Question

Mimi conducted a survey on a random sample of 100 adults. 58 adults in the sample chose banana as his / her favorite fruit. Construct a 95% confidence interval estimate of the proportion of adults whose favorite fruit is banana. Show all work. Just the answer, without supporting work, will receive no credit.

Answer 1

Solution :

Given that,

n = 100

x = 58

= x / n = 58 / 100 = 0.580

1 - = 1 - 0.580 = 0.420

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z_/2 = Z_0.025 = 1.96

Margin of error = E = Z _{/ 2} * (( * (1 - )) / n)

= 1.96 * (((0.580 * 0.420) / 100)

= 0.097

A 95% confidence interval for population proportion p is ,

- E < P < + E

0.580 - 0.097 < p < 0.580 + 0.097

0.483 < p < 0.677

(0.483 , 0.677)