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derek35 derek35
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10 years ago
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?
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10 years ago
We have n = 1280, X = 575 and α = 1 – 0.95 = 0.05
Proportion, p = X/n = 575/1280 = 0.45
Using Z-tables,
Z (α/2) =Z (0.05/2) = Z (0.025) = 1.96
95% confidence interval for the proportion is given by:-
p ± Z x √ p (1 – p)/n
0.45 ± 1.96 x √0.45 x 0.55/1280
0.45 ± 1.96 x 0.0139
0.45 ± 0.0273
0.4227, 0.4773
We have n = 2560, X = 1150 and α = 1 – 0.95 = 0.05
Proportion, p = X/n = 1150/2560 = 0.45
Using Z-tables,
Z (α/2) =Z (0.05/2) = Z (0.025) = 1.96
95% confidence interval for the proportion is given by:-
p ± Z x √ p (1 – p)/n
0.45 ± 1.96 x √0.45 x 0.55/2560
0.45 ± 1.96 x 0.0098
0.45 ± 0.0193
0.4307, 0.4693
Increase in the sample size would decrease the width of the confidence interval.
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