A particular report included the following table classifying 713 fatal bicycle a

Question

A particular report included the following table classifying 713 fatal bicycle accidents according to time of day the accident occurred.

(a) Assume it is reasonable to regard the 713 bicycle accidents summarized in the table as a random sample of fatal bicycle accidents in that year. Do these data support the hypothesis that fatal bicycle accidents are not equally likely to occur in each of the 3-hour time periods used to construct the table? Test the relevant hypotheses using a significance level of .05. (Round your 2 value to two decimal places, and round your P-value to three decimal places.)

What can you conclude?

There is sufficient evidence to reject H0.There is insufficient evidence to reject H0.    

(b) Suppose a safety office proposes that bicycle fatalities are twice as likely to occur between noon and midnight as during midnight to noon and suggests the following hypothesis: H0: p1 = 1/3, p2 = 2/3, where p1 is the proportion of accidents occurring between midnight and noon and p2 is the proportion occurring between noon and midnight. Do the given data provide evidence against this hypothesis, or are the data consistent with it? Justify your answer with an appropriate test. (Hint: Use the data to construct a one-way table with just two time categories. Use = 0.05. Round your 2 value to two decimal places, and round your P-value to three decimal places.)

What can you conclude?

There is sufficient evidence to reject H0.There is insufficient evidence to reject H0.

Time of Day Number of Accidents Midnight to 3 a.m. 38 3 a.m. to 6 a.m. 29 6 a.m. to 9 a.m. 67 9 a.m. to Noon 77 Noon to 3 p.m. 97 3 p.m. to 6 p.m. 126 6 p.m. to 9 p.m. 164 9 p.m. to Midnight 115

Explanation / Answer

a) applying chi square goodness of fit test on above:

for above 2 =163.40

and p value =0.000

There is sufficient evidence to reject H0

b)applying chi square goodness of fit on above:

X2 =4.49

p value =0.034

There is sufficient evidence to reject H0

observed Expected Chi square Probability O E=total*p =(O-E)^2/E Midnight to 3 a.m. 1/8 38      89.13 29.33 3 a.m. to 6 a.m. 1/8 29      89.13 40.56 6 a.m. to 9 a.m. 1/8 67      89.13 5.49 9 a.m. to Noon 1/8 77      89.13 1.65 Noon to 3 p.m. 1/8 97      89.13 0.70 3 p.m. to 6 p.m. 1/8 126      89.13 15.26 6 p.m. to 9 p.m. 1/8 164      89.13 62.90 9 p.m. to Midnight 1/8 115      89.13 7.51 1 713 713 163.3983

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