Question

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 =928 and x=581 who said​ "yes." Use a 99% confidence level.

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

B. Identify the value of the margin of error E. (round to five decimal places as needed.)

C. Construct the confidence interval. _ < p <_ round to four decimal places.

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

Answer 1

Solution :

Given that,

n = 9258

x = 581

= x / n = 581 / 928 = 0.626

1 - = 1 - 0.626 = 0.374

At 99% confidence level the z is ,

  = 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z_/2 = Z_0.005 = 2.576

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

= 1.96 * (((0.626 * 0.374) / 928)

= 0.0409

A 99% confidence interval for population proportion p is ,

- E < P < + E

0.626 - 0.0409 < p < 0.626 + 0.0409

0.5851 < p < 0.6669

The 95% confidence interval for the population proportion p is : (0.5851 , 0.6669)