Question

In this activity, you will test if the distribution of colors from a sample bag of M&Ms candy is the same as the Mars company claims. You will need to purchase a small bag of M&Ms from your local store. M&M Mars Co. claims the following distribution of the number of different colors of M&M's in each bag: Brown 13% Yellow 14% Red 13% Orange 20% Green 16% Blue 24% Suppose we are only interested in the number of Orange, Green, Blue and "Other" colors, which would then have the following claimed distribution: Orange 20% Green 16% Blue 24% Other 40% Take a sample of 35 M & M's and run a hypothesis test to see if you accept Mar's claim. Use α = .05.

Answer 1

The claim made by Mar's is correct.

Explanation:

In this case we need to test whether M&M Mars Co.'s claim about the proportion of different colored M&Ms in a small bag is correct or not.

The hypothesis is:

H₀: The claim made by Mar's is not correct.

Hₐ: The claim made by Mar's is correct.

The proportions claimed are as follows:

Orange: 20%

Green: 16%

Blue: 24%

Other: 40%

A sample of 35 M&M's are selected.

The observed frequencies are:

Observed = {O = 4, G = 8, B = 5 and Ot = 18}

Compute the expected frequencies as follows:

`E(O)=4* 20\%=0.80\\\\E(G) = 8* 16\%=1.28\\\\E(B)=5* 24\%=1.20\\\\E(Ot)=18* 40\%=7.20`

Compute the Chi-square statistic as follows:

`\chi^(2)=\sum[((O-E)^(2)))/(E)]`

`=[((4-0.80)^(2))/(0.80)]+[((8-1.28)^(2))/(1.28)]+[((5-1.20)^(2))/(1.20)]+[((18-7.20)^(2))/(7.20)]\\\\=12.80+35.28+12.03+16.20\\\\=76.31`

The Chi-square test statistic value is 76.31.

The degrees of freedom of the test is:

df = n - 1

= 4 - 1

= 3

Compute the p-value as follows:

`p-value=P(\chi^(2)_(0.05,3)<76.31)<0.00001`

The p-value < 0.05.

The null hypothesis will be rejected at 5% level of significance.

Thus, concluding that the claim made by Mar's is correct.