In: Statistics and Probability
Emerald Golf Club in Thailand manages five courses in the Phuket area. The main boss of Emerald Golf Club wishes to study the number of rounds of golf played per weekday at the five courses. He gathered the following sample information.
Day | Rounds | ||
Monday | 120 | ||
Tuesday | 105 | ||
Wednesday | 100 | ||
Thursday | 150 | ||
Friday | 150 | ||
At the 0.01 significance level, is there a difference in the number of rounds played by day of the week?
H0: Rounds played is the same for each
day.
H1: Rounds played is not the same.
Solution:
Here, we have to use chi square test for goodness of fit.
Null hypothesis: H0: Rounds played is the same for each day.
Alternative hypothesis: Ha: Rounds played is not the same.
We are given level of significance = α = 0.01
We are given
Number of categories/days = N = 5
Degrees of freedom = df = N - 1 = 4
α = 0.01
Critical value = 13.27670414
(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 |
Monday |
120 |
125 |
0.2 |
Tuesday |
105 |
125 |
3.2 |
Wednesday |
100 |
125 |
5 |
Thursday |
150 |
125 |
5 |
Friday |
150 |
125 |
5 |
Total |
625 |
625 |
18.4 |
Test Statistic = Chi square = ∑[(O – E)^2/E] = 18.4
χ2 = 18.4
P-value = 0.001030602
(By using Chi square table or excel)
P-value < α = 0.01
So, we reject the null hypothesis
There is sufficient evidence to conclude that there is a difference in the number of rounds played by day of the week.