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:

A) 0.0179
B) 0.0003
C) 0.1600
D) 0.0256
E) 0.1482

Question

Taryn Gutter-Spence · Accepted Answer

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

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