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In a recent poll of 200 households, it was found that 152 had at least one...

In a recent poll of 200 households, it was found that 152 had at least one computer and one television. A 95% confidence interval to estimate the population proportion was calculated to be .701 to .819. What sample size would be needed to change the margin of error to .030?

Math   Statistics-And-Probability   
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Answer #1

Solution :

Given that,

n = 200

x = 152

= x / n = 152 / 200 = 0.76

1 - = 1 - 0.76 = 0.24

Margin of error = E = 0.030

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z/2 = Z0.025 = 1.96

Sample size = (  Z/2 / E)2 * * (1 - )

= (1.96 / 0.030)2 * 0.76 * 0.24

= 778.56 = 779

Sample size = n = 779

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