Question

The U.S. federal ban on assault weapons expired in September 2004, which meant that after 10 years (since the ban was instituted in 1994) there were certain types of guns that could be manufactured legally again. A poll asked a random sample of 1,200 eligible voters (among other questions) whether they were satisfied with the fact that the law had expired. The datafile linked below contains the results of this poll (Data were generated based on a poll conducted by NBC News/Wall Street Journal Poll). We would like to estimate p, the proportion of U.S. eligible voters who were satisfied with the expiration of the law, with a 95% confidence interval.

M.E.= 2.9%

(a) How many of the 1,200 sampled voters were satisfied?

(b) What is the sample proportion (p-hat) of those who were satisfied?

(c) What is the 95% confidence interval for p? Interpret this interval.

In question 1 you found that the margin of error of this poll was about 2.9%. What is the margin of error of the confidence interval you found in question 2?

Answer 1

a.
Compute Sample Size ( n ) = n=(Z/E)^2*p*(1-p)
Z a/2 at 0.05 is = 1.96
Sample Proportion = 0.5
ME = 0.029
n = ( 1.96 / 0.029 )^2 * 0.5*0.5
= 1141.974 ~ 1142          
b.
possible chances (x)=1142
sample size(n)=1200
success rate ( p )= x/n = 0.952
c.
TRADITIONAL METHOD
given that,
possible chances (x)=1142
sample size(n)=1200
success rate ( p )= x/n = 0.952
I.
sample proportion = 0.952
standard error = Sqrt ( (0.952*0.048) /1200) )
= 0.006
II.
margin of error = Z a/2 * (standard error)
where,
Za/2 = Z-table value
level of significance, α = 0.05
from standard normal table, two tailed z α/2 =1.96
margin of error = 1.96 * 0.006
= 0.012
III.
CI = [ p ± margin of error ]
confidence interval = [0.952 ± 0.012]
= [ 0.94 , 0.964]
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DIRECT METHOD
given that,
possible chances (x)=1142
sample size(n)=1200
success rate ( p )= x/n = 0.952
CI = confidence interval
confidence interval = [ 0.952 ± 1.96 * Sqrt ( (0.952*0.048) /1200) ) ]
= [0.952 - 1.96 * Sqrt ( (0.952*0.048) /1200) , 0.952 + 1.96 * Sqrt ( (0.952*0.048) /1200) ]
= [0.94 , 0.964]
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interpretations:
1. We are 95% sure that the interval [ 0.94 , 0.964] contains the true population proportion
2. If a large number of samples are collected, and a confidence interval is created
for each sample, 95% of these intervals will contains the true population proportion
Answer:
marin of error is 0.012