Question

A researcher wishes to estimate the proportion of adults in the city of Darby who are vegetarian. In a random sample of 1,613 adults from this city, the proportion that are vegetarian is 0.075. Find a 90% confidence interval for the proportion of all adults in the city of Darby that are vegetarians. Group of answer choices

Answer 1

Answer:

90% confidence interval for the proportion of all adults in the city of Darby that are vegetarians

(0.0643075 , 0.0856925)

Step-by-step explanation:

Explanation

Size of the random sample  'n' = 1613

The proportion of the  vegetarian p = 0.075

Level of significance = 0.10

90% confidence interval for the proportion of all adults in the city of Darby that are vegetarians

$(p^{-} - Z_{0.10} \sqrt{\frac{p(1-p)}{n} } , (p^{-} + Z_{0.10} \sqrt{\frac{p(1-p)}{n} })$

$(0.075 - 1.645 \sqrt{\frac{0.075(1-0.075)}{1613} } , (0.075 + 1.645 \sqrt{\frac{0.075(1-0.075)}{1613} })$

( 0.075 - 1.645 × 0.0065 ,  0.075 +1.645 × 0.0065)

( 0.075 -0.0106925 , 0.075+0.0106925)

(0.0643075 , 0.0856925)

Final answer:-

90% confidence interval for the proportion of all adults in the city of Darby that are vegetarians

(0.0643075 , 0.0856925)