Correlation

Solution-14:

a)

Step-1: Select the appropriate test

We shall use 1 sample z-test to test the given hypothesis.

Step-2: State the research hypothesis and null hypothesis

Null Hypothesis: The mean SAT score for men is equal to 500.

Research Hypothesis: The mean SAT score for men is different from 500.

Step-3: Describe the null distribution

We assume that is true.

We shall use significance level, α = 0.05 to test the null hypothesis.

Standard Error of Estimate:

Step-4: Determination of Critical Values

As the alternative hypothesis isthe given test is a two-tailed z-test.

The critical value for a two-tailed z-test for is given as

Decision Rule: Reject if

Step-5: Calculation of test statistic and Cohen’s d

Step-6: Conclusion

This hypothesis test shows that we should retain the null hypothesis and conclude that the mean SAT score for men is equal to the population mean. The test was not statistically significant with p > .05. The size of the effect was small, Cohen’s d = 0.11.

b)

Step-1: Select the appropriate test

We shall use 1 sample z-test to test the given hypothesis.

Step-2: State the research hypothesis and null hypothesis

Null Hypothesis: The mean SAT score for women is either less than or equal to 500.

Research Hypothesis: The mean SAT score for women is greater than 500.

Step-3: Describe the null distribution

We assume that is true.

We shall use significance level, α = 0.05 to test the null hypothesis.

Standard Error of Estimate:

Step-4: Determination of Critical Values

As the alternative hypothesis isthe given test is an upper-tailed z-test.

The critical value for an upper-tailed z-test for is given as

Decision Rule: Reject if

Step-5: Calculation of test statistic and Cohen’s d

Step-6: Conclusion

This hypothesis test shows that we should reject the null hypothesis and conclude that the mean SAT score for women is significantly greater than the population mean. The test was statistically significant with p < .05. The size of the effect was medium, Cohen’s d = 0.41.

c)

Step-1: Select the appropriate test

We shall use 2 sample z-test to test the given hypothesis.

Step-2: State the research hypothesis and null hypothesis

Null Hypothesis: The mean SAT score for men is equal to the mean SAT score for women.

Research Hypothesis: The mean SAT score for men is different from the mean SAT score for women.

Step-3: Describe the null distribution

We assume that is true.

We shall use significance level, α = 0.05 to test the null hypothesis.

Standard Error of Estimate: The standard error of estimate is SE = 20.

Step-4: Determination of Critical Values

As the alternative hypothesis isthe given test is a two-tailed z-test.

The critical value for a two-tailed z-test for is given as

Decision Rule: Reject if

Step-5: Calculation of test statistic and Cohen’s d

Step-6: Conclusion

This hypothesis test shows that we should retain the null hypothesis and conclude that the mean SAT score for men is not different from the mean SAT score for women. The test was not statistically significant with p > .05.

Solution-16:

a)

Step-1: Select the appropriate test

We shall use 1 sample t-test to test the given hypothesis.

Step-2: State the research hypothesis and null hypothesis

Null Hypothesis: The mean number of cigarettes smoked per week by college bound students is either less than or equal to 3.5.

Research Hypothesis: The mean number of cigarettes smoked per week by college bound students is higher than to 3.5.

Step-3: Describe the null distribution

We assume that is true.

We shall use significance level, α = 0.05 to test the null hypothesis.

Standard Error of Estimate:

Step-4: Determination of Critical Values

As the alternative hypothesis isthe given test is an upper-tailed t-test.

Degree of freedom,

From the t-distribution table, the critical value for an upper-tailed t-test for and is given as

Decision Rule: Reject if

Step-5: Calculation of test statistic and Cohen’s d

Step-6: Conclusion

This hypothesis test shows that we should reject the null hypothesis and conclude that the mean number of cigarettes smoked per week by college bound students is higher than to 3.5. The test was statistically significant with p < .05. The size of the effect was medium, Cohen’s d = 0.25.

b)

Step-1: Select the appropriate test

We shall use 1 sample t-test to test the given hypothesis.

Step-2: State the research hypothesis and null hypothesis

Null Hypothesis: The mean number of cigarettes smoked per week by non college bound students is either less than or equal to 3.5.

Research Hypothesis: The mean number of cigarettes smoked per week by non college bound students is higher than to 3.5.

Step-3: Describe the null distribution

We assume that is true.

We shall use significance level, α = 0.05 to test the null hypothesis.

Standard Error of Estimate:

Step-4: Determination of Critical Values

As the alternative hypothesis isthe given test is an upper-tailed t-test.

Degree of freedom,

From the t-distribution table, the critical value for an upper-tailed t-test for and is given as

Decision Rule: Reject if

Step-5: Calculation of test statistic and Cohen’s d

Step-6: Conclusion

This hypothesis test shows that we should reject the null hypothesis and conclude that the mean number of cigarettes smoked per week by non college bound students is higher than the population mean of 3.5. The test was statistically significant with p < .05. The size of the effect was large, Cohen’s d = 1.83.

Solution-17:

Step-1: Select the appropriate test

We shall use 2 sample t-test to test the given hypothesis.

Step-2: State the research hypothesis and null hypothesis

Null Hypothesis: The mean time taken to complete a four year college degree for men is same as the mean time taken to complete a four year college degree for women.

Research Hypothesis: The mean time taken to complete a four year college degree for men is different from the mean time taken to complete a four year college degree for women.

Step-3: Describe the null distribution

We assume that is true.

We shall use significance level, α = 0.05 to test the null hypothesis.

