written 6.2 years ago by
teamques10
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modified 6.0 years ago
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The null hypothesis $H_0$ : Accidents occur equally on all months.
Alternative hypothesis $H_a$ : Accidents do not occur equally on all months.
(ii) Calculation of test statistic: On the basis of this hypothesis,the number of accidents per month = $\frac{total}{10}$ = $\frac{(20 +17+12+6+7+15+8+5+16+14)}{10}$ = $\frac{120}{10}$ = 12
$\chi^2$ = $\sum$ $\frac{(O-E)^2}{E}$ = $\frac{(20-12)^2}{12}$ + $\frac{(17-12)^2}{12}$ + $\frac{(12-12)^2}{12}$ + $\frac{(6-12)^2}{12}$ + $\frac{(7-12)^2}{12}$ + $\frac{(15-12)^2}{12}$ + $\frac{(8-12)^2}{12}$ + $\frac{(5-12)^2}{12}$ + $\frac{(16-12)^2}{12}$ + $\frac{(14-12)^2}{12}$ = $\frac{244}{12}$ = 20.33
(iii) Level of significance : $α =0.05$
(iv) Critical value : the table value of χ$^2$ at 5% level of significance for $ν = 10 - 1 = 9$ degrees of freedom is $16.92$
(v) Decision : since the calculated value of $|t| =20.33$ is greater than the table value $χ^2=16.92$, the null hypothesis is rejected.
∴ Accidents do not occur equally on all months.