Using the table above, determine the contents of the following sets and express them in list-notation
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question.docxMath 104 Fall '14Lab Assignment #4
Math 104 Fall '14Lab Assignment #4
[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.
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Z Test for Differences in Two Proportions |
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Data |
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Hypothesized Difference |
0 |
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Level of Significance |
0.05 |
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Group 1 |
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Number of Items of Interest |
7 |
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Sample Size |
94 |
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Group 2 |
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Number of Items of Interest |
27 |
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Sample Size |
270 |
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Intermediate Calculations |
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Group 1 Proportion |
0.074468085 |
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Group 2 Proportion |
0.1 |
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Difference in Two Proportions |
-0.02553191 |
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Average Proportion |
0.0934 |
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Z Test Statistic |
-0.7326 |
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Two-Tail Test |
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Lower Critical Value |
-1.9600 |
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Upper Critical Value |
1.9600 |
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p-Value |
0.4638 |
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Do not reject the null hypothesis |
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[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?
Math 104 Fall '14Lab Assignment #4
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,
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Z Test for Differences in Two Means |
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Data |
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Hypothesized Difference |
0 |
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Level of Significance |
0.01 |
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Population 1 Sample |
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Sample Size |
40 |
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Sample Mean |
594.8 |
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Population Standard Deviation |
12.2 |
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Population 2 Sample |
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Sample Size |
40 |
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Sample Mean |
785.2 |
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Population Standard Deviation |
16.3 |
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Intermediate Calculations |
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Difference in Sample Means |
-190.4 |
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Standard Error of the Difference in Means |
3.2192 |
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Z Test Statistic |
-59.1451 |
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Lower-Tail Test |
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Lower Critical Value |
-2.3263 |
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p-Value |
0.0000 |
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Reject the null hypothesis |
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.
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F Test for Differences in Two Variances |
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Data |
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Level of Significance |
0.05 |
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Larger-Variance Sample |
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Sample Size |
186 |
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Sample Variance |
484416 |
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Smaller-Variance Sample |
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Sample Size |
190 |
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Sample Variance |
416025 |
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Intermediate Calculations |
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F Test Statistic |
1.1644 |
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Population 1 Sample Degrees of Freedom |
185 |
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Population 2 Sample Degrees of Freedom |
189 |
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Two-Tail Test |
|
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Upper Critical Value |
1.3329 |
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p-Value |
0.2986 |
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Do not reject the null hypothesis |
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Here the p-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.
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Pooled-Variance t Test for the Difference Between Two Means |
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(assumes equal population variances) |
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Data |
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Hypothesized Difference |
0 |
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Level of Significance |
0.05 |
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Population 1 Sample |
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Sample Size |
186 |
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Sample Mean |
3103 |
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Sample Standard Deviation |
696 |
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Population 2 Sample |
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Sample Size |
190 |
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Sample Mean |
2700 |
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Sample Standard Deviation |
645 |
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Intermediate Calculations |
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Population 1 Sample Degrees of Freedom |
185 |
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Population 2 Sample Degrees of Freedom |
189 |
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Total Degrees of Freedom |
374 |
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Pooled Variance |
449854.7727 |
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Standard Error |
69.1826 |
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Difference in Sample Means |
403.0000 |
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t Test Statistic |
5.8252 |
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Two-Tail Test |
|
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Lower Critical Value |
-1.9663 |
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Upper Critical Value |
1.9663 |
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p-Value |
0.0000 |
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Reject the null hypothesis |
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Confidence Interval Estimate |
|
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for the Difference Between Two Means |
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Data |
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Confidence Level |
95% |
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Intermediate Calculations |
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Degrees of Freedom |
374 |
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t Value |
1.9663 |
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Interval Half Width |
136.0356 |
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Confidence Interval |
|
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Interval Lower Limit |
266.9644 |
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Interval Upper Limit |
539.0356 |
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.
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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.
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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.
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Here the population variances are not equal so we should run a separate variance t test. The results are as follows,
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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 =
