Question

a researcher wishes to estimate, with 90% confidence, the population proportion of likely u.s. voters who think congress is doing a good or excellent job. her estimate must be accurate within 1% of the true proportion. (a) no preliminary estimate is available. find the minimum sample size needed. (b) find the minimum sample size needed, using a prior study that found that 40% of the respondents said they think congress is doing a good or excellent job. (c) compare the results from parts (a) and (b). (a) what is the minimum sample size needed assuming that no prior information is available? n = 6766 (round up to the nearest whole number as needed.) (b) what is the minimum sample size needed using a prior study that found that 40% of the respondents said they think congress is doing a good or excellent job? n = (round up to the nearest whole number as needed.)

Explanation:

Step1: Determine the z - value

For a 90% confidence level, the significance level \(\alpha=1 - 0.90 = 0.10\), and \(\alpha/2=0.05\). The \(z\) - value \(z_{\alpha/2}=z_{0.05}\). From the standard normal table, \(z_{0.05} = 1.645\). The margin of error \(E = 0.01\).

Step2: Use the formula for sample size when a prior estimate \(p\) is available

The formula for the sample size \(n\) when estimating a proportion is \(n=\dfrac{z_{\alpha/2}^{2}\times p\times(1 - p)}{E^{2}}\). Given \(p = 0.40\) (from the prior study), \(z_{\alpha/2}=1.645\), and \(E = 0.01\).
Substitute the values into the formula:

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Answer:

\(n = 6495\)