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 equals 1082n=1082 and x equals 502x=502 who said​ "yes." Use a 99 %99% confidence level

Answer 1

Solution:

Given:

n = total number of respondents = 1082

x = Number of respondents felt vulnerable to identity theft = 502

c = confidence level = 99%

We have to find 99% confidence interval for population proportion of people who felt vulnerable to identity theft.

Thus point estimate of true population proportion of people who felt vulnerable to identity theft is :

Formula for confidence interval for proportion is:

where

Z_c is z critical value for c = 0.99 confidence level.

Find Area = ( 1+c)/2 = ( 1 + 0.99 ) / 2 = 1.99 /2 = 0.9950

Thus look in z table for Area = 0.9950 or its closest area and find corresponding z critical value.

From above table we can see area 0.9950 is in between 0.9949 and 0.9951 and both are at same distance from 0.9950, Hence corresponding z values are 2.57 and 2.58

Thus average of both z values is 2.575

Thus Z_c = 2.575

Thus margin of error is :

Thus limits of confidence interval are:

Thus we are 99% confident that population proportion of people who felt vulnerable to identity theft is between the limits: