The manager of the local county fair believes that no more than 30 percent of the adults in the county would object to a fee increase to attend the fair if it meant that better entertainment could be secured. To estimate the true proportion, he has selected a random sample of 200 adults. The manager will use a 90 percent confidence level. Assuming his assumption about the 30 percent holds, the margin of error for the estimate will be approximately ±.169.

Question

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The margin of error for a proportion at a given confidence level is calculated using the formula $ME = Z \times \sqrt{\frac{p(1-p)}{n}}$, where $Z$ is the Z-score corresponding to the confidence level, $p$ is the estimated proportion, and $n$ is the sample size. Given the manager's assumption of $p = 0.30$ and $n = 200$, and knowing that the Z-score for a 90% confidence level is approximately 1.645, the actual margin of error would be smaller than ±.169.

The Manager of the Local County Fair Believes That No