A simple random sample of 60 households in City 1 is taken.In the sample,there are 45 households that decorate their houses with lights for the holidays.A simple random sample of 50 households is also taken from the neighboring City 2.In the sample,there are 40 households that decorate their houses.We wish to estimate the difference in proportions of households that decorate their houses with lights for the holidays,with a 95% confidence interval.Under the null hypothesis of equality,of the proportions of households that decorate their houses in the two neighboring towns,what is the standard error of the difference in sample proportions?
A)0.040
B)0.0795
C)0.0802
D)0.1125

Question

Quizplus · Accepted Answer

The standard error of the difference in sample proportions can be calculated as $SE = \sqrt{\frac{p_1(1-p_1)}{n_1}+\frac{p_2(1-p_2)}{n_2}}$, where $p_1$ and $p_2$ are the sample proportions, and $n_1$ and $n_2$ are the sample sizes. Plugging in the values, we get $SE=\sqrt{\frac{0.75(0.25)}{60}+\frac{0.8(0.2)}{50}}\approx0.0802$. Therefore, the answer is (C).

A Simple Random Sample of 60 Households in City 1