The 2004 General Social Survey found that 78.9% of the adult Americans interview
ID: 3154968 • Letter: T
Question
The 2004 General Social Survey found that 78.9% of the adult Americans interviewed claimed to have gien a financial contribution to charity in the previous 12 months. A 99% confidence interval for this poulation proportion of interest is (0.760,0.818)
(a) Interpret this confidence interval in the context of the problem.
(b) What's the margin of error for the confidence interval? What is the value of the sample proportion?
(c) Compare the margin of error and your formula. Solce for the sample size used to construct this confidence interval.
(d) A similar survey is to be done in 2013. The organization performing the survery wants to construct a 99% confidence interval for the new population proportion and guarantee that the margin of error is no more than 2.5%. Using the proportion found in 2004 as an estimate, how many people should be sampled?
Explanation / Answer
A 99% confidence interval for this poulation proportion of interest is (0.760,0.818)
(a) Interpret this confidence interval in the context of the problem.
We are 99% confident that the true proportion of population is between 0.760 and 0.818
(b) What's the margin of error for the confidence interval? What is the value of the sample proportion?
0.760 = 0.789 - margin of error
margin of error = 0.029
value of sample proportion is 0.789
(c) Compare the margin of error and your formula. Solce for the sample size used to construct this confidence interval.
0.789 +/- 0.029
0.76 < p < 0.818
(d) A similar survey is to be done in 2013. The organization performing the survery wants to construct a 99% confidence interval for the new population proportion and guarantee that the margin of error is no more than 2.5%. Using the proportion found in 2004 as an estimate, how many people should be sampled?
alpha / 2 = 0.005 Z=2.57
n = ( 2.57^2 * 0.789 * 0.211 / 0.025^2 )
n = 1759.32
n = 1760
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