Question

A consumer advocate group selects a random sample of 205 snow foods in find 174 of them contain more what is indicated on the snacks nutritional information. Create a 90% confidence interval for the population proportion for a snack foods that contain more fat grams and was indicated on the snacks nutritional information

Answer 1

Answer: (0.809, 0.891)

Explanation:

Let p = population proportion for a snack foods that contain more fat grams and was indicated on the snacks nutritional information .

As per given , we have

Sample size : n= 205

Sample proportion:
`\hat{p}=(174)/(205)\approx0.85`

z-value for 90% confidence = 1.645

Confidence interval for p:
`\hat{p}\pm z^* \sqrt{(p(1-p))/(n)}`

`0.85\pm (1.645)\sqrt{((0.85)(1-0.85))/(205)}\\\\=0.85\pm (1.645)\sqrt{(0.85*0.15)/(205)}\\\\= 0.85\pm (1.645)√(0.000621951219512)\\\\=0.85\pm0.041\\\\=(0.85-0.041,\ 0.85+0.041)\\\\= (0.809,\ 0.891)`

Hence, the required confidence interval = (0.809, 0.891)