8:52 . LTE 7. A researcher in exercise physiology wanted to know whether body composition differs among prepubescent males and females. To test the null hypothesis, she measured skinfolds (in millimeters) on 5 males and 6 females and obtained the following results. Dominant Nondominant X, = 40 pounds X , =35 pounds SD, = 12.7 r=.83 SD, = 14.2 a. Is the difference in the means significant? If so, at what level of confidence? The means are not significant. Males - 2 b. Do you accept or reject the null hypothesis? c. Check your hand calculations on a computer. See the hint in problem 4. 8. A sport psychologist wondered whether motivation could Improve aerobic capacity. To answer the question, he measured 10 subjects on O2max (ml/kg/min) on a treadmill. Students were verbally instructed to do their very best. One week later, the subjects were measured again and were told they would receive $100 if their O2max was higher than it was the first time. Following are the data. First Second 45 54 33 50 59 58 32 38 30 42 27 35 29 38 59 66 44 48 40 49 a. Did the money cause them to significantly increase their Ozmax values? b. If so, what is the probability of error in your conclusion? Should you accept or reject the null hypothesis? c. Check your hand calculations on a computer. Hint: In the computer database, enter values for each subject on the same row. Use column 1 for the first test and column 2 for the second test. 9. An aerobic dance teacher wanted to know whether two workouts per week of 30 minutes each were enough to increase Ozmax in sedentary middle-aged women. The teacher proposed to compare Ozmax of some women who had been in the aerobic dance classes for 6 months with Ozmax of a group of sedentary women. Based on related literature, the teacher estimated that O2max would be about 7 milliliters per kilogram per minute higher in the aerobic dancers, with a standard deviation of 4.5. If this is a fair estimate of the data to be expected, how many women must she test in each group to produce a power coefficient of .90 at a = .05