Question

A few years ago, Mars sold M&Ms inspired by candy corn. The M&Ms in those bags came in three colors: white, yellow, and orange. The following table shows the distributions of those colors in a bag of 90 M&Ms purchased at a local store.A few years ago, Mars sold M&Ms inspired by candy corn.

Using an alpha of 0.05, carry out a goodness-of-fit test for the null hypothesis that the three colors appear equally often.

(Hint: We will not be able to reject the null hypothesis, but you need to show the formal steps and calculations that lead to that decision.)

Color of M&Ms	Count
White	32
Yellow	33
Orange	25

Part (b)            (4 points)

State one example (there are many different possible answers) for the distribution of the three colors of M&Ms that would lead us to reject the null hypothesis that the three colors appear equally often. You again need to show the formal steps and calculations that lead to that decision.

Answer 1

Solution:

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

Null hypothesis: H₀: the three colors appear equally often.

Alternative hypothesis: H_a: the three colors do not appear equally often.

We are given 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 = 3

Degrees of freedom = df = N – 1 = 3 – 1 = 2

α = 0.05

Critical value = 5.991464547

(by using Chi square table or excel)

Calculation tables for test statistic are given as below:

Color	O	E	(O - E)^2/E
White	32	30	0.1333
Yellow	33	30	0.3000
Orange	25	30	0.8333
Total	90	90	1.2667

Chi square = ∑[(O – E)^2/E] = 1.2667

P-value = 0.530819452

(By using Chi square table or excel)

P-value > α = 0.05

So, we do not reject the null hypothesis

There is sufficient evidence to conclude that the three colors appear equally often.

Now, we have to show one example in which we reject the claim that the three colors appears equally often.

Required example is given as below:

Color	O	E	(O - E)^2/E
White	34	30	0.5333
Yellow	38	30	2.1333
Orange	18	30	4.8000
Total	90	90	7.4667

Chi square = ∑[(O – E)^2/E] = 7.4667

P-value = 0.023912993

P-value < α = 0.05

So, we reject the null hypothesis

There is insufficient evidence to conclude that the three colors appear equally often.