Question

Use the following prompt to answer the next 6 questions. Suppose we want to test the color distribution claim on the M&M’s website that a bag of plain M&M’s is made up of 10% blue, 10% orange, 10% green, 20% red, 20% yellow, and 30% brown. We select a sample of 400 plain M&M’s and found the following: Color Blue Orange Green Red Yellow Brown Frequency 30 48 55 66 70 131 Is there evidence to doubt the color distribution claimed by the website? Use ????=0.05

Answer 1

The claim on the M&M’s website is not true.

Explanation:

A Chi-square test for goodness of fit will be used in this case.

The hypothesis can be defined as:

H₀: The observed frequencies are same as the expected frequencies.

Hₐ: The observed frequencies are not same as the expected frequencies.

The test statistic is given as follows:

`\chi^(2)=\sum\limits^(n)_(i=1){((O_(i)-E_(i))^(2))/(E_(i))}`

Here,

`O_(i)` = Observed frequencies

`E_(i)=N* p_(i)` = Expected frequency.

The chi-square test statistic value is, 14.433.

The degrees of freedom is:

df = k - 1 = 6 - 1 = 5

Compute the p-value as follows:

`p-value=P(\chi^(2)_(k-1) >14.433) =P(\chi^(2)_(5) >14.433) =0.013`

*Use a Chi-square table.

The significance level is, α = 0.05.

p-value = 0.013 < α = 0.05.

So, the null hypothesis will be rejected at 5% significance level.

Thus, concluding that the claim on the M&M’s website is not true.