Author: Christine Farr


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 = 11.84 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 decision 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)/72 = 5/49 = .10 D) (40  35)/1.632 = 5/2.66 = 1.88 5. What effect size did Professor Juarez find? A) (40  35)/7 = .71 B) (40  35)/1.63 = 3.07 C) (40  35)/72 = 5/49 = .10 D) (40  35)/1.632 = 5/2.66 = 1.88 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 

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