| written 6.9 years ago by | modified 5.7 years ago by |
| Day | Sun | Mon | Tue | Wed | Thu | Fri | Sat | Total |
|---|---|---|---|---|---|---|---|---|
| No. of Accident | 13 | 15 | 9 | 11 | 12 | 10 | 14 | 84 |
| written 6.9 years ago by | modified 5.7 years ago by |
| Day | Sun | Mon | Tue | Wed | Thu | Fri | Sat | Total |
|---|---|---|---|---|---|---|---|---|
| No. of Accident | 13 | 15 | 9 | 11 | 12 | 10 | 14 | 84 |
| written 6.9 years ago by |
i. Null hypothesis $\ (H_0)$: Accidents are equally distributed over all the days of week.
Alternative Hypothesis $\ (H_a)$ : Accidents do hot occur equally.
ii. Calculation of test statics: If the accidents occur equally on all days of a week, than there will be $ \frac {84}{7} = 12$ accident per day, i.e E=12
| Day | Observed frequency (O) | Expected Frequency (E) | $\ (O-E)^2$ | $\ X^2 = \frac {(O-E)^2}{E}$ |
|---|---|---|---|---|
| Sun | 13 | 12 | 1 | 0.0833 |
| Mon | 15 | 12 | 9 | 0.75 |
| Tue | 11 | 12 | 1 | 0.0833 |
| Wed | 9 | 12 | 9 | 0.75 |
| Thu | 12 | 12 | 0 | 0 |
| Fri | 10 | 12 | 4 | 0.3333 |
| Sat | 14 | 12 | 4 | 0.3333 |
| Total | $\sum{x^2}$ = 2.33 |
iii. Level of significance: $ \alpha$ = 0.05
Degree of freedom = h-1
=7-1
=6
iv Critical value $ \Rightarrow$ For 6 degrees of freedom at 5% level of significance table value.
$\ x^2$ is 12.59
v. Decision $ \Rightarrow$ Since the calculated value of $\ x^2$ is less than the table value. The hypothesis is accepted.
$\therefore$ The accidents occur equally on all working days.