Question

3. 22.13 with ove 50% of adults spending morethan an hour a day on the Internet, the number e adults, aged 18 and older, reached by random digt dialing found 113 aduits in the sample who said that they or a household member was a victim of a computer or teternet crime on their home computer In the past year e end g co uter or Internet based on e cont n es to rise. A survey in 2010 of a random sa ple of 1025 Step 1: eve the 95% larg sample crnderce interval for the popotion p of all households that heve experienced co puter or Inter et enering the year befor the aney was earaeed Answer L to Answe 2 (Round your answers to 4 decimal places) Step 2 ave the phas four 95% confidence interval for p, write answer in decimal form. Answer 1 to Answer 2 (Round your answers to 4 decimal places.)

Answer 1

Solution.

Step 1.

n=1025, x=113

the sample proportion and 95% confidence interval is given by:

1130.110 n 1025 21-p) 0.110(1 - 0.110) 1025 1.96 n_ = 0.110士1.961/

=$0.110 \pm 0.0191$

Here Z=critical value at 95% confidence interval=1.96(using standard normal table)

b.)using plus four

Sample Proportion=

$\hat{p}=\frac{x+2}{n+4}=\frac{113+2}{1025+4}=0.112$

p(1-p) 0.112t 1.96y + 1 6./0.112(1 -0.112) 1029

$=0.112\pm 0.01926$