Political polls typically sample randomly from the U.S population to investigate the percentage of voters who favor some candidate or issue. The number of people polled is usually on the order of 1000. Suppose that one such poll asks voters how they feel about the President’s handling of the crisis in the financial markets. The results show that 575 out of the 1280 people polled say they either “approve” or “strongly approve” of the President’s handling of this matter. Based on the sample referenced above, find a 95% confidence interval estimate for the proportion of the entire voter population who “approve” or “strongly approve” of the President’s handling of the crisis in the financial markets.
Now, here’s an interesting twist. If the same sample proportion was found in a sample twice as large—that is, 1150 out of 2560—how would this affect the confidence interval?
AnswerSince our sample is large (1280), we can use the normal approximation for the confidence limits. The formula is
p ± zSQRT[p(1- p)/n]
where p is the observed proportion (575/1280 = 0.44921875), n is the sample size (1280), and z is the value from a standard normal distribution for 95% confidence (z = 1.96). So our interval is
0.44921875 ± 1.96SQRT[(0.44921875)(0.55078125)/1280]
= 0.44921875 ± 1.96SQRT0.000193298
= 0.44921875 ± 1.96(0.01390316)
= 0.44921875 ± 0.027250194
= (0.421968556, 0.476468944)
answer
If we multiply the limits by 1280, the interval is roughly
(540/1280, 610/1280)
For the second part, it's easy to see from the formula for the interval,
p ± zSQRT[p(1- p)/n]
that p and z will stay the same, and n will be twice as large, so the new interval will be shorter by a factor of 1/SQRT2 = 0.707. In other words, the new interval will be roughly 71% as long as the original interval.
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