Suppose the true proportion of voters in the county who support a new fire district is 0.42 . Consider the sampling distribution for the proportion of supporters with sample size $n=183$.
What is the mean of this distribution?
What is the standard error of this distribution?
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Final Answer: The mean of the distribution is \(\boxed{0.42}\) and the standard error of the distribution is approximately \(\boxed{0.0365}\).
Step 1 ：The problem is asking for the mean and standard error of the sampling distribution for the proportion of voters who support a new fire district.
Step 2 ：The mean of the sampling distribution of the proportion is equal to the true population proportion. In this case, the mean is \(0.42\).
Step 3 ：The standard error of the sampling distribution of the proportion is calculated using the formula: \[SE = \sqrt{ \frac{p(1-p)}{n} }\] where p is the population proportion and n is the sample size. In this case, p = 0.42 and n = 183.
Step 4 ：Substituting the given values into the formula, we get: \[SE = \sqrt{ \frac{0.42(1-0.42)}{183} }\]
Step 5 ：After calculating, we find that the standard error of the distribution is approximately 0.0365.
Step 6 ：Final Answer: The mean of the distribution is \(\boxed{0.42}\) and the standard error of the distribution is approximately \(\boxed{0.0365}\).
