Question

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

A research institute poll asked respondents if they felt vulnerable to identity theft. In the poll, n=1072 and x=569 who said​ "yes." Use a 95% confidence level.

​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 95​% chance that the true value of the population proportion will fall between the lower bound and the upper bound.

B. One has 95​% confidence that the sample proportion is equal to the population proportion.

C. 95​% of sample proportions will fall between the lower bound and the upper bound.

D. One has 95​% 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 = 1072

x = 569

(a)

Point estimate = sample proportion = = x / n = 569 / 1072 = 0.531

1 - = 1 - 0.531 = 0.469

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

(b)

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

= 1.96 * (((0.531 * 0.469) / 1072)

= 0.030

Margin of error = E = 0.030

(c)

A 95% confidence interval for population proportion p is ,

- E < p < + E

0.531 - 0.030 < p < 0.531 + 0.030

0.501 < p < 0.561

D. One has 95​% confidence that the interval from the lower bound to the upper bound actually does contain the true value of the population proportion.