Question

The Mars Company, the maker of M&M’s, recently claimed that M&M’s were so much fun because they were a perfect rainbow, that is, each bag they made contained equal numbers of each of the five colors. An enterprising student decided to test this and bought one small bag, chosen at random.

The number of M&M’s found for each color was:
Brown: 19
Blue: 5
Green: 5
Orange: 9
Yellow: 10

Perform a hypothesis test and state your conclusions about Mars’ claim.

If the colors were in equal proportions, what would be the probability (in percent form) that a random sample of 48 M&M’s will have the above proportion of brown ones? (Hint: Go back to the Normal Distribution Curve and look at the probabilities under the curve.)

No need to perform another hypothesis test here.

Answer 1

Perform a hypothesis test and state your conclusions about Mars’ claim.

Solution:

Here, we have to use chi square test for goodness of fit.

Null hypothesis: H₀: Each bag of M&M contained equal numbers of each of the five colors.

Alternative hypothesis: H_a: Each bag of M&M does not contain equal numbers of each of the five colors.

We assume level of significance = α = 0.05

Test statistic formula is given as below:

Chi square = ∑[(O – E)^2/E]

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

We are given

N = 5

Degrees of freedom = df = N – 1 = 4

α = 0.05

Critical value = 9.487729

(by using Chi square table or excel)

Calculation tables for test statistic are given as below:

	%	O	E	(O - E)^2/E
Brown	0.2	19	9.6	9.2042
Blue	0.2	5	9.6	2.2042
Green	0.2	5	9.6	2.2042
Orange	0.2	9	9.6	0.0375
Yellow	0.2	10	9.6	0.0167
Total	1	48	48	13.6667

Chi square = ∑[(O – E)^2/E] = 13.6667

P-value = 0.0084

(By using Chi square table or excel)

P-value < α = 0.05

So, we reject the null hypothesis

There is insufficient evidence to conclude that each bag of M&M contained equal numbers of each of the five colors.

If the colors were in equal proportions, what would be the probability (in percent form) that a random sample of 48 M&M’s will have the above proportion of brown ones?

We have to find P(X>19)

Mean = np = 48*0.20 = 9.6

SD = sqrt(npq) = sqrt(48*0.20*0.80) = 2.771281

Z = (X – mean)/SD

Z = (19 - 9.6)/ 2.771281

Z = 3.391933189

P(Z>3.391933189) = 1 – P(Z<3.391933189) = 1 - 0.999652993 = 0.000347007

(by using z-table or excel)

Required probability = 0.03%