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.
Accepted Solution
A:
Answer: (0.1239,0.2721)Step-by-step 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 15Total M&Ms = 24+22+20+15+15+15=111The confidence interval for population proportion is given by :-[tex]\hat{p}\pm z^*\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}[/tex], [tex]\hat{p}[/tex] = sample proportion.where n= sample size.[tex]z^*[/tex] = 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 haven= 111[tex]\hat{p}=\dfrac{\text{Number of red M&Ms}}{\text{Total M&Ms}}[/tex][tex]=\dfrac{22}{111}\approx0.198[/tex]Two -tailed z-value for 95% confidence level [tex]z^*=1.96[/tex]Then, the 95% confidence interval for the proportion of red M&Ms in that bag will be :-[tex]0.198\pm (1.96)\sqrt{\dfrac{0.198(1-0.198)}{111}}\\\\0.198\pm0.0741\\\\=(0.198-0.0741,\ 0.198+0.0741)=(0.1239,\ 0.2721) [/tex]Hence, the 95% confidence interval for the proportion of red M&Ms in that bag=(0.1239,0.2721)