Question
Jan Feb Mar Apr MayJunAug Sep Oct Nov Dec Fatal Accidents 1710 12 199 10 10 121218 17 12
Answer
Step 1 of 10:Ho: distribution of fatal accidents are uniformly distributed across months
Ha: distribution of fatal accidents are not uniformly distributed across months
Step 2 of 10: null hypothesis indicate that the proportions of fatal accidents during each month are equal
Step 3 of 10: ho:for each month Pi =1/12
YHa:at least one month Pi 1/12
Step 4 of 10: expected value for the number of fatal accidents that occurred in January. =np=158/12=13.17
Step 5 of 10: expected value for the number of fatal accidents that occurred in April =13.17
Step 6 of 10:
| applying chi square goodness of fit test: |
| relative | observed | Expected | residual | Chi square | likelihood ratio | |
| category | frequency(p) | Oi | Ei=total*p | R2i=(Oi-Ei)/√Ei | R2i=(Oi-Ei)2/Ei | G2 =2*Oi*ln(Oi/Ei) |
| 1 | 1/12 | 17.000 | 13.167 | 1.06 | 1.116 | 8.6878 |
| 2 | 1/12 | 10.000 | 13.167 | -0.87 | 0.762 | -5.5021 |
| 3 | 1/12 | 12.000 | 13.167 | -0.32 | 0.103 | -2.2268 |
| 4 | 1/12 | 19.000 | 13.167 | 1.61 | 2.584 | 13.9365 |
| 5 | 1/12 | 9.000 | 13.167 | -1.15 | 1.319 | -6.8483 |
| 6 | 1/12 | 10.000 | 13.167 | -0.87 | 0.762 | -5.5021 |
| 7 | 1/12 | 10.000 | 13.167 | -0.87 | 0.762 | -5.5021 |
| 8 | 1/12 | 12.000 | 13.167 | -0.32 | 0.103 | -2.2268 |
| 9 | 1/12 | 12.000 | 13.167 | -0.32 | 0.103 | -2.2268 |
| 10 | 1/12 | 18.000 | 13.167 | 1.33 | 1.774 | 11.2566 |
| 11 | 1/12 | 17.000 | 13.167 | 1.06 | 1.116 | 8.6878 |
| 12 | 1/12 | 12.000 | 13.167 | -0.32 | 0.103 | -2.2268 |
| total | 1.000 | 158 | 158 | 10.6076 | 10.3072 | |
| test statistic X2 = | 10.608 | |||||
Step 7 of 10 :
| degree of freedom =categories-1= | 11 | ||
Step 8 of 10:
| for 0.1 level and 11 df :crtiical value X2 = | 17.275 | |||
Step 9 of 10: fail to reject the null hypothesis at the 0.10 level
Step 10 of 10: we can not conclude that distribution of fatal accidents are not uniformly distributed across months
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