Reference no: EM13842968

Question 1:

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) Set up the null and alternative hypotheses to test the claim that both samples are from the populations having the same standard deviation.

Null Hypothesis (Ho): σ1 = σ2

Alternative Hypothesis (Ha): σ1 ≠ σ2

Where σ1 = population standard deviation of mothers who did not use cocaine during their pregnancies and σ2 = population standard deviation of mothers who used cocaine during their pregnancies

b) Perform the hypothesis test you set up in a) with the significance level a = 0.05.

F Test for Differences in Two Variances
Data
Level of Significance	0.05
Larger-Variance Sample
Sample Size	186
Sample Variance	484416
Smaller-Variance Sample
Sample Size	190
Sample Variance	416025
Intermediate Calculations
F Test Statistic	1.1644
Population 1 Sample Degrees of Freedom	185
Population 2 Sample Degrees of Freedom	189
Two-Tail Test
Upper Critical Value	1.3329
p-Value	0.2986
Do not reject the null hypothesis

c) Using the results of b), does cocaine use appear to affect the birth weight of a baby? Substantiate you conclusion.

Since the p-value of the test is bigger than 0.05 level of significance so we will not be able to reject the null hypothesis and conclude that cocaine use does not appear to affect the birth weight of a baby.

Question 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.

Null Hypothesis (Ho): µ1 = µ2

Alternative Hypothesis (Ha): µ 1 < µ2

Where µ1 = population mean FICO score of borrowers with high-interest mortgage and µ2 = population mean FICO score of borrowers with low-interest mortgage

b) Does the FICO credit rating score appear to affect mortgage payments? If so, how?

Pooled-Variance t Test for the Difference Between Two Means
(assumes equal population variances)
Data
Hypothesized Difference	0
Level of Significance	0.01
Population 1 Sample
Sample Size	40
Sample Mean	594.8
Sample Standard Deviation	12.2
Population 2 Sample
Sample Size	40
Sample Mean	785.2
Sample Standard Deviation	16.3
Intermediate Calculations
Population 1 Sample Degrees of Freedom	39
Population 2 Sample Degrees of Freedom	39
Total Degrees of Freedom	78
Pooled Variance	207.2650
Standard Error	3.2192
Difference in Sample Means	-190.4000
t Test Statistic	-59.1451
Lower-Tail Test
Lower Critical Value	-2.3751
p-Value	0.0000
Reject the null hypothesis

Question 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.

Null Hypothesis (Ho): σ1 = σ2

Alternative Hypothesis (Ha): σ1 ≠ σ2

Where σ1 = population standard deviation of mothers who did not use cocaine during their pregnancies and σ2 = population standard deviation of mothers who used cocaine during their pregnancies

a) Using your finding in (a), construct a 90% 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.

F Test for Differences in Two Variances
Data
Level of Significance	0.05
Larger-Variance Sample
Sample Size	186
Sample Variance	484416
Smaller-Variance Sample
Sample Size	190
Sample Variance	416025
Intermediate Calculations
F Test Statistic	1.1644
Population 1 Sample Degrees of Freedom	185
Population 2 Sample Degrees of Freedom	189
Two-Tail Test
Upper Critical Value	1.3329
p-Value	0.2986
Do not reject the null hypothesis

b) Using your finding in (b), does cocaine use appear to affect the birth weight of a baby? Substantiate you conclusion.

Confidence Interval Estimate
for the Difference Between Two Means
Data
Confidence Level	90%
Intermediate Calculations
Degrees of Freedom	374
t Value	1.6489
Interval Half Width	114.0778
Confidence Interval
Interval Lower Limit	288.9222
Interval Upper Limit	517.0778

c) Using your finding in (b), does cocaine use appear to affect the birth weight of a baby? Substantiate you conclusion.

We can see that 0 does not lie inside the 90% confidence interval so we can conclude that cocaine use appear to affect the birth weight of a baby. We can see that the mean weight of the baby is high whose mother does not use cocaine as compare to mothers who used cocaine during their pregnancies.

Question 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 a =0.05.

Null Hypothesis (Ho): σ1 = σ2

Alternative Hypothesis (Ha): σ1 ≠ σ2

Where σ1 = population standard deviation of Ohio and σ2 = population standard deviation of Illinois

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

b) Test with a =0.05 whether there is a difference between the population mean dollar amounts owed by consumers with delinquent charge accounts in Ohio and Illinois.

Null Hypothesis (Ho): µ1 = µ2

Alternative Hypothesis (Ha): µ 1 ≠ µ2

Where µ1 = population mean dollar amounts owed by consumers with delinquent charge accounts in Ohio and µ2 = population mean dollar amounts owed by consumers with delinquent charge accounts in Illinois

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

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?

Confidence Interval Estimate
for the Difference Between Two Means
Data
Confidence Level	95%
Intermediate Calculations
Degrees of Freedom	33
t Value	2.0345
Interval Half Width	20.7463
Confidence Interval
Interval Lower Limit	30.2537
Interval Upper Limit	71.7463