Question

a pollster wishes to estimate the proportion of united states voters who favor capital punishment. how large a sample is needed in order to be 95% confident that the sample proportion will not differ from the true proportion by more than 5%?

Answer 1

The sample size must be 543 so the sample proportion is will not differ by 5%

We are given the following in the question:

Margin of error = 5%

Confidence interval:

p ±z√(p'(1 - p')/n)

Margin of error =

p = ±z√(p'(1 - p')/n)

Since no particular proportion is given, we take

p' = 0.5

Z critical at α 0.02 is ± 2.33

Putting values, we get,

p = ±z√(p'(1 - p')/n)

2.33 x √(0.5(1 - 0.5)/n)

√n = 2.33x 0.5/0.05

n = 542.89 = 543

Thus, the sample size must be 543 so the sample proportion is will not differ by 5%

To learn more about confidence interval refer here

brainly.com/question/15712887

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