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)

Step-by-step 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: $\hat{p}=\dfrac{174}{205}\approx0.85$

z-value for 90% confidence = 1.645

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

$0.85\pm (1.645)\sqrt{\dfrac{(0.85)(1-0.85)}{205}}\\\\=0.85\pm (1.645)\sqrt{\dfrac{0.85\times0.15}{205}}\\\\= 0.85\pm (1.645)\sqrt{0.000621951219512}\\\\=0.85\pm0.041\\\\=(0.85-0.041,\ 0.85+0.041)\\\\= (0.809,\ 0.891)$

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