Question

A random sample of 100 voters found that 46% were going to vote for a certain candidate. Find the 90% confidence interval for the population proportion of voters who will vote for that candidate.

A. 38.7% < p < 53.3%

B. 37.8% < p < 54.2%

C. 41.9% < p < 50.1%

D. 39.6% < p < 52.4%

Answer 1

TRADITIONAL METHOD
given that,
possible chances (x)=46
sample size(n)=100
success rate ( p )= x/n = 0.46
I.
sample proportion = 0.46
standard error = Sqrt ( (0.46*0.54) /100) )
= 0.05
II.
margin of error = Z a/2 * (standard error)
where,
Za/2 = Z-table value
level of significance, α = 0.1
from standard normal table, two tailed z α/2 =1.645
margin of error = 1.645 * 0.05
= 0.082
III.
CI = [ p ± margin of error ]
confidence interval = [0.46 ± 0.082]
= [ 0.378 , 0.542]
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DIRECT METHOD
given that,
possible chances (x)=46
sample size(n)=100
success rate ( p )= x/n = 0.46
CI = confidence interval
confidence interval = [ 0.46 ± 1.645 * Sqrt ( (0.46*0.54) /100) ) ]
= [0.46 - 1.645 * Sqrt ( (0.46*0.54) /100) , 0.46 + 1.645 * Sqrt ( (0.46*0.54) /100) ]
= [0.378 , 0.542]
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interpretations:
1. We are 90% sure that the interval [ 0.378 , 0.542] contains the true population proportion
2. If a large number of samples are collected, and a confidence interval is created
for each sample, 90% of these intervals will contains the true population proportion
90% sure that the interval [ 0.378 , 0.542] = 37.8%,54.2%
option:B