The Consumer Reports National Research Center conducted a telephone about the ma

Question

The Consumer Reports National Research Center conducted a telephone about the major economic concerns for the future. The survey results showed that 1640 of the respondents think the future health of Social Security is a major economic concern. survey of 2100 adults to learn Develop a 90% confidence interval for the population proportion of adults who think the future health of Social Security is a major economic concern. b. Develop a 93% confidence interval for this population proportion. Should the Social Security department assume that a majority of adults feel that the future Social Security is a major concern? Explain.

Explanation / Answer

PART A.
TRADITIONAL METHOD
given that,
possibile chances (x)=1640
sample size(n)=2100
success rate ( p )= x/n = 0.781
I.
sample proportion = 0.781
standard error = Sqrt ( (0.781*0.219) /2100) )
= 0.009
II.
margin of error = Z a/2 * (stanadard error)
where,
Za/2 = Z-table value
level of significance, = 0.1
from standard normal table, two tailed z /2 =1.645
margin of error = 1.645 * 0.009
= 0.0148
III.
CI = [ p ± margin of error ]
confidence interval = [0.781 ± 0.0148]
= [ 0.7661 , 0.7958]
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DIRECT METHOD
given that,
possibile chances (x)=1640
sample size(n)=2100
success rate ( p )= x/n = 0.781
CI = confidence interval
confidence interval = [ 0.781 ± 1.645 * Sqrt ( (0.781*0.219) /2100) ) ]
= [0.781 - 1.645 * Sqrt ( (0.781*0.219) /2100) , 0.781 + 1.645 * Sqrt ( (0.781*0.219) /2100) ]
= [0.7661 , 0.7958]
-----------------------------------------------------------------------------------------------
interpretations:
1. We are 90% sure that the interval [ 0.7661 , 0.7958] 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

PART B.
AT 93% CI
given that,
possibile chances (x)=1640
sample size(n)=2100
success rate ( p )= x/n = 0.781
CI = confidence interval
confidence interval = [ 0.781 ± 1.812 * Sqrt ( (0.781*0.219) /2100) ) ]
= [0.781 - 1.812 * Sqrt ( (0.781*0.219) /2100) , 0.781 + 1.812 * Sqrt ( (0.781*0.219) /2100) ]
= [0.7646 , 0.7973]

PART C.
Yes, since the lower confidence has the percentage above 50%

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