For the following question(s): A school counselor tests the level of depression in fourth graders in a particular class of 20 students. The counselor wants to know whether the kind of students in this class differs from that of fourth graders in general at her school. On the test, a score of 10 indicates severe depression, while a score of 0 indicates no depression. From reports, she is able to find out about past testing. Fourth graders at her school usually score 5 on the scale, but the variation is not known. Her sample of 20 fifth graders has a mean depression score of 4.4. Use the .01 level of significance.

1. The counselor calculates the unbiased estimate of the population’s variance to be 15. What is the variance of the distribution of means?

A) 15/20 = 0.75

B) 15/19 = 0.79

C) 152/20 = 11.25

D) 152/19 = 11.

2. Suppose the counselor tested the null hypothesis that fourth graders in this class were less depressed than those at the school generally. She figures her t score to be -.20. What decision should she make regarding the null hypothesis?

A) Reject it

B) Fail to reject it

C) Postpone any decisions until a more conclusive study could be conducted

D) There is not enough information given to make a

3. Suppose the standard deviation she figures (the square root of the unbiased estimate of the population variance) is .85. What is the effect size?

A) 5/.85 = 5.88

B) .85/5 = .17

C) (5 – 4.4)/.85 = .71

D) .85/(5 – 4.4) = 1.42

For the following question(s): Professor Juarez thinks the students in her statistics class this term are more creative than most students at this university. A previous study found that students at this university had a mean score of 35 on a standard creativity test. Professor Juarez finds that her class scores an average of 40 on this scale, with an estimated population standard deviation of 7. The standard deviation of the distribution of means comes out to 1.63.

4. What is the t score?

A) (40 – 35)/7 = .71

B) (40 – 35)/1.63 = 3.07

C) (40 – 35)/7^{2}= 5/49 = .10

D) (40 – 35)/1.63^{2}= 5/2.66 = 1.

5. What effect size did Professor Juarez find?

A) (40 – 35)/7 = .71

B) (40 – 35)/1.63 = 3.07

C) (40 – 35)/7^{2}= 5/49 = .10

D) (40 – 35)/1.63^{2}= 5/2.66 = 1.

6. If Professor Juarez had 30 students in her class, and she wanted to test her hypothesis using the 5% level of significance, what cutoff t score would she use? (You should be able to figure this out without a table because only one answer is in the correct region.)

A) 304.11

B) 1.699

C) -.113

D) -2.500

For the following question(s): A school counselor claims that he has developed a technique to reduce prestudying 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.

7. Suppose the counselor wants to examine whether there is a change of any kind (either an increase or decrease) in procrastination after attending his workshop. What would be the appropriate description of “Population 2” (the population to which the population his sample represents is being compared)?

A) People whose posttest scores will be lower than their pretest scores

B) People whose change scores will be greater than 0

C) People whose change scores will be 0

D) People whose change scores will be less than their pretest scores

8. Presume the counselor wants to examine whether there is a change (either an increase or decrease) in procrastination after attending his workshop. If the counselor tests 10 students using the .05 level of significance, what cutoff t score(s) will he use? (You should be able to figure this out without a table.)

A) -2.62, 0, +2.62

B) +2.262

C) -2.262, 0

D) -2.262, +2.262

9. 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 be the estimated population variance?

A) 135/10 = 13.5

B) 135/9 = 15.0

C) 10/135 = .074

D) 9/135 = .067

10. A researcher conducts a study of perceptual illusions under two different lighting conditions. Twenty participants were each tested under both of the two different conditions. The experimenter reported: “The mean number of effective illusions was 6.72 under the bright conditions and 6.85 under the dimly lit conditions, a difference that was not significant, *t*(19) = 1.62.”

Explain this result to a person who has never had a course in statistics. Be sure to use sketches of the distributions in your answer.

The statement “The mean number of effective illusions was 6.72 under the bright conditions and 6.85 under the dimly lit conditions, a difference that was not significant, t(19) = 1.62.” implies that under the bright conditions we can expect on an average effective illusions would be 6.72 and under dimly light condition it is 6.85. As we also have t(19) = 1.62 which is less than the critical value so the conclusion is the perceptual illusions under two different lighting conditions does not differ significantly.

The graph above shows the test statistic and critical values. The red area combined together gives the p-value which is 0.1226.

The test statistic does not lie in the critical region (right of 2,093 and left of -2.093) and thus we have the above conclusion.