Question

Do men and women select different breakfasts? The breakfasts ordered by randomly selected men and women at a popular breakfast place are shown below. Frequencies of Breakfasts for Men and Women French Toast Pancakes Waffles Omelets Men 100 46 89 77 Women 99 72 101 51 What can be concluded at the a = 0.01 significance level? a. What is the correct statistical test to use? Goodness-of-Fit O Independence O Paired t-test Homogeneity b. What are the null and alternative hypotheses? Ho: The distribution of breakfast ordered for men is not the same as it is for women. The distribution of breakfast ordered for men is the same as it is for women. O Breakfast ordered and gender are independent. O Breakfast ordered and gender are dependent

Answer 1

Part a

Answer:

Independence

For the given scenario we have to test whether the two categorical variables breakfast ordered and gender are independent or not. So, for checking this independence between two categorical variables we need to use chi-square test for independence between two categorical variables.

Part b

The null and alternative hypotheses for this test are given as below:

Null hypothesis:

H₀: Breakfast ordered and gender are independent.

Alternative hypothesis:

H₁: Breakfast ordered and gender are dependent.

Part c

Test statistic formula is given as below:

Chi square = ∑[(O – E)^2/E]

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

E = row total * column total / Grand total

α = 0.01

Critical value = 11.34487

(by using Chi square table or excel)

Calculation tables for test statistic are given as below:

Observed Frequencies
	Column variable
Row variable	French Toast	Pancakes	Waffles	Omelets	Total
Men	100	46	89	77	312
Women	99	72	101	51	323
Total	199	118	190	128	635

Expected Frequencies
	Column variable
Row variable	French Toast	Pancakes	Waffles	Omelets	Total
Men	97.77638	57.97795	93.35433	62.89134	312
Women	101.2236	60.02205	96.64567	65.10866	323
Total	199	118	190	128	635

Calculations
(O - E)
2.223622	-11.978	-4.35433	14.10866
-2.22362	11.97795	4.354331	-14.1087

(O - E)^2/E
0.050569	2.474585	0.203099	3.165052
0.048847	2.390311	0.196183	3.057263

Test Statistic = Chi square = ∑[(O – E)^2/E] = 11.58591

χ² test-statistic = 11.586

Part e

We are given

Number of rows = r = 2

Number of columns = c = 4

Degrees of freedom = df = (r – 1)*(c – 1) = 1*3 = 3

χ² test-statistic = 11.586

So, we have

P-value = 0.008945

(By using Chi square table or excel)

P-value = 0.0089

Part e

P-value < α = 0.05

The p-value is less than α.

So, we reject the null hypothesis

Part f

Reject the null

Part g

Conclusion:

There is sufficient evidence to conclude that breakfast ordered and gender are dependent.