Question

A particular report included the following table classifying 712 fatal bicycle accidents according to time of day the accident occurred.

Time of Day	Number of Accidents
Midnight to 3 a.m.	36
3 a.m. to 6 a.m.	29
6 a.m. to 9 a.m.	65
9 a.m. to Noon	75
Noon to 3 p.m.	97
3 p.m. to 6 p.m.	128
6 p.m. to 9 p.m.	167
9 p.m. to Midnight	115

(a) Assume it is reasonable to regard the 712 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 χ² value to two decimal places, and round your P-value to three decimal places.)

χ2	=
P-value	=

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 α = 0.05. Round your χ² value to two decimal places, and round your P-value to three decimal places.)

χ2	=
P-value	=

What can you conclude?

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

You may need to use the appropriate table in Appendix A to answer this question.

Answer 1

a)

Ho:  fatal bicycle accidents are equally likely to occur in each of the 3-hour time periods

H1:  fatal bicycle accidents are not equally likely to occur in each of the 3-hour time periods

Chi square test for Goodness of fit  
  
expected frequncy,E = expected proportions*total frequency =1/8*712
total frequency=   712

observed frequencey, O	expected proportion	expected frequency,E			(O-E)²/E
36	0.125	89.00			31.562
29	0.125	89.00			40.449
65	0.125	89.00			6.472
75	0.125	89.00			2.202
97	0.125	89.00			0.719
128	0.125	89.00			17.090
167	0.125	89			68.360
115	0.125	89			7.596

chi square test statistic,X² = Σ(O-E)²/E =   174.45

level of significance, α=   0.05              
Degree of freedom=k-1=   8   -   1   =   7
                  
P value =   0.0000   [ excel function: =chisq.dist.rt(test-stat,df) ]          
Decision: P value < α, Reject Ho                  
There is sufficient evidence to reject H₀.

==========================

b)

observed frequencey, O	expected proportion	expected frequency,E			(O-E)²/E
205	1/3	237.33			4.405
507	2/3	474.67			2.202

chi square test statistic,X² = Σ(O-E)²/E =   6.61

level of significance, α=   0.05              
Degree of freedom=k-1=   2   -   1   =   1
                  
P value =   0.010 [ excel function: =chisq.dist.rt(test-stat,df) ]          
Decision: P value < α, Reject Ho
There is sufficient evidence to reject H₀.