Suppose an actual census showed that 20% of the households in Michigan have incomes in excess of $60,000. Assume that a random sample of 500 households in Michigan is taken. Then, the standard error of the sampling distribution of sample proportion of households who have incomes in excess of $60,000 will be:

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Quizplus · Accepted Answer

The standard error of the sampling distribution of the sample proportion is calculated using the formula: $\sqrt{\frac{p(1-p)}{n}}$, where $p$ is the population proportion (0.20 in this case) and $n$ is the sample size (500). Plugging in the values, we get $\sqrt{\frac{0.20(1-0.20)}{500}} = \sqrt{\frac{0.20(0.80)}{500}} = \sqrt{\frac{0.16}{500}} = \sqrt{0.00032} = 0.0179$.

Suppose an Actual Census Showed That 20% of the Households