A Pew Research Center survey of 800 adults nationwide conducted December 6–19, 2
ID: 3132457 • Letter: A
Question
A Pew Research Center survey of 800 adults nationwide conducted December 6–19, 2011, found that 39% of 18- to 34-year-olds lived with their parents or had moved back in temporarily because of the economy in the past few years. On the basis of this sample, we want to estimate (at the time of the survey) the proportion p in the population of all 18- to 34-year-olds in the United States who lived with their parents or had moved back in temporarily because of the economy in the past few years. Are the conditions for inference met? The Pew Research Center sample was not an SRS but is described as a "nationally representative sample." We proceed with some caution, but will treat the Pew Research Center sample as an SRS from the population of all 18- to 34-year-olds in the United States. The number of "successes" in the sample is npˆ=800 0.39=312, and the number of "failures" is n(1pˆ)=488. Both are much larger than 10, so we regard the sample size as sufficiently large for using the Normal approximation.
1) Give three confidence intervals (±0.0001) for the proportion p of all 18- to 34-year-olds in the United States who lived with their parents or had moved back in temporarily because of the economy in the past few years, using 90%, 95%, and 99% confidence.
2) What are the margins of error (±0.0001) for 90%, 95%, and 99% confidence?
Explanation / Answer
1.
FOR 90% CONFIDENCE:
Note that
p^ = point estimate of the population proportion = x / n = 0.39
Also, we get the standard error of p, sp:
sp = sqrt[p^ (1 - p^) / n] = 0.017244564
Now, for the critical z,
alpha/2 = 0.05
Thus, z(alpha/2) = 1.644853627
Thus,
Margin of error = z(alpha/2)*sp = 0.028364784
lower bound = p^ - z(alpha/2) * sp = 0.361635216
upper bound = p^ + z(alpha/2) * sp = 0.418364784
Thus, the confidence interval is
( 0.361635216 , 0.418364784 ) [ANSWER]
********************
FOR 95% CONFIDENCE:
Note that
p^ = point estimate of the population proportion = x / n = 0.39
Also, we get the standard error of p, sp:
sp = sqrt[p^ (1 - p^) / n] = 0.017244564
Now, for the critical z,
alpha/2 = 0.025
Thus, z(alpha/2) = 1.959963985
Thus,
Margin of error = z(alpha/2)*sp = 0.033798725
lower bound = p^ - z(alpha/2) * sp = 0.356201275
upper bound = p^ + z(alpha/2) * sp = 0.423798725
Thus, the confidence interval is
( 0.356201275 , 0.423798725 ) [ANSWER]
*******************
FOR 99% CONFIDENCE:
Note that
p^ = point estimate of the population proportion = x / n = 0.39
Also, we get the standard error of p, sp:
sp = sqrt[p^ (1 - p^) / n] = 0.017244564
Now, for the critical z,
alpha/2 = 0.005
Thus, z(alpha/2) = 2.575829304
Thus,
Margin of error = z(alpha/2)*sp = 0.044419054
lower bound = p^ - z(alpha/2) * sp = 0.345580946
upper bound = p^ + z(alpha/2) * sp = 0.434419054
Thus, the confidence interval is
( 0.345580946 , 0.434419054 ) [ANSWER]
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2.
90%:
Margin of error = z(alpha/2)*sp = 0.028364784
95%
Margin of error = z(alpha/2)*sp = 0.033798725
99%:
Margin of error = z(alpha/2)*sp = 0.044419054
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