the proportion of smokers in the population with a margin of error of .02 (to the nearest whole number)? Use 95% confidence.
b. Assume that the study uses your sample size recommendation in part (a) and finds 520 smokers. What is the point estimate of the proportion of smokers in the population (to 4 decimals)?
c. What is the 95% confidence interval for the proportion of smokers in the population (to 4 decimals)?

The margin of error is the 1.96(sqrt(pq/n) and that has to equal .02. We know p =.26, so q = .74
So we have 1.96(sqrt(.26*.74/n) = .02
3.8416(.1924)/n = .0004
0.73912384/n = .0004
n = 1848
b) point estimate is 524/1848 = .2835
c) .26 +/- .02 = (.24, .28)

## Wednesday, October 7, 2015

In this recession, yours truly, CEO of the Outrageous Products Enterprise, would like to make extra money to support my frequent filet-mignon-and-double-lobster-tail dinner habit. A promising enterprise is to mass-produce tourmaline wedding rings for brides. Based on my diligent research, I have found out that women's ring size normally distributed with a mean of 6.0, and a standard deviation of 1.0. I am going to order 5000 tourmaline wedding rings from my reliable Siberian source. They will manufacture ring size from 4.0, 4.5, 5.0, 5.5, 6.0, 6.5, 7.0, 7.5, 8.0, 8.5, 9.0, and 9.5. How many wedding rings should I order for each of the ring size should I order 5000 rings altogether?

what we have to do is notice that the ring sizes or 4, 4.5, 5, 5.5, 6, 6.5, 7, 7.5, 8, 8.5, 9 and 9.5 are a certain number of standard deviations from the mean. This is how we will solve this
4 is -2 st dev obtained by taking (4 - 6)/1 = -2
4.5 is -1.5 st dev by taking (4.5 - 6)/1 = -1.5 etc
5 is -1 st dev
5.5 is -0.5 st dev
6 is the mean , so 0 st dev
6.5 is 0.5 st dev
7 is 1 st dev
7.5 is 1.5 st dev
8 is 2 st dev
8.5 is 2.5 st dev
9 is 3 st dev
9.5 is 3.5 st dev.
This is how we will figure out what percentage of each ring should be bought. We can't just say we look at standard normal distribution probabilities and take P(Z = -2) for size 4.0 because remember in a continuous distribution, there are no probabilities for exact values. So we take P(-2.5 < Z < -2.0) = .0166 (i did half standard deviations because each ring size is half st dev apart) . Now take .0166(5000) = 83
The same method is applied to each size
size 4.5 = P(-2 < Z < -1.5) = .044, multiplied by 5000 = 220
size 5 = P(-1.5 < Z < -1) = .0919, multiplied by 5000 = 460
size 5.5 = P(-1 < Z < -0.5) = .1498, multiplied by 5000 = 749
size 6 = P(-0.5 < Z < 0) = .1915, multiplied by 5000 = 958
Since this is the normal distribution, this is symmetrical, so size 6.5 = 958, size 7 = 749, size 7.5 = 460, size 8 = 220, size 8.5 = 83
For size 9, same idea = P(2.5 < Z < 3) = .0049 multiplied by 5000 = 25
size p.5 = P(3 < Z < 3.5) = .0011 times 5000 = 6
Now that gives a total of 4971, the rest are either bigger than 9.5 or less than 4, since no rings will be produced bigger than size 9.5, the other 29 should be of size 4, making size 4 total 112

## Friday, October 2, 2015

Suppose you know that the heights of elementary school students in southern Kenya follow a normal probability distribution with a mean of 40.1 inches and a (population) standard deviation of 2.2 inches.

What is the probability that an elementary school student in southern Kenya is greater than or equal to 41.3 inches tall? (Round to the nearest percent.)

z score is (x - mean)/standard deviation

so (41.3 - 40.1)/2.2
• and look up that on a standard normal distribution chart
• z-chart
• to that value for z (.55) is .7088
•   that is the probability of LESS than z(.55)
So you take 1- .7088 to get .2912