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 2674 subjects randomly selected from an online group involved with ears. 1160 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 the confidence interval.

___ < p < ___​ (Round to three decimal places as​needed.)

​d) Write a statement that correctly interprets the confidence interval. Choose the correct answer below.

A. There is a 90​% chance that the true value of the population proportion will fall between the lower bound and the upper bound.

B. 90% of sample proportions will fall between the lower bound and the upper bound.

C. One has 950% confidence that the interval from the lower bound to the upper bound actually does contain the true value of the population proportion.

D. One has 90​% confidence that the sample proportion is equal to the population proportion.

Answer 1

Solution :

Given that,

(a)

Point estimate = sample proportion = = x / n = 1160 / 3674 = 0.434

(b)

Z_/2 = 1.645

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

= 1.645 * (((0.434 * 0.566) / 2674)

= 0.019

(c)

A 90% confidence interval for population proportion p is ,

- E < p < + E

0.434 - 0.019 < p < 0.434 + 0.019

0.415 < p < 0.453

(d)

C. One has 90% confidence that the interval from the lower bound to the upper bound actually does contain the true value of the population proportion.