In: Statistics and Probability
The HR department of a certain corporation wants to know if some days of the week had more workers call in sick than others. 100 sick day reports are randomly selected. Test the claim that sick days occur with the same frequency on every day of the week using a 0.05 significance level.
Day Mon Tue Wed Thu Fri
# of sick calls 22 15 17 16 30
We are testing for the goodness of fit. That is we want to know if the observed values match the distribution given . Here the distribution is uniform that is all the days have same frequency. So if there are 100 reports and 5 days, the expected equal frequency would be 20 reports each da (100 / 5).
Day | Observed | Exp | Oi- Ei | (Oi- Ei)2 | |
Mon | 22 | 20 | 2 | 4 | 0.2 |
Tue | 15 | 20 | -5 | 25 | 1.25 |
Wed | 17 | 20 | -3 | 9 | 0.45 |
Thu | 16 | 20 | -4 | 16 | 0.8 |
Fri | 30 | 20 | 10 | 100 | 5 |
Total | 100 | 7.7 |
Test
Null: The freq of the reports is same for all days.
Alternative: The freq of the reports is same for all days.
Chi -square test STat =
Test Stat = 7.7
p-value = P( > Test stat) ....................df = n-1
= P(4 > 7.7)
p-value = 0.1032 .................using chi-square tables
. level of significance = 0.05
Since p-value > 0.05
We reject the null hypothesis at 5%. There is sufficient evidence to reject the claim that sick days occur with the same frequency on every day of the week.