A researcher wanted to determine whether certain accidents were uniformly distributed over the days of the week. The data show the day of the week for n=303 randomly selected accidents. Is there reason to believe that the accidents occur with equal frequency with respect to the day of the week at the α=0.05 level of significance?Compute the expected counts for each day of the week.Day of the WeekObserved CountExpected Count Sunday43? Monday41? Tuesday28? Wednesday39? Thursday43? Friday50? Saturday59? (Round to three decimal places)2) then calculate P value and the range that the P value corresponds to 3) Compare the P-value with the givenα=0.05 level of significance. If the P-value is less than α,reject the null hypothesis. Then, draw a conclusion in the context of the problem.

The observed counts for each day of the week are listed below Day of the Week Observed Count Sunday 43 Monday 41 Tuesday 28 Wednesday 39 Thursday 43 Friday 50 Saturday 59 It is required to test whether the accidents occur with equal frequency with respect to the day of the week at the α=0.05 level of significance.To test the above claim we will chi-squared goodness of fit test. H0: accidents occur with equal frequency with respect to the day of the week H1: accidents do not occur with the equal frequency with respect to the day of the week The formula for the expected count is E = N* P Where P: The proportion of accidents occur with respect to the day of the week = 17 N: Total counts = 303 The expected count is given as Day of the Week Observed Count Expected count Sunday 43 43.28571429 Monday 41 43.28571429 Tuesday 28 43.28571429 Wednesday 39 43.28571429 Thursday 43 43.28571429 Friday 50 43.28571429 Saturday 59 43.28571429 N =303 The formula for the test statistic is χ2= ∑Oi2Ei - N Calculating the Oi2Ei Day of the Week Observed Count (Oi) Expected count (Ei) Oi2Ei Sunday 43 43.286 42.716 Monday 41 43.286 38.835 Tuesday 28 43.286 18.112 Wednesday 39 43.286 35.138 Thursday 43 43.286 42.716 Friday 50 43.286 57.755 Saturday 59 43.286 80.419 χ2= ∑Oi2Ei - N = 315.6931 - 303 = 12.693 The degree of freedom is df = k - p - 1 = 6 P-value: The p-value for test statistic 12.6931 with df = 6 is 0.04818 Using MS-Excel command CHISQ.DIST.RT(12.6931,6). The range for the p-value is 0.04 to 0.05 Decision rule: At 5% level of significance Reject the H0, if p-value < 0.05 Failed to reject otherwise. Since p-value = 0.048178 < 0.05 , reject the H0. Conclusion : There is sufficient evidence to conclude that the accidents do not occur with the equal frequency with respect to the day of the week

Answered: Compute the expected counts for each…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

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A researcher wanted to determine whether certain accidents were uniformly distributed over the days of the week. The data show the day of the week for n=303 randomly selected accidents. Is there reason to believe that the accidents occur with equal frequency with respect to the day of the week at the α=0.05 level of significance?

Compute the expected counts for each day of the week.

Day of the Week	Observed Count	Expected Count
Sunday	43	?
Monday	41	?
Tuesday	28	?
Wednesday	39	?
Thursday	43	?
Friday	50	?
Saturday	59	?
		(Round to three decimal places)

2) then calculate P value and the range that the P value corresponds to

3) Compare the P-value with the given

α=0.05 level of significance. If the P-value is less than α,

reject the null hypothesis. Then, draw a conclusion in the context of the problem.

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