Question

The table below shows the number of students absent on the particular day of the week:

Day	M	Tu	W	Th	F
Number	111	80	85	94	99

Test if the distribution of students absent is uniform through the week with the significance level of 5%.

Answer 1

Solution:

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

Null hypothesis: H0: the distribution of students absent is uniform through the week.

Alternative hypothesis: Ha: the distribution of students absent is not uniform through the week.

We assume/given level of significance = α = 0.05

Test statistic formula is given as below:

Chi square = ∑[(O – E)^2/E]

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

We are given

N = 5

Degrees of freedom = df = N - 1 = 5 - 1 = 4

α = 0.05

Critical value = 9.487729037

(by using Chi square table or excel)

Calculation tables for test statistic are given as below:

Day	O	E	(O - E)^2/E
M	111	93.8	3.153944563
Tu	80	93.8	2.030277186
W	85	93.8	0.825586354
Th	94	93.8	0.000426439
F	99	93.8	0.288272921
Total	469	469	6.298507463

Test Statistic = Chi square = ∑[(O – E)^2/E] = 6.298507463

χ2 statistic = 6.298507463

P-value = 0.177937087

(By using Chi square table or excel)

P-value > α = 0.05

So, we do not reject the null hypothesis

There is sufficient evidence to conclude that the distribution of students absent is uniform through the week.