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A newspaper conducted a statewide survey concerning a proposal to raise taxes to prevent budget cuts to education. The newspaper took a random sample (assume it is an SRS) of 1200 registered voters and found that 580 would vote to raise taxes. Let p represent the proportion of registered voters in the state that would vote to raise taxes.Reference: Ref 20-4How large a sample n would you need to estimate p with margin of error 0.01 with 90% confidence

User AtliB
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Answer:

A sample of 6758 would be needed.

Explanation:

In a sample with a number n of people surveyed with a probability of a success of
\pi, and a confidence level of
1-\alpha, we have the following confidence interval of proportions.


\pi \pm z\sqrt{(\pi(1-\pi))/(n)}

In which

z is the zscore that has a pvalue of
1 - (\alpha)/(2).

The margin of error is:


M = z\sqrt{(\pi(1-\pi))/(n)}

The newspaper took a random sample (assume it is an SRS) of 1200 registered voters and found that 580 would vote to raise taxes.

This means that
\pi = (580)/(1200) = 0.4833

90% confidence level

So
\alpha = 0.1, z is the value of Z that has a pvalue of
1 - (0.1)/(2) = 0.95, so
Z = 1.645.

How large a sample n would you need to estimate p with margin of error 0.01 with 90% confidence?

This is n for which M = 0.01. So


M = z\sqrt{(\pi(1-\pi))/(n)}


0.01 = 1.645\sqrt{(0.4833*0.5167)/(n)}


0.01√(n) = 1.645√(0.4833*0.5167)


√(n) = (1.645√(0.4833*0.5167))/(0.01)


(√(n))^2 = ((1.645√(0.4833*0.5167))/(0.01))^2


n = 6757.5

Rounding up:

A sample of 6758 would be needed.

User Gavy
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