Question

Classic Golf Inc. manages five courses in the Jacksonville, Florida, area. The director of golf 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		100
Tuesday		125
Wednesday		110
Thursday		145
Friday		145

At the 0.05 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.

State the decision rule, using the 0.05 significance level. (Round your answer to 3 decimal places.)

Classic Golf Inc. manages five courses in the Jacksonville, Florida, area. The director of golf 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		100
Tuesday		125
Wednesday		110
Thursday		145
Friday		145

At the 0.05 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.

State the decision rule, using the 0.05 significance level. (Round your answer to 3 decimal places.)

Reject H0 if chi-square >

Compute the value of chi-square. (Round your answer to 3 decimal places.)

Chi-square value

Compute the value of chi-square. (Round your answer to 3 decimal places.)

Answer 1

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: H_a: Rounds played is not the same.

We assume 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 days

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

α = 0.05

Critical value = 9.487729037

(by using Chi square table or excel)

Decision rule: Reject H0 if test statistic Chi square > 9.488

Calculation tables for test statistic are given as below:

Day	O	E	(O - E)^2/E
Monday	100	125	5
Tuesday	125	125	0
Wednesday	110	125	1.8
Thursday	145	125	3.2
Friday	145	125	3.2
Total	625	625	13.2

Test Statistic = Chi square = ∑[(O – E)^2/E] = 13.2

χ² statistic = 13.200

P-value = 0.010338797

(By using Chi square table or excel)

P-value < α = 0.05

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.