Question

1. A branch manager of McDonalds is responsible for the profitability of restaurants in Florida, Georgia, Alabama, and Tennessee. She wants to see if different types of sandwiches (hamburger, cheeseburger, filet-o-fish, and big mac) are purchased similarly across her four assigned states. She takes a random sample of McDonalds restaurants from Florida, Georgia, Alabama, and Tennessee, and calculates a chi square test statistic of 2.589 and a p-value = 0.7682.

What conclusion can she make with alpha = 0.05?

Group of answer choices

a. With a p-value greater than 0.05, we have no evidence that sandwiches are not purchased similarly across the four different states.

b. With a p-value greater than 0.05, we have very strong evidence that sandwiches are not purchased similarly across the four different states.

c. We are 95% confident that the true population distribution is in the interval (0.7682, 2.589).

d. With a p-value less than 0.05, we have moderate evidence that sandwiches are not purchased similarly across the four different states.

2. A CEO of a chain dollar store wants to know if 4 different products sold in his stores (cookies, diapers, hair brushes, combs) in different states are purchased similarly. He takes a sample of 4 of his stores, one in each of the four following states: Florida, Georgia, Alabama, and South Carolina. The data is presented in the table below.

	cookies	diapers	hair brushes	combs	Total
Florida	768	534	623	234	2159
Georgia	685	478	589	354	2106
Alabama	824	598	570	316	2308
South Carolina	738	522	603	289	2152
Total	3015	2132	2385	1193	8725

Find the contribution to the chi square test statistic for combs sold in South Carolina.

Group of answer choices

a. 0.169

b. 0.009

c. 0.937

d. 0.094

Answer 1

Ans:

1)

Option a is correct.

With a p-value greater than 0.05, we have no evidence that sandwiches are not purchased similarly across the four different states.

(As p-value>0.05,we fail to reject the null hypothesis and results are not significant.)

2)

Expected count=row sum*column sum/overall sum

Expected count for the cell (combs sold in South Carolina)=1193*2152/8725=294.2505

Observed count=289

Chi square contribution=(289-294.2505)^2/294.2505=0.094

Option d is correct(0.094)