Question

Randomly selected deaths of motorcycle riders in a region of the northern hemisphere are summarized in the accompanying table. Use a 0.05 significance level to test the claim that such fatalities occur with equal frequency in the different months. How might the results be explained? Month Jan. Feb. March April May June Number 9 13 8 24 28 Month July Aug. Sept. Oct. Nov. Dec. Number 2829 24 15 10 Determine the null and altenative hypotheses. Ho: Fatalities occur with the same frequency in the different months of the year. H: At least one month has a different frequency of fatalities than the other months. Cacuiale the testsai (Round to three decimal places as needed.)

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Answer #1

Let's make table in excel:

2 0.083333 3 0.083333 4 0.083333 5 0.083333 6 0.083333 7 0.083333 8 0.083333 9 0.083333 10 0.083333 11 0.083333 12 0.083333 13 0.083333 14 15 degrees of freedom n -1 12- 1 11 16 p-value0.000141 17 Decision rule: 18 1) If p-value < level of significance (alpha) then we reject null hypothesis 19 2) lf p-value > level of significance (alpha) then we fail to reject null hypothesis 20 Here p value 0.0001410.05 so we used first rule 21 That is we reject the null hypothesis 22 Conclusion: At 5% level of significance there are sufficient evidence to say that at least one month has a different frequency of fatalities than the other months. 23 9 18.33333 87.11111 4.75151!5 11 18.33333 53.77778 2.933333 13 18.33333 28,44444 1.551515 24 18.33333 32.11111 1.751515 28 18.33333 93.44444 5.09697 28 18.33333 93.444445.09697 29 18.33333 113.7778 6.206061 24 18.33333 32.11111 1.751515 15 18.33333 11.11111 0.606061 11 18.33333 53.77778 2.933333 10 18.33333 69.44444 3.787879 220 36.473 This is the chi -square test statistic value

The formulae used on the above excel-sheet are as follows:

((O - E )A2)/E D2/C2 D3/C3 (o -E)A2 (B2-C2)A2 (B3-C3)A2 (B4-C4) 2 (B5-C5)A2 (B6-C6)A2 (B7-C7)A2 (B8-C8)A2 F(B9-C9)A2 (B10-C10)42 (B11-C11)A2 (B12-C12)A2 (B13-C13)A2 2 1/12 3 1/12 4 1/12 3A2 * 220 A3*220 -A4*220 3A5*220 A6*220 FA7 220 FA8*220 A9*220 A10* 220 A11220 A12 220 A13 220 9 13 18 24 28 28 29 24 15 D5/C5 D6/C6 D7/C7 D8/C8 D9/C9 -D10/C10 D11/C11 -D12/C12 D13/C13 SUM(E2:E13)This is the chi -squar 6 1/12 7 1/12 8 1/12 12-1/12 10 14 15 degrees of freedom 16 -SUM(B2:B13) p-value- 17 Decision rule 18 1) If p-value < level o 19 2) If p-value > level o 20 Here p value 0.000 21 That is we reject the 22 Conclusion: At 5% le 23 CHIDIST(36.473,11)

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