Question

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 1020 adults, aged 18 and older, reached by random digit dialing found 122 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 90 % 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? Using p-hat rounded to 4 decimals:

The 90 % confidence interval (±±0.001) is from  to

When use plus four method the 90 % confidence interval (±±0.001) is from  to

Answer 1

Solution :

Given that,

n = 1020

x = 122

Point estimate = sample proportion = = x / n = 122/ 1020 = 0.1196

1 - = 1 - 0.1196 = 0.8804

At 90% confidence level

= 1-0.90% =1-0.90 =0.10

/2 =0.10/ 2= 0.05

Z/2 = Z_{0.05 = 1.645}

Z_/2 = 1.645

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

= 1.645 * ((0.1196*(0.8804) /1020 )

= 0.017

A 90% confidence interval for population proportion p is ,

- E < p < + E

0.1196 - 0.017 < p < 0.1196 + 0.017

0.103< p < 0.126

The 90% confidence interval for the population proportion p is : 0.103 , 0.126