Answer
Step 1:
$H_0$ : The office workers show no preference for ways to combat midday drowsiness.
$H_1$: The office workers show a preference.
Step 2:
Since α=0.10, and the degrees of freedom are 5-1=4 , hence the critical value is 7.779.
Step 3:
Expected Value:E = n/ k = 60/5 = 12.
Test Value :
χ2 = Σ $\frac{(O-E)^{2}}{E}$ =$\frac{(21-12)^{2}}{12}$ + $\frac{(16-12)^{2}}{12}$ +$\frac{(10-12)^{2}}{12}$ +$\frac{(18-12)^{2}}{12}$ + $\frac{(5-12)^{2}}{12}$
=6.750+1.333+0.333+1.333+4.083
=13.833
Step 4:
Since 13.833 > 7.779, the decision is to reject the null hypothesis.
Step 5:
There is enough evidence to reject the claim that the office workers show no preference for ways to combat midday drowsiness.
