Question

Day Customers Monday 73 Tuesday 78 Wednesday Thursday 61 88 Friday 74 Saturday 55 Sunday 60 The table shows a recent random sample of customers at a cafe. At a 10% Level of Significance, test to see if customers are distributed uniformly throughout the week. 1. The hypotheses are : O Ho:Customers are not distributed uniformly through the week Ha:Customers are distributed uniformly through the week Ho:Customers are distributed uniformly through the week Ha:Customers are not distributed uniformly through the week 2. This is a lefto two right tailed test; d.f. = 3a. The STS (to 2 decimals) is x = 3b. The P-value (to 3 decimals) = 4a. The decision at a = 10% is O Reject Ho O Do not reject Ho 4b. The conclusion is There is insufficient evidence to conclude that customers are distributed uniformly through the week There is sufficient evidence to conclude that customers are distributed uniformly through the week There is sufficient evidence to conclude that customers are not distributed uniformly through the week There is insufficient evidence to conclude that customers are not distributed uniformly through the week

Answer 1

Part 1

The hypotheses are:

H₀: Customers are distributed uniformly through the week.

H_a: Customers are not distributed uniformly through the week.

Part 2

df = n - 1 = 7 - 1 = 6

This is a right tailed test; d.f. = 6

Part 3a

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	73	69.85714	0.141396436
Tuesday	78	69.85714	0.949167397
Wednesday	61	69.85714	1.122991528
Thursday	88	69.85714	4.711948583
Friday	74	69.85714	0.245690914
Saturday	55	69.85714	3.159801344
Sunday	60	69.85714	1.390885188
Total	489	489	11.72188139

Test Statistic = Chi square = ∑[(O – E)^2/E] = 11.72188139

χ² = 11.72

Part 3b

P-value = 0.069

(By using Chi square table or excel)

Part 4a

We have

α = 0.10

P-value = 0.069

So, P-value < α

So we reject the null hypothesis H₀

Reject H₀

Part 4b

Conclusion:

There is insufficient evidence to conclude that customers are distributed uniformly through the week.