Question

Use the sample data and confidence level given below to complete parts​ (a) through​ (d).

In a study of cell phone use and brain hemispheric​ dominance, an Internet survey was​ e-mailed to 2356 subjects randomly selected from an online group involved with ears. 1161

surveys were returned. Construct a 90​% confidence interval for the proportion of returned surveys.

a) Find the best point estimate of the population proportion p.

​(Round to three decimal places as​ needed.)

​b) Identify the value of the margin of error E.

​(Round to three decimal places as​ needed.)

C. Construct a confidence intervel

Answer 1

Solution :

Given that,

(a)

Point estimate = sample proportion = = x / n = 1161 / 2356 = 0.493

Z_/2 = 1.645

(b)

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

= 1.645 * (((0.493 * 0.507) / 2356)

= 0.017

A 90% confidence interval for population proportion p is ,

- E < p < + E

0.493 - 0.017 < p < 0.493 + 0.017

0.476 < p < 0.510

(0.476 , 0.510)