Supposed you are interested in examining the extent to which police look for evi
ID: 3336132 • Letter: S
Question
Supposed you are interested in examining the extent to which police look for evidence from social networking sites in their cases. You surveyed 150 officers in Florida and found 80% (0.80) of those surveyed indicated that they do use information from social networking sites to help build their cases (SEM is 0.38). Construct a 65 percent confidence interval AND a 90 percent confidence interval around this point estimate (10 points). Supposed you are interested in examining the extent to which police look for evidence from social networking sites in their cases. You surveyed 150 officers in Florida and found 80% (0.80) of those surveyed indicated that they do use information from social networking sites to help build their cases (SEM is 0.38). Construct a 65 percent confidence interval AND a 90 percent confidence interval around this point estimate (10 points).Explanation / Answer
Q1.
AT 65% CI,
TRADITIONAL METHOD
given that,
sample size(n)=150
success rate ( p )= x/n = 0.8
I.
sample proportion = 0.8
standard error = Sqrt ( (0.8*0.2) /150) )
= 0.033
II.
margin of error = Z a/2 * (stanadard error)
where,
Za/2 = Z-table value
level of significance, = 0.35
from standard normal table, two tailed z /2 =0.935
margin of error = 0.935 * 0.033
= 0.031
III.
CI = [ p ± margin of error ]
confidence interval = [0.8 ± 0.031]
= [ 0.769 , 0.831]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
possibile chances (x)=120
sample size(n)=150
success rate ( p )= x/n = 0.8
CI = confidence interval
confidence interval = [ 0.8 ± 0.935 * Sqrt ( (0.8*0.2) /150) ) ]
= [0.8 - 0.935 * Sqrt ( (0.8*0.2) /150) , 0.8 + 0.935 * Sqrt ( (0.8*0.2) /150) ]
= [0.769 , 0.831]
-----------------------------------------------------------------------------------------------
interpretations:
1. We are 65% sure that the interval [ 0.769 , 0.831] contains the true population proportion
2. If a large number of samples are collected, and a confidence interval is created
for each sample, 65% of these intervals will contains the true population proportion
Q2.
AT 90% CI
given that,
sample size(n)=150
success rate ( p )= x/n = 0.8
CI = confidence interval
confidence interval = [ 0.8 ± 1.645 * Sqrt ( (0.8*0.2) /150) ) ]
= [0.8 - 1.645 * Sqrt ( (0.8*0.2) /150) , 0.8 + 1.645 * Sqrt ( (0.8*0.2) /150) ]
= [0.746 , 0.854]
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.