Questions: Suppose that a particular candidate for public office is favored by 48% of all registered voters in the district. A polling organization will take a random sample of 500 of these voters and will use p̂, the sample proportion, to estimate p. a. Calculate σp̂, the standard deviation of p̂. σp̂=0.48 b. If for a different sample size, σp̂=0.071, would you expect more or less sample-to-sample variability in the sample proportions than when n=500 ? c. Is the sample size that resulted in σp̂=0.071 larger than 500 or smaller than 500 ? Explain your reasoning.

Solution Steps

To find the standard deviation of the sample proportion \( \sigma_{\widehat{p}} \), we use the formula:

\[ \sigma_{\widehat{p}} = \sqrt{\frac{p(1 - p)}{n}} \]

Substituting the values \( p = 0.48 \) and \( n = 500 \):

\[ \sigma_{\widehat{p}} = \sqrt{\frac{0.48 \times (1 - 0.48)}{500}} = \sqrt{\frac{0.48 \times 0.52}{500}} \approx 0.0223 \]

For a different sample size, we have \( \sigma_{\widehat{p}} = 0.071 \). We compare this with the previously calculated \( \sigma_{\widehat{p}} \):

Since \( 0.071 > 0.0223 \), we conclude:

\[ \text{I would expect more sample-to-sample variability in the sample proportions because } \sigma_{\widehat{p}} \text{ is greater than } \sigma_{\widehat{p}} \text{ when } n = 500. \]

To determine if the sample size that resulted in \( \sigma_{\widehat{p}} = 0.071 \) is larger or smaller than 500, we note that:

The standard deviation of the sample proportion is inversely related to the sample size \( n \). Therefore, if \( \sigma_{\widehat{p}} \) increases, the sample size must decrease. Since \( 0.071 > 0.0223 \), we conclude:

\[ \text{The sample size should be smaller than 500 because in order for } \sigma_{\widehat{p}} \text{ to be greater, its denominator, value of } n, \text{ should be smaller.} \]

Final Answer