Test for differences in two variances

Assignment Help Basic Statistics
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

Reference no: EM13842968

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