[1] Lipitor is a drug used to control cholesterol. In clinical trials of Lipitor, 94 subjects were treated with Lipitor and 270 subjects were given a placebo. Among those who were treated with Lipitor, 7 developed infections. Among those given a placebo, 27 developed infections. Use a 0.05 significance level to test the claim that the rate of infections was the same for those treated with Lipitor and those given a placebo.
[2] Simple random samples of high-interest (5.36%) mortgages and low-interest (3.77%) mortgage were obtained. For the 40 high-interest mortgages, the borrowers had a mean FICO credit score of 594.8 and standard deviation of 12.2. For the 40 low-interest mortgages, the borrowers had a mean FICO credit score of 785.2 and standard deviation of 16.3.
a) Use a 0.01 significance level to test the claim that the mean FICO score of borrowers with high-interest mortgage is lower than the mean FICO score of borrowers with low-interest mortgage.
b) Does the FICO credit rating score appear to affect mortgage payments? If so, how?

The test of interest here is,

As here the sample size for both the samples are large enough so we can assume the normality using CLT and thus a Z-test would be appropriate. And the hypotheses also show that it’s a left tail test.
The output is given below,

Here the lower p-value suggests that we should reject the null hypothesis and conclude that the mean FICO score of borrowers with high-interest mortgage is lower than the mean FICO score of borrowers with low-interest mortgage.

Note that the above analysis showed that the interest rate is higher for the group with lower FICO scores. Thus the FICO credit rating scores appear to affect mortgage payments. If the FICO score is high then the interest rate is expected to be low so the mortgage payments would be low. And for low FICO score the opposite is expected to be observed.[3] A random sample of the birth weights of 186 babies has a mean of 3103g and a standard deviation of 696g (based on data from “Cognitive Outcomes of Preschool Children with Prenatal Cocaine Exposure,” by Singer et al., Journal of the American Medical Association, Vol. 291, No. 20). These babies were born to mothers who did not use cocaine during their pregnancies. Further, a random sample of the birth weights of 190 babies born to mothers who used cocaine during their pregnancies has a mean of 2700g and a standard deviation of 645g.
a) The birth weights of babies are known normally distributed. Use a 0.05 significance level to test the claim that both samples are from populations having the same standard deviation.
b) Using your finding in (a), construct a 95% confidence interval estimate of the difference between the mean birth weight of a baby born to mothers who did not use cocaine and that of a baby born to mothers who used cocaine during their pregnancies.
c) Using your finding in (b), does cocaine use appear to affect the birth weight of a baby? Substantiate you conclusion.

a) Here we need to test for the equality of the standard deviation (same as variance) so the hypotheses of interest are,

As this is two sample test for standard deviation or variance so the test can be performed using a F test. The obtained output is given below.
-value is larger than the significance level so we are failing to reject the null hypothesis and concluding that the data does not suggest that both samples are from populations having different standard deviation.
b) Here we found out that the variations are not different in two groups so we should use a pooled variance t-test.

From the values from above table,
95% confidence interval for difference =

c) From the above calculated confidence interval we can see that the confidence interval does not contain the value 0. So based on the findings we can conclude that cocaine use appears to affect the birth weight of a baby.[4] A large discount chain compares the performance if its credit managers in Ohio and Illinois by comparing the mean dollars amounts owed by customers with delinquent charge accounts in these two states. Here a small mean dollar amount owed is desirable because it indicates that bad credit risks are not being extended large amounts of credit. Two independent, random samples of delinquent accounts are selected from the populations of delinquent accounts in Ohio and Illinois, respectively. The first sample, which consists of 15 randomly selected delinquent accounts in Ohio, givens a mean dollar amount of $524 with a standard deviation of $38. The second sample, which consists of 20 randomly selected delinquent accounts in Illinois, gives a mean dollar amount of $473 with a standard deviation of $22.
a) Assuming that the normality assumption, test to determine if the population variances are equal with  =0.05.
Here we need to test for the equality of the variance so the hypotheses of interest are,

As this is two sample test for variance and normality assumption is valid so the test can be performed using a F test. The obtained output is given below.
F Test for Differences in Two Variances

Data
Level of Significance
0.05
Larger-Variance Sample
 
Sample Size
15
Sample Variance
1444
Smaller-Variance Sample
 
Sample Size
20
Sample Variance
484

Intermediate Calculations
F Test Statistic
2.9835
Population 1 Sample Degrees of Freedom
14
Population 2 Sample Degrees of Freedom
19
 
 
Two-Tail Test
 
Upper Critical Value
2.6469
p-Value
0.0283
Reject the null hypothesis
 
Here the p-value is smaller than the significance level so we are rejecting the null hypothesis and concluding that the data suggests that the population variances are not equal.

b) Test with  =0.05whether there is a difference between the population mean dollar amounts owed by consumers with delinquent charge accounts in Ohio and Illinois.
Here the population variances are not equal so we should run a separate variance t test. The results are as follows,
Separate-Variances t Test for the Difference Between Two Means
(assumes unequal population variances)
Data
Hypothesized Difference
0
Level of Significance
0.05
Population 1 Sample
 
Sample Size
15
Sample Mean
524
Sample Standard Deviation
38.0000
Population 2 Sample
 
Sample Size
20
Sample Mean
473
Sample Standard Deviation
22.0000

Intermediate Calculations
Numerator of Degrees of Freedom
14512.2178
Denominator of Degrees of Freedom
692.7711
Total Degrees of Freedom
20.9481
Degrees of Freedom
20
Standard Error
10.9757
Difference in Sample Means
51.0000
Separate-Variance t Test Statistic
4.6466

Two-Tail Test
 Lower Critical Value
-2.0860
Upper Critical Value
2.0860
p-Value
0.0002
Reject the null hypothesis
 

Here based on the result we are rejecting the null hypothesis and thus the conclusion is based in the data there is a difference between the population mean dollar amounts owed by consumers with delinquent charge accounts in Ohio and Illinois at 0.05 significance level.

c) Assuming that the normality assumption and the condition you checked in a) hold, calculate a 95 percent confidence interval for the difference between the mean dollar amounts owed in Ohio and Illinois. Based on this interval, do you think that these mean dollar amounts differ in a practically important way?

Here the 95% confidence interval is =

As we can see the computed confidence interval has values not so high as lowest is 28.10 so based on this interval, I don’t think that these mean dollar amounts differ in a practically important way.