Question

A bag of M&Ms was randomly selected from the grocery store shelf, and the color counts were: Brown 24 Red 22 Yellow 20 Orange 15 Green 15 Blue 15 .Find the 95% confidence interval for the proportion of red M&Ms in that bag.

Answer 1

Answer: (0.1239,0.2721)

Explanation:

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

Brown 24 Red 22 Yellow 20 Orange 15 Green 15 Blue 15

Total M&Ms = 24+22+20+15+15+15=111

The confidence interval for population proportion is given by :-

`\hat{p}\pm z^*\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}`

,
`\hat{p}` = sample proportion.

where n= sample size.

`z^*` = Two -tailed z-value for confidence level of c.

Let p be the population proportion of red M&Ms in that bag.

As per given , we have

n= 111

`\hat{p}=\frac{\text{Number of red M&Ms}}{\text{Total M&Ms}}`

`=(22)/(111)\approx0.198`

Two -tailed z-value for 95% confidence level

`z^*=1.96`

Then, the 95% confidence interval for the proportion of red M&Ms in that bag will be :-

`0.198\pm (1.96)\sqrt{(0.198(1-0.198))/(111)}\\\\0.198\pm0.0741\\\\=(0.198-0.0741,\ 0.198+0.0741)=(0.1239,\ 0.2721)`

Hence, the 95% confidence interval for the proportion of red M&Ms in that bag=(0.1239,0.2721)