Answer to Question #219974 in Statistics and Probability for sai

Question

A researcher wishes to estimate the proportion of adults in the city of Darby who are vegetarian. In a random sample of 682 adults from this​ city, the proportion that are vegetarian is 0.072. Find a 95​% confidence interval for the proportion of all adults in the city of Darby that are vegetarians. Round to three decimal places.

​A. (0.034​, 0.110​)

B.​(0.053​, 0.091​)

C.(0.058​, 0.086​)

D. ​(0.062​, 0.082​)

Answer 1

Solution:

Given, $p=0.072,n=682$

Required 95% CI$=(p\pm1.96\sqrt{\dfrac{p(1-p)}{n}})$

$=(0.072\pm1.96\sqrt{\dfrac{0.072\times0.928}{682}})$

$=(0.072\pm0.0194) \\=(0.0526,0.0914)$

$=(0.053,0.091)$

Hence, option B is correct.