In a recent poll of 600 homeowners in the United States, one in five homeowners
ID: 2922478 • Letter: I
Question
In a recent poll of 600 homeowners in the United States, one in five homeowners reports having a home equity loan that he or she is currently paying off. Using a confidence coefficient of 0.95, derive an interval estimate for the proportion of all homeowners in the United States that hold a home equity loan. Use Table 1. (Round intermediate calculations to 4 decimal places. Round "z-value" and final answers to 3 decimal places.)
In a recent poll of 600 homeowners in the United States, one in five homeowners reports having a home equity loan that he or she is currently paying off. Using a confidence coefficient of 0.95, derive an interval estimate for the proportion of all homeowners in the United States that hold a home equity loan. Use Table 1. (Round intermediate calculations to 4 decimal places. Round "z-value" and final answers to 3 decimal places.)
Explanation / Answer
TRADITIONAL METHOD
given that,
possibile chances (x)=120
sample size(n)=600
success rate ( p )= x/n = 0.2
I.
sample proportion = 0.2
standard error = Sqrt ( (0.2*0.8) /600) )
= 0.0163
II.B9
margin of error = Z a/2 * (stanadard error)
where,
Za/2 = Z-table value
level of significance, = 0.05
from standard normal table, two tailed z /2 =1.96
margin of error = 1.96 * 0.0163
= 0.032
III.
CI = [ p ± margin of error ]
confidence interval = [0.2 ± 0.032]
= [ 0.168 , 0.232]
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DIRECT METHOD
given that,
possibile chances (x)=120
sample size(n)=600
success rate ( p )= x/n = 0.2
CI = confidence interval
confidence interval = [ 0.2 ± 1.96 * Sqrt ( (0.2*0.8) /600) ) ]
= [0.2 - 1.96 * Sqrt ( (0.2*0.8) /600) , 0.2 + 1.96 * Sqrt ( (0.2*0.8) /600) ]
= [0.168 , 0.232]
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interpretations:
1. We are 95% sure that the interval [ 0.168 , 0.232] 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
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