Question

a senator wishes to estimate the proportion of united states voters who favor new road construction what size sample should be obtained in order to be 99% confident that the sample proportion will not differ from the true proportion by more than 4%? round up to the nearest whole number.

a. 17
b. 2,073
c. 1,037
d. 849

Explanation:

Step1: Determine the confidence level and margin of error

The confidence level is \(99\%\), so the significance level \(\alpha = 1 - 0.99=0.01\). Then \(\alpha/2 = 0.005\). The \(z\) - value \(z_{\alpha/2}\) corresponding to a probability of \(1-\alpha/2 = 0.995\) in the standard normal distribution is \(z_{0.005}= 2.576\) (from standard normal tables). The margin of error \(E = 0.04\).

Step2: Use the formula for sample size when estimating a proportion

When no prior estimate of the proportion \(p\) is given, we use \(p = 0.5\) (this maximizes the value of \(p(1 - p)\)). The formula for the sample size \(n\) is \(n=\dfrac{z_{\alpha/2}^{2}\times p(1 - p)}{E^{2}}\).
Substitute \(z_{\alpha/2}=2.576\), \(p = 0.5\), \(1-p = 0.5\) and \(E = 0.04\) into the formula:

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Step3: Round up the sample size

Since the sample size \(n\) must be an integer, and we want to ensure the margin of error requirement is met, we round up \(1036.84\) to \(n = 1037\).

Answer:

C. 1,037