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.

## Thursday, April 28, 2016

## Saturday, April 23, 2016

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.

## Saturday, April 16, 2016

### 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.

## Sunday, April 10, 2016

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.

## Monday, April 4, 2016

### 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

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

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