The safety director of a large steel mill took samples at random from company records of minor work-related accidents and classified them according to the time the accident took place.
Number of | Number of | |||||||
Time | Accidents | Time | Accidents | |||||
8 up to 9 a.m. | 8 | 1 up to 2 p.m. | 8 | |||||
9 up to 10 a.m. | 4 | 2 up to 3 p.m. | 10 | |||||
10 up to 11 a.m. | 8 | 3 up to 4 p.m. | 9 | |||||
11 up to 12 p.m. | 11 | 4 up to 5 p.m. | 12 | |||||
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Using the goodness-of-fit test and the 0.02 level of significance, determine whether the accidents are evenly distributed throughout the day.
H0: The accidents are evenly distributed
throughout the day.
H1: The accidents are not evenly distributed
throughout the day.
State the decision rule, using the 0.02 significance level. (Round your answer to 3 decimal places.)
Compute the value of chi-square. (Round your answer to 3 decimal places.)
What is your decision regarding H0?
degree of freedom =categories-1= | 7 | ||||
for 0.02 level and 7 df :crtiical value X2 = | 16.622 | from excel: chiinv(0.02,7) | |||
Decision rule: reject Ho if value of test statistic X2>16.622 |
applying chi square goodness of fit test: |
relative | observed | Expected | Chi square | ||
Category | frequency(p) | Oi | Ei=total*p | R2i=(Oi-Ei)2/Ei | |
1 | 1/8 | 8 | 8.75 | 0.064 | |
2 | 1/8 | 4 | 8.75 | 2.579 | |
3 | 1/8 | 8 | 8.75 | 0.064 | |
4 | 1/8 | 11 | 8.75 | 0.579 | |
5 | 1/8 | 8 | 8.75 | 0.064 | |
6 | 1/8 | 10 | 8.75 | 0.179 | |
7 | 1/8 | 9 | 8.75 | 0.007 | |
8 | 1/8 | 12 | 8.75 | 1.207 | |
total | 1.00 | 70 | 70 | 4.74 | |
test statistic X2= | 4.743 |
decision regarding H0 :
Fail to reject Ho ; since test statistic < critical value
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