In: Statistics and Probability
A highway department executive claims that the number of fatal accidents which occur in her state does not vary from month to month. The results of a study of 161 fatal accidents were recorded. Is there enough evidence to reject the highway department executive's claim about the distribution of fatal accidents between each month?
| Month | Jan | Feb | Mar | Apr | May | Jun | Jul | Aug | Sep | Oct | Nov | Dec |
|---|---|---|---|---|---|---|---|---|---|---|---|---|
| Fatal Accidents | 16 | 8 | 10 | 13 | 13 | 11 | 16 | 16 | 13 | 13 | 18 | 14 |
Step 1 of 10:
State the null and alternative hypothesis.
Step 2 of 10:
What does the null hypothesis indicate about the proportions of fatal accidents during each month?
Step 3 of 10:
State the null and alternative hypothesis in terms of the expected proportions for each category.
Step 4 of 10:
Find the expected value for the number of fatal accidents that occurred in January. Round your answer to two decimal places.
Step 5 of 10:
Find the expected value for the number of fatal accidents that occurred in February. Round your answer to two decimal places.
Step 6 of 10:
Find the value of the test statistic. Round your answer to three decimal places.
Step 7 of 10:
Find the degrees of freedom associated with the test statistic for this problem.
Step 8 of 10:
Find the critical value of the test at the 0.005 level of significance. Round your answer to three decimal places.
Step 9 of 10:
Make the decision to reject or fail to reject the null hypothesis at the 0.005 level of significance.
Step 10 of 10:
State the conclusion of the hypothesis test at the 0.005 level of significance.
null hypothesis: all proportion are equal
Alternate hypothesis: Not all proportion are equal
step 3: Ho: pjan=pfeb=pmarch =.....p i=1/12
Ha: pj is not equal to 1/12
step4:
expected value for the number of fatal accidents that occurred in January =np=161*1/12= 13.4
5)
expected value for the number of fatal accidents that occurred in February =13.4
6)
| applying chi square goodness of fit test: |
| relative | observed | Expected | residual | Chi square | |
| category | frequency(p) | Oi | Ei=total*p | R2i=(Oi-Ei)/√Ei | R2i=(Oi-Ei)2/Ei |
| 1 | 0.08 | 16.0 | 13.42 | 0.71 | 0.497 |
| 2 | 0.08 | 8.0 | 13.42 | -1.48 | 2.187 |
| 3 | 0.08 | 10.0 | 13.42 | -0.93 | 0.870 |
| 4 | 0.08 | 13.0 | 13.42 | -0.11 | 0.013 |
| 5 | 0.08 | 13.0 | 13.42 | -0.11 | 0.013 |
| 6 | 0.08 | 11.0 | 13.42 | -0.66 | 0.435 |
| 7 | 0.08 | 16.0 | 13.42 | 0.71 | 0.497 |
| 8 | 0.08 | 16.0 | 13.42 | 0.71 | 0.497 |
| 9 | 0.08 | 13.0 | 13.42 | -0.11 | 0.013 |
| 10 | 0.08 | 13.0 | 13.42 | -0.11 | 0.013 |
| 11 | 0.08 | 18.0 | 13.42 | 1.25 | 1.566 |
| 12 | 0.08 | 14.0 | 13.42 | 0.16 | 0.025 |
| total | 1.000 | 161 | 161 | 6.6273 | |
| test statistic X2 = | 6.627 | ||||
Step 7 of 10:
| degree of freedom =categories-1= | 11 | ||
Step 8 of 10:
| for 0.005 level and 11 df :crtiical value X2 = | 26.757 | |||
Step 9 of 10:
fail to reject the null hypothesis
Step 10 of 10:
conclude that number of fatal accidents which occur in her state does not vary from month to month.