Question

A bag of M&Ms was randomly selected from the grocery store shelf, and the color counts were:

Brown 22
Red 22
Yellow 22
Orange 12
Green 15
Blue 15
Find the 95% confidence interval for the proportion of yellow M&Ms in that bag

Answer 1

Answer: 95% confidence interval for the proportion of yellow is (0.125,0.275).

Explanation:

Since we have given that

n = 22+22+22+12+15+15=108

x = yellow = 22

So,
`\hat{p}=(22)/(108)=0.20`

We need to find the 95% confidence interval.

So, z = 1.96

So, Interval would be

`\hat{p}\pm z\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\\\\=0.20\pm 1.96* \sqrt{(0.2* 0.8)/(108)}\\\\=0.20\pm 0.075\\\\=(0.20-0.075, 0.20+0.075)\\\\=(0.125, 0.275)`

Hence, 95% confidence interval for the proportion of yellow is (0.125,0.275).