Question

(22.13) With over 50% of adults spending more than an hour a day on the Internet, the number experiencing computer- or Internet-based crime continues to rise. A survey in 2010 of a random sample of 1019 adults, aged 18 and older, reached by random digit dialing found 102 adults in the sample who said that they or a household member was a victim of a computer or Internet crime on their home computer in the past year.

What is a 98 % large-sample confidence interval for the proportion p of all households that have experienced computer or Internet crime during the year before the survey was conducted?

The 98 % confidence interval (±±0.001) is from  to

When use plus four method the 98 % confidence interval (±±0.001) is from  to

Answer 1

Solution :

Given that,

n = 1019

x = 102

Point estimate = sample proportion = = x / n = 102/1019 = 0.100

1 - = 1-0.100 = 0.900

Z_/2 = 2.326

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

= 2.326* ((0.100*(0.900) /1019 )

= 0.022

A 98% confidence interval for population proportion p is ,

- E < p < + E

0.100 -0.022 < p < 0.100 +0.022

0.078< p < 0.122

The 98% confidence interval for the population proportion p is : 0.078,0.122