Question

10. A school administrator thinks that teacher absences are equally probable for each day of the week. A co-worker does not agree and gathers data from last year and reports the absences for the day of the week in this table: Monday (46) Tuesday (47) Wednesday (40 )Thursday (50) Friday (67) Test that the days of the week have an equal probability of absences at the 0.10 level of significance.

Answer 1

Solution:

Here, we have to use chi square test for goodness of fit.

Null hypothesis: H0: The teacher absences are equally probable for each day of the week.

Alternative hypothesis: Ha: The teacher absences are not equally probable for each day of the week.

We assume/given level of significance = α = 0.10

We are given

Number of categories = N = 5

Degrees of freedom = df = N - 1 = 4

α = 0.10

Critical value = 7.77944034

(by using Chi square table or excel)

Test statistic formula is given as below:

Chi square = ∑[(O – E)^2/E]

Where, O is observed frequencies and E is expected frequencies.

Calculation tables for test statistic are given as below:

Day	O	E	(O - E)^2/E
Mon	46	50	0.32
Tue	47	50	0.18
Wed	40	50	2
Thu	50	50	0
Fri	67	50	5.78
Total	250	250	8.28

Test Statistic = Chi square = ∑[(O – E)^2/E] = 8.28

χ2 = 8.28

P-value = 0.081843457

(By using Chi square table or excel)

P-value < α = 0.10

So, we reject the null hypothesis

There is not sufficient evidence to conclude that teacher absences are equally probable for each day of the week.