Question

-20 > Assessment Day Escalante 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 Wales Omelets 83 60 81 73 Women 97 71 114 49 Men What can be concluded at the a-0.05 significance level? a. What is the correct statistical test to use! Goodness-of-Fit Paired t-test Independence Homogeneity b. What are the null and alternative hypotheses! H: The distribution of breakfast ordered for men is not the same as it is for women Breakfast ordered and gender are dependent, O Breakfast ordered and gender are independent The distribution of breakfast ordered for men is the same as it is for women The distribution of breakfastdered 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 Breakfast ordered and gender e dependent Breakfast ordered and gender are independent The test statistie for this data Please show your answer to three decimal places & The p-value for this sample Please show your answer to four decimal places)

Answer 1

Observed Frequencies
	C1	C2	C3	C4	Total
R1	83	60	81	73	297
R2	97	71	114	49	331
Total	180	131	195	122	628
Expected Frequencies
	C1	C2	C3	C4	Total
R1	180 * 297 / 628 = 85.1274	131 * 297 / 628 = 61.9538	195 * 297 / 628 = 92.2213	122 * 297 / 628 = 57.6975	297
R2	180 * 331 / 628 = 94.8726	131 * 331 / 628 = 69.0462	195 * 331 / 628 = 102.7787	122 * 331 / 628 = 64.3025	331
Total	180	131	195	122	628
	(fo-fe)²/fe
R1	(83 - 85.1274)²/85.1274 = 0.0532	(60 - 61.9538)²/61.9538 = 0.0616	(81 - 92.2213)²/92.2213 = 1.3654	(73 - 57.6975)²/57.6975 = 4.0585
R2	(97 - 94.8726)²/94.8726 = 0.0477	(71 - 69.0462)²/69.0462 = 0.0553	(114 - 102.7787)²/102.7787 = 1.2251	(49 - 64.3025)²/64.3025 = 3.6417

a)

test: Homogeneity

b)

Null and Alternative hypothesis:  

Ho: The distribution of breakfast ordered for men is the same as it is for women.   

H1: The distribution of breakfast ordered for men is not the same as it is for women.   

c)

Test statistic:

χ² = ∑ ((fo-fe)²/fe) = 10.509

d)

df = (r-1)(c-1) = 3

p-value = CHISQ.DIST.RT(10.509, 3) = 0.0147