Question

A research institute poll asked respondents if they felt vulnerable to identity theft. In the​ poll, n equals 942 and x equals 520 who said​ "yes." Use a 90 % confidence level.

A. Find the best point of estimate of the population of portion 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.

D. Write a statement that correctly interprets the confidence interval.

Answer 1

Solution :

Given that,

n = 942

x = 520

A.

Point estimate = sample proportion = = x / n = 520/942=0.552

1 - = 1-0.552=0.448

At 90% confidence level

= 1 - 90%  

= 1 - 0.90 =0.10

/2 = 0.05

Z/2 = Z_{0.05 = 1.645}

B.

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

= 1.645 (((0.552*0.448) / 942)

E = 0.027

C

A 90% confidence interval for population proportion p is ,

- E < p < + E

0.552-0.027 < p <0.552+0.027

0.525< p < 0.579

The 90% confidence interval for the population proportion p is : 0.525 , 0.579

D.:

LOWER LIMIT CI: 0.529

UPPER LIMIT CI : 0.579