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: [tex]\hat{p}=\dfrac{174}{205}\approx0.85[/tex]
z-value for 90% confidence = 1.645
Confidence interval for p: [tex]\hat{p}\pm z^* \sqrt{\dfrac{p(1-p)}{n}}[/tex]
[tex]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)[/tex]
Hence, the required confidence interval = (0.809, 0.891)
