A bicycle safety organization claims that fatal bicycle accidents are uniformly

Question

A bicycle safety organization claims that fatal bicycle accidents are uniformly distributed throughout the week. The table on the right shows the day of the week for which 779 randomly selected fatal bicycle accidents occurred. At alpha = 0.01, can you reject the claim that the distribution is uniform? Complete parts a through d below. H_0: The distribution of fatal bicycle accidents throughout the week is H_a: The distribution of fatal bicycle accidents throughout the week is Which hypothesis is the claim? H_0 H_a (b) Determine the critical value, chi^2_0, and the rejection region. chi^2_0 = (Round to three decimal places as needed.) Choose the correct rejection region below. A. chi^2 lessthanorequalto chi^2_0 B. chi^2 chi^2_0 D. chi^2 greaterthanorequalto chi^2_0 (c) Calculate the test statistic. chi^2 = (Round to three decimal places as needed.) chi^2 = 0 (Round to three decimal places as needed.) (d) Decide whether to reject or fail to reject the null hypothesis. Then interpret the decision in the context of the original claim. H_0. At the 1% significance level, there enough evidence to reject the claim that the distribution of fatal bicycle accidents throughout the week is uniform.

Explanation / Answer

Part (a)

H0 : The distribution of fatal accidents throughout the week is uniform.

Ha : The distribution of fatal accidents throughout the week is not uniform.

So, claim is H0ANSWER option 1

Part (b)

The test statistic 2 is distributed as chi-square distribution with 6 [number of classes - 1] degrees of freedom. Further, given = 0.01, the critical value 2o is the upper 1% point of 26= 16.812 ANSWER 1

Rejection region is 2 > 2o ANSWER option c

Part (c)

Test statistic = 2 = [i = 1,7]{(Oi - Ei)2/Ei}, where Oi and Ei are respectively the observed and expected frequencies. Under H0, Ei = total frequency/7 [7 days of the week]

The above formula can be simplified into [i = 1,7](Oi2/Ei) - [i = 1,7]Oi. Further, in the present case, Ei is the same for all i [i.e., 779/7 = 111.2857] and hence,

2 = (1/111.2857)[i = 1,7]Oi2 – 779 = (86899/111.2857) – 779 = 780.8640 – 779 = 1.864 ANSWER

Part (d)

Since 2cal (1.864) < 2o (16.812), H0 is accepted.

So, to fill in the blanks,

Accept H0. At 1% significance, there is not enough evidence to reject the claim that the distribution of fatal accidents throughout the week is uniform. ANSWER

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