# Test of hypothesis using ANOVA, Chi Square

11.2 One particular morning, the length of time spent in the examination rooms is recorded for each patient seen by each physician at an orthopedic clinic. Does the data prove a significant difference in mean times?

Time in Examination Rooms (Minutes)FIGURE 11.12

Physician 1 Physician 2 Physician 3 Physician 4

34 33 17 28

25 35 30 33

27 31 30 31

31 31 26 27

26 42 32 32

34 33 28 33

21 26 40

29

Instructions: For each data set: (a) State the hypotheses. (Please show 5 steps hypothesis and use the critical value for your hypothesis testing

15.6 Advertisers need to know which age groups are likely to see their ads. Purchasers of 120 copies of Cosmopolitan are shown by age group.

Perform the chi-square test for a uniform distribution. At alpha = .01, does this sample contradict the assumption that readership is uniformly distributed among these six age groups?

Please show 5 steps hypothesis and use the critical value for your hypothesis testing.

Purchaser Age Units Sold

18-24 38

25-34 28

35-44 19

45-54 16

55-64 10

65+ 9

Total 120

Perform the chi-square test for a uniform distribution. At alpha = .01, does this sample contradict the assumption that readership is uniformly distributed among these six age groups?

10.56 .A sample of 25 concession stand purchases at the October 22 matinee of Bride of Chucky showed a mean purchase of $5.29 with a standard deviation of $3.02. For the October 26 evening showing of the same movie, for a sample of 25 purchases the mean was $5.12 with a standard deviation of $2.14. The means appear to be very close, but not the variances. At alpha = .05, is there a difference in variances? Show all steps clearly, including an illustration of the decision rule. Please show 5 steps hypothesis and use the critical value for your hypothesis testing

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#### Solution Summary

Questions on testing hypothesis using ANOVA, Chi- square etc have been answered. Hypothesis regarding differences between means and variances have been tested. Also, a chi-square test for a uniform distribution has been performed.