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

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

(0.0643075 , 0.0856925)

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{(p(1-p))/(n) } , (p^(-) + Z_(0.10) \sqrt{(p(1-p))/(n) })`

`(0.075 - 1.645 \sqrt{(0.075(1-0.075))/(1613) } , (0.075 + 1.645 \sqrt{(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)