If 15 subjects take a pre-test and a post-test with a mean change score of 1.5, and if the standard deviation of the comparison distribution is calculated to be .5, what is the t score?

A. 1.5/(15-1)=0.11

B. 1.5/15=0.10

C. 1.5/.5=3.00

D. 1.5/(.5(15-1)=0.21

A school counselor claims that he has developed a technique to reduce pre-studying procrastination in students. He has students time their procrastination for a week and uses this as a pretest (before) indicator of procrastination. Students then attend a workshop in which they are instructed to do a specific warming up exercise for studying by focusing on a pleasant activity. For the next week, students again time their procrastination. The counselor then uses the time from this week as the posttest (after) measure.Suppose the counselor found the sum of squared deviations from the mean of the sample to be 135. Given that he tested 10 people, what would the estimated population variance be?

A. 135/10=13.5

B. 135/9=15.0

C. 10/135=0.70

D. 9/135=0.70

A researcher conducts a t test for dependent means with 10 subjects. If the estimated population variance of the change scores is 20, what would the variance of the distribution of the means of change scores be?

A. 20/10=2.0

B. 20/9=2.2

C. 400/10=40.0

D. 400/9=44.4

Suppose a researcher conducts a t test for dependent means in which it is predicted that there will be a decrease in unemployment from before to after a particular job-skills training program. The cutoff t needed is -1.833. The standard deviation of the distribution of means of change scores is 2.0 and the mean change score for the sample studied is an increase of 5.2. What is the t score?

A. 5.2/-1.833=-2.84

B. 2/5.2=.38

C. 5.2/2=2.60

D. 22

In the scenario from the question above, what is the appropriate conclusion?

A. Reject the null hypothesis.

B. Do not reject null hypothesis.

C. It cannot be determined without also knowing the population mean.

D. It cannot be determined without also knowing the population standard deviation or its estimate.

A. 1.5/(15-1)=0.11

B. 1.5/15=0.10

C. 1.5/.5=3.00

D. 1.5/(.5(15-1)=0.21

A school counselor claims that he has developed a technique to reduce pre-studying procrastination in students. He has students time their procrastination for a week and uses this as a pretest (before) indicator of procrastination. Students then attend a workshop in which they are instructed to do a specific warming up exercise for studying by focusing on a pleasant activity. For the next week, students again time their procrastination. The counselor then uses the time from this week as the posttest (after) measure.Suppose the counselor found the sum of squared deviations from the mean of the sample to be 135. Given that he tested 10 people, what would the estimated population variance be?

A. 135/10=13.5

B. 135/9=15.0

C. 10/135=0.70

D. 9/135=0.70

A researcher conducts a t test for dependent means with 10 subjects. If the estimated population variance of the change scores is 20, what would the variance of the distribution of the means of change scores be?

A. 20/10=2.0

B. 20/9=2.2

C. 400/10=40.0

D. 400/9=44.4

Suppose a researcher conducts a t test for dependent means in which it is predicted that there will be a decrease in unemployment from before to after a particular job-skills training program. The cutoff t needed is -1.833. The standard deviation of the distribution of means of change scores is 2.0 and the mean change score for the sample studied is an increase of 5.2. What is the t score?

A. 5.2/-1.833=-2.84

B. 2/5.2=.38

C. 5.2/2=2.60

D. 22

In the scenario from the question above, what is the appropriate conclusion?

A. Reject the null hypothesis.

B. Do not reject null hypothesis.

C. It cannot be determined without also knowing the population mean.

D. It cannot be determined without also knowing the population standard deviation or its estimate.