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 2330 subjects randomly selected from an online group involved with ears. 1176 surveys were returned. Construct a 99​% 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. One has 99% confidence that the sample proportion is equal to the population proportion.

B. 99​% of sample proportions will fall between the lower bound and the upper bound.

C. There is a 99% chance that the true value of the population proportion will fall between the lower bound and the upper bound.

D. One has 99​% confidence that the interval from the lower bound to the upper bound actually does contain the true value of the population proportion.

Answer 1

Solution :

Given that,

n = 2330

x = 1176

a) Point estimate = sample proportion = = x / n = 1176 / 2330 = 0.505

1 - = 1 - 0.505 = 0.495

At 99% confidence level

= 1 - 99%

=1 - 0.99 =0.01

/2 = 0.005

Z/2 = Z0.005  = 2.576

b) Margin of error = E = Z / 2 * (( * (1 - )) / n)

= 2.576 (((0.505 * 0.495) / 2330)

= 0.027

c) A 99% confidence interval for population proportion p is ,

- E < p < + E

0.505 - 0.027 < p < 0.505 + 0.027

( 0.478 < p < 0.532 )

d) correct option is D

D. One has 99​% confidence that the interval from the lower bound to the upper bound actually does contain the true value of the population proportion.