Descriptive Statistics:	Men	Women
Sample Mean	4.500	5.000
Sample Size	6	6
Sample Standard Deviation	0.837	1.414

Standard Error of Estimate:

Step-4: Determination of Critical Values

As the alternative hypothesis isthe given test is a two-tailed t-test.

Degree of freedom,

From the t-distribution table, the critical value for a two-tailed t-test for and is given as

Decision Rule: Reject if

Step-5: Calculation of test statistic and Cohen’s d

Step-6: Conclusion

This hypothesis test shows that we should retain the null hypothesis and conclude that men and women do not differ in mean time taken to complete a four year college degree. The test was not statistically significant with p > .05. The size of the effect was medium, Cohen’s d = -0.43.

Solution-20:

Step-1: Select the appropriate test

We shall use 2 sample t-test to test the given hypothesis.

Step-2: State the research hypothesis and null hypothesis

Null Hypothesis: The mean prison sentence length for white defendants is same as the mean prison sentence length for black defendants.

Research Hypothesis: The mean prison sentence length for white defendants is different from the mean prison sentence length for black defendants.

Step-3: Describe the null distribution

We assume that is true.

We shall use significance level, α = 0.05 to test the null hypothesis.

Descriptive Statistics:	Whites	Blacks
Sample Mean	4.300	5.000
Sample Size	10	8
Sample Standard Deviation	1.494	1.690

Standard Error of Estimate:

Step-4: Determination of Critical Values

As the alternative hypothesis isthe given test is a two-tailed t-test.

Degree of freedom,

From the t-distribution table, the critical value for a two-tailed t-test for and is given as

Decision Rule: Reject if

Step-5: Calculation of test statistic and Cohen’s d

Step-6: Conclusion

This hypothesis test shows that we should retain the null hypothesis and conclude that white and black defendants do not differ in mean prison sentence length. The test was not statistically significant with p > .05. The size of the effect was medium, Cohen’s d = -0.44.

Solution-22:

Step-1: Select the appropriate test

We shall use 2 sample t-test to test the given hypothesis.

Step-2: State the research hypothesis and null hypothesis

Null Hypothesis: The mean number of smiles for males is same as the mean number of smiles for females.

Research Hypothesis: The mean number of smiles for males is different from the mean number of smiles for females .

Step-3: Describe the null distribution

We assume that is true.

We shall use significance level, α = 0.05 to test the null hypothesis.

Descriptive Statistics:	Males	Females
Sample Mean	7.600	15.200
Sample Size	5	5
Sample Standard Deviation	4.615	3.347

Standard Error of Estimate:

Step-4: Determination of Critical Values

As the alternative hypothesis isthe given test is a two-tailed t-test.

Degree of freedom,

From the t-distribution table, the critical value for a two-tailed t-test for and is given as

Decision Rule: Reject if

Step-5: Calculation of test statistic and Cohen’s d

Step-6: Conclusion

This hypothesis test shows that we should reject the null hypothesis and instead conclude that females smile at others more than males do. This conclusion is statistically significant with p < .05. The size of the effect was large, Cohen’s d = -1.88.

Solution-34:

Step-1: Select the appropriate test

We shall use Dependent sample t-test to test the given hypothesis.

Step-2: State the research hypothesis and null hypothesis

Null Hypothesis: There is no difference in the mean number of 911 calls to the police before and after the neighborhood watch program.

Research Hypothesis: There is significant difference in the mean number of 911 calls to the police before and after the neighborhood watch program.

Step-3: Describe the null distribution

We assume that is true.

We shall use significance level, α = 0.05 to test the null hypothesis.

Block	Before Neighborhood Watch	After Neighborhood Watch	Difference, d
A	11	6	5
B	15	11	4
C	17	13	4
D	10	8	2
E	16	12	4
F	21	17	4
Mean, M =			3.833
Standard Deviation, s			0.9832

Step-4: Determination of Critical Values

As the alternative hypothesis isthe given test is a two-tailed t-test.

Degree of freedom,

From the t-distribution table, the critical value for a two-tailed t-test for and is given as

Decision Rule: Reject if

Step-5: Calculation of test statistic and Cohen’s d

Step-6: Conclusion

This hypothesis test shows that we should reject the null hypothesis and instead conclude that there is significant difference in the number of 911 calls to the police before and after the neighborhood watch program. This conclusion is statistically significant with p < .05. The size of the effect was large, Cohen’s d = 3.90.

Solution-35:

Step-1: Select the appropriate test

We shall use Dependent sample t-test to test the given hypothesis.

Step-2: State the research hypothesis and null hypothesis

Null Hypothesis: The film did not result in a reduction of racist attitudes.

Research Hypothesis: The film resulted in a reduction of racist attitudes.

Step-3: Describe the null distribution

We assume that is true.

We shall use significance level, α = 0.05 to test the null hypothesis.

	Before	After	Difference, d
A	36	24	12
B	25	20	5
C	26	26	0
D	30	27	3
A	31	18	13
B	27	19	8
C	29	27	2
D	31	30	1
Mean, M			5.500
Standard Deviation, s			4.9857

Step-4: Determination of Critical Values

As the alternative hypothesis isthe given test is an upper-tailed t-test.

Degree of freedom,

From the t-distribution table, the critical value for an upper-tailed t-test for and is given as

Decision Rule: Reject if

Step-5: Calculation of test statistic and Cohen’s d

Step-6: Conclusion

This hypothesis test shows that we should reject the null hypothesis and instead conclude that the film resulted in a significant reduction of racist attitudes. This conclusion is statistically significant with p < .05. The size of the effect was large, Cohen’s d = 1.10.