WEEK 5 EXERCISESEnter your answers in the spaces provided. Save the file using your last name as thebeginning of the file name (e.g., ruf_week5_exercises) and submit via Assignments.When appropriate, show your work. You can do the work by hand, scan/take a digitalpicture, and attach that file with your work.For the following question(s): A school counselor tests the level of depression in fourth graders ina particular class of 20 students. The counselor wants to know whether the kind of students inthis class differs from that of fourth graders in general at her school. On the test, a score of 10indicates severe depression, while a score of 0 indicates no depression. From reports, she is ableto find out about past testing. Fourth graders at her school usually score 5 on the scale, but thevariation 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 populations variance to be 15. What isthe variance of the distribution of means?A) 15/20 = 0.75B) 15/19 = 0.79C) 152/20 = 11.25D) 152/19 = 11.842.Suppose the counselor tested the null hypothesis that fourth graders in this class were lessdepressed than those at the school generally. She figures her t score to be .20. What decisionshould she make regarding the null hypothesis?A) Reject itB) Fail to reject itC) Postpone any decisions until a more conclusive study could be conductedD) There is not enough information given to make a decision3.Suppose the standard deviation she figures (the square root of the unbiased estimate of thepopulation variance) is .85. What is the effect size?A) 5/.85 = 5.88B) .85/5 = .17C) (5 4.4)/.85 = .71D) .85/(5 4.4) = 1.42For the following question(s): Professor Juarez thinks the students in her statistics class this termare more creative than most students at this university. A previous study found that students atthis university had a mean score of 35 on a standard creativity test. Professor Juarez finds thather class scores an average of 40 on this scale, with an estimated population standard deviationof 7. The standard deviation of the distribution of means comes out to 1.63.4.What is the t score?A) (40 35)/7 = .71B) (40 35)/1.63 = 3.07C) (40 35)/72 = 5/49 = .10D) (40 35)/1.632 = 5/2.66 = 1.885.What effect size did Professor Juarez find?A) (40 35)/7 = .71B) (40 35)/1.63 = 3.07C) (40 35)/72 = 5/49 = .10D) (40 35)/1.632 = 5/2.66 = 1.886.If Professor Juarez had 30 students in her class, and she wanted to test her hypothesis usingthe 5% level of significance, what cutoff t score would she use? (You should be able to figurethis out without a table because only one answer is in the correct region.)A) 304.11B) 1.699C) .113D) 2.500For the following question(s): A school counselor claims that he has developed a technique toreduce prestudying procrastination in students. He has students time their procrastination for aweek and uses this as a pretest (before) indicator of procrastination. Students then attend aworkshop in which they are instructed to do a specific warming-up exercise for studying byfocusing on a pleasant activity. For the next week, students again time their procrastination. Thecounselor 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 anincrease or decrease) in procrastination after attending his workshop. What would be theappropriate description of Population 2 (the population to which the population his samplerepresents is being compared)?A) People whose posttest scores will be lower than their pretest scoresB) People whose change scores will be greater than 0C) People whose change scores will be 0D) People whose change scores will be less than their pretest scores8.9.10.Presume the counselor wants to examine whether there is a change (either an increase ordecrease) in procrastination after attending his workshop. If the counselor tests 10 studentsusing the .05 level of significance, what cutoff t score(s) will he use? (You should be able tofigure this out without a table.)A) 2.62, 0, +2.62B) +2.262C) 2.262, 0D) 2.262, +2.262Suppose the counselor found the sum of squared deviations from the mean of the sample to be135. Given that he tested 10 people, what would be the estimated population variance?A) 135/10 = 13.5B) 135/9 = 15.0C) 10/135 = .074D) 9/135 = .067A researcher conducts a study of perceptual illusions under two different lighting conditions.Twenty participants were each tested under both of the two different conditions. Theexperimenter reported: The mean number of effective illusions was 6.72 under the brightconditions 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 sketchesof the distributions in your answer.

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