Question

accidents were uniformly distributed over the days of the week. The data n 317 randomly selected accidents. Is there reason to believe that the accident occurs with un 1. A researcher wanted to determine whether certain for equal frequenay with respect to the day of the week at the a 0.05 level of significance? Click the icon to view the table Let p, the proportion of accidents on day i. where i 1 for Sunday, 2 for Monday, etc. What are the null and alternative hypotheses? O A. H,: More accidents occur later in the week than earlier Ho: At least one proportion is different from, the others B. с. H,: At least one proportion is different from the others O D. H,: More accidents occur earlier in the week than later Compute the expected counts for day of the week Day of the Week Observed Count Expected Count Sunday Monday Tuesday Wednesday Thursday Friday Saturday 45 43 30 42 52 61 (Round to two decimal places as needed.) What is the test statistic? (Round to three decimal places as needed.) What is the P-value of the test? P-value Based on the results, do the accidents follow a uniform distribution? Round to three decimal places as needed.) O A. Do not reject Ho. because the calculated P-value is greater than the given a level of significance. O B. Reject Ho. because the calculated P.value is less than the given a level of significance. O c. Reject Ho, because the calculated P-value is greater than the given α level of significance 0 D. Do not reject Ho, because the calculated P-value is less than the given α level of significance.

Answer 1

Null and Alternative hypothesis:

Answer C.

$H_0:p_1=p_2=p_3=p_4=p_5=p_6=p_7 =\frac{1}{7}$

$H_1:\textup{At least one proportion is different from the others.}$

Expected count:

Day of the Week	Observed Count	Expected Count
Sunday	45	= 317*1/7 = 45.29
Monday	43	= 317*1/7 = 45.29
Tuesday	30	= 317*1/7 = 45.29
Wednesday	42	= 317*1/7 = 45.29
Thursday	44	= 317*1/7 = 45.29
Friday	52	= 317*1/7 = 45.29
Saturday	61	= 317*1/7 = 45.29

Test statistic:

$\chi^2_0 = \sum_{i=1}^n \frac{(O_i-E_i)^2}{E_i}$

$= \frac{(45-45.29)^2}{45.29} + \frac{(43-45.29)^2}{45.29} +....+ \frac{(61-45.29)^2}{45.29}$

$= 0.002 + 0.116+ 5.162 +0.239 + 0.037 + 0.994+ 5.449 = \textbf{11.999}$

p-value = 0.062

Conclusion:

A. Do not reject H_o. Because the calculated P-value is greater than the given $\alpha$ level of significance