Question

Thank you for your email regarding M&M'S® Chocolate Candies. Our color blends were selected by conducting consumer preference tests, which indicate the assortment of colors that pleased the greatest number of people and created the most attractive overall effect. On average, our mix of colors for M&M'S Dark CHOCOLATE CANDIES is:

•   M&M'S DARK: 17% cyan blue, 16% orange, 16% green, 17% bright yellow, 17% red, 17% brown.

Using a sample of size 150 from Tedd's bag, He found 41 blue, 34 orange, 19 green, 21 yellow, 14 red, 21 brown
is there evidence to support the consumer group’s claim?
Is there any other statistic test besides Chi square to test alternative/null hypothesis?

Answer 1

(i)

Question:

is there evidence to support the consumer group’s claim?

Answer:
H0:Null Hypothesis: The data is as per consumer preference (Claim)

HA: Alternative Hypothesis: The data is not as per consumer preference

The Test Statistic ($\chi ^{2}$) is calculated as follows:

Observed (O)	Expected (E)	(O - E)2/E
41	150X0.17=25.5	9.422
34	150X0.16=24	4.167
19	150X0.16=24	1.042
21	150X0.17=25.5	0.794
14	150X0.17=25.5	5.186
21	150X0.17=25.5	0.794
Total = $\chi ^{2}$ =		21.404

df = 6 - 1 = 5

Take $\alpha$ = 0.05

From Table, critical value of $\chi ^{2}$ = 11.07

Since calculated value of $\chi ^{2}$ = 21.404 is greater than critical value of $\chi ^{2}$ = 11.07, the difference is significant. Reject null hypothesis.

Conclusion:
The data do not support the claim that the data is as per consumer preference.

(ii)

Question:

Is there any other statistic test besides Chi square to test alternative/null hypothesis ?

Answer:

Kolmolgorov - Smirov (KS) Test