Question

help 10

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 2340 subjects randomly selected from an online group involved with ears, 1013 surveys were returned. Construct a 90% confidence interval for the proportion of returned surveys B1: Click the icon to view a table of z scores 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 EN (Round to three decimal places as needed

Answer 1

Solution :

Given that,

a)

Point estimate = sample proportion = $\hat p$ = x / n = 1013 / 2340 = 0.433

1 - $\hat p$ = 1 - 0.433 = 0.567

Z$\alpha$/2 = 1.646

b)

Margin of error = E = Z$\alpha$ / 2 * $\sqrt$ (($\hat p$ * (1 - $\hat p$ )) / n)

= 1.645 * ($\sqrt$((0.433 * 0.567) / 2340)

Margin of error = E = 0.017

c)

A 90% confidence interval for population proportion p is ,

$\hat p$ - E < p < $\hat p$ + E

0.433 - 0.017 < p < 0.433 + 0.017

0.416 < p < 0.450

d)

option B. is correct