Researchers found that 16.7% of Americans age 12 or older had been a victim of c
ID: 3276206 • Letter: R
Question
Researchers found that 16.7% of Americans age 12 or older had been a victim of crime. The size of the sample was 144,400.
a) Estimate the percentage of Americans age 12 or older who were victims at the 95% confidence level. State in words the meaning of that result.
b) Estimate the percentage of victims at the 99% confidence level. Imagine that the Sample size was cut in half but the survey found the same value of 16.7% for the percentage of victims.
c) By how much would the 95% confidence level increase?
d) By how much would the 99% confidence interval increase?
Please write clearly and legibly
Explanation / Answer
(a)
n = 144400
p = 0.167
% = 95
Standard Error, SE = {p(1 - p)/n} = (0.167(1 - 0.167))/144400 = 0.000981515
z- score = 1.959963985
Width of the confidence interval = z * SE = 1.95996398454005 * 0.000981515448771585 = 0.00192373
Lower Limit of the confidence interval = P - width = 0.167 - 0.00192373492986198 = 0.16507627
Upper Limit of the confidence interval = P + width = 0.167 + 0.00192373492986198 = 0.16892373
The 95% confidence interval is [0.165, 0.169]
This means the true proportion lies within the above interval 95% of the time
(b)
n = 144400
p = 0.167
% = 99
Standard Error, SE = {p(1 - p)/n} = (0.167(1 - 0.167))/144400 = 0.000981515
z- score = 2.575829304
Width of the confidence interval = z * SE = 2.57582930354892 * 0.000981515448771585 = 0.00252822
Lower Limit of the confidence interval = P - width = 0.167 - 0.00252821625483181 = 0.16447178
Upper Limit of the confidence interval = P + width = 0.167 + 0.00252821625483181 = 0.16952822
The 99% confidence interval is [0.164, 0.170]
(c)
n = 72200
p = 0.167
% = 95
Standard Error, SE = {p(1 - p)/n} = (0.167(1 - 0.167))/72200 = 0.001388072
z- score = 1.959963985
Width of the confidence interval = z * SE = 1.95996398454005 * 0.00138807245933149 = 0.00272057
Lower Limit of the confidence interval = P - width = 0.167 - 0.00272057202822166 = 0.16427943
Upper Limit of the confidence interval = P + width = 0.167 + 0.00272057202822166 = 0.16972057
The 95% confidence interval is [0.164, 0.170]
The confidence interval increased by 0.003 - 0.0019 = 0.0011
(d)
n = 72200
p = 0.167
% = 99
Standard Error, SE = {p(1 - p)/n} = (0.167(1 - 0.167))/72200 = 0.001388072
z- score = 2.575829304
Width of the confidence interval = z * SE = 2.57582930354892 * 0.00138807245933149 = 0.00357544
Lower Limit of the confidence interval = P - width = 0.167 - 0.00357543771619526 = 0.16342456
Upper Limit of the confidence interval = P + width = 0.167 + 0.00357543771619526 = 0.17057544
The confidence interval increased by 0.0035 - 0.0025 = 0.0010
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