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Statistical Analysis of Seasonal Differences in Frequency of Suicide-Related Phone Calls

Questions:

Use the information below for questions 44. - 48.

You are interested in whether there are seasonal differences in the frequency of suicide-related phone calls to a 'hot-line' center. Your tally of calls for the past year is:

Table 3 Frequency of Phone Calls Per Season Fall Winter Spring Summer Observed Frequency 27 58 71 44 The null hypothesis is that there are no seasonal differences. Test the 2-tailed alternative hypothesis that there are seasonal differences. Answer the questions below.

44A. The statistical analysis you are using is the chi square test of : a) goodness of fit b) test of independence c) 2-way (2 factor)

44B. The df of the test = 2 3 4 Other ____________ 4

4C. Critical values of chi square at .05 and .01 are: 1.96 ; 2.58 3.84 ; 6.63 7.81 ; 11.34 0.950 ; 0.990

44D. The chi square formula is: MS Between / MS Within SS / df ? [ ]

45. The expected frequencies are as follows:

a) 50 for each season

b) 25 for each season

c) 33.3 for each season

d) 100 for each season

46A. Chi Square for Fall is approximately: 0.72 1.28 8.82 10.58

46B. Chi Square for Winter is approximately: 0.72 1.28 8.82 10.58

46C. Chi Square for Spring is approximately: 0.72 1.28 8.82 10.58

46D. Chi Square for Summer is approximately: 0.72 1.28 8.82 10.58

47. The overall (total) Chi Square for Seasonal Differences is approximately:

a) 19.00

b) 11.34

c) 21.4

d) Other

48. Conclusion: The Chi square

a) is significant at the .05 level

b) is significant at the .01 level

c) is not significant

d) is almost significant

49. Seniors in college have higher hardiness scores than freshmen in college.This statement is a __________________ for a ___________________. Null hypothesis ; t for independent samples Null hypothesis ; two-way (2-factor ANOVA) 1-tailed alternative hypothesis ; t for independent samples 2-tailed alternative hypothesis ; one-way (1 factor) ANOVA

50. Dr. T. did a study measuring test performance. Table 4 presents the results. Table 4 Mean Test Performance as a Function of Noise and Temperature What value for the missing mean would result in no interaction of Noise x Temperature? _____ 50 60 70 80 The End. (Bonus questions – optional - start on the next page.)

NOTE – I will be posting the exam grades on BB. I will also post an answer key.

## Table 3 Frequency of Phone Calls Per Season

Bonus: Use the information below to solve Questions 51-52. (You will find the instructions for completing the analysis on page 15. (We have not done this analysis in class, but the instructions will tell you what to do.)

Dr. Whynot has carried out a study in which subjects' anxiety level was measured before and after administration of a tranquilizing drug. Dr. W. hypothesized that the drug alters anxiety.

Test the hypothesis with a sign test following the instructions on the next page (Page 15). The data are shown in Table 5 below. Table 5 Anxiety Scores Subject Number Before Drug After Drug 1 7 3 2 5 1 3 4 4 4 8 2 5 6 5 6 5 3 7 2 2 8 9 4 9 8 3 10 6 4 11 7 9 12 9 10 13 5 6 14 6 5 15 4 6 16 3 2

Sign Test: Nonparametric Test Of Difference Between Two Matched Samples For each of the pairs of scores, calculate the D (difference between Score 1 – Before Drug - and Score 2 – After Drug).

Then count the number of positive Difference scores (i.e., where ‘Before Drug’ is greater than ‘After Drug’), and the number of negative Difference scores (i.e., where ‘After Drug’ is greater than ‘Before Drug.’). If D = 0, drop that pair, and reduce the n accordingly. As you can see, this procedure focuses on the sign of the differences.

To complete the sign test, you need to compute the following (see definitions below): Sign Test = (fp - fn)2 / n where fp = the number of positive D's fn = the number of negative D's n = the number of participants / subjects (reduced by the number of pairs in which D = 0) To get the critical value, assume that df = 1, and look in the chi square table: Table B.7. The sign test is significant if the obtained value is larger than the critical value.

51A - number of positive D’s : 16 10 12 Other

51B - number of negative D’s : 10 4 6 Other

51C - number of participants (reduced by number of pairs in which D = 0) : 16 14 12 Other

52A - Sign test = approximately 2.57 0.428 4.57 Other

52B. Critical Value at .05 level: 3.84 6.63 0.95 1.96

52C. Conclusion: The difference between anxiety scores before and after the Drug administration was significant was not significant was slightly significant was extremely significant Note. There are more bonus questions on the next page.

More Bonus Questions.

53. Label each of the following as a) True or b) False:

53A. A significant result must have a high effect size

53B. A high effect size tells you that a result is definitely replicable

53C. Pearson’s r and ANOVA test the same hypotheses

54. Dr. Psychologist planned to assess the relationship between two variables: grammar test scores and vocabulary test scores. Because of the simplicity of the grammar test (and the high I.Q. of the students), all of the students scored 100 on the grammar test.

The Pearson’s correlation between grammar test scores and vocabulary test scores can be expected to approximate: a. a high positive correlation b. a high negative correlation c. a zero correlation d. either a high positive or a high negative correlation THE END (again). Have a great summer.