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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

Respuesta :

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

Step-by-step explanation:

Since we have given that

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

x = yellow = 22

So, [tex]\hat{p}=\dfrac{22}{108}=0.20[/tex]

We need to find the 95% confidence interval.

So, z = 1.96

So, Interval would be

[tex]\hat{p}\pm z\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}\\\\=0.20\pm 1.96\times \sqrt{\dfrac{0.2\times 0.8}{108}}\\\\=0.20\pm 0.075\\\\=(0.20-0.075, 0.20+0.075)\\\\=(0.125, 0.275)[/tex]

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

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