Question

Suppose that 25% of all males in a population are carriers of a certain infectious disease \\( (p_1= 0.25) \\) and that 28% of all females in the population are carriers \\( (p_2= 0.28). \\) Researchers who don’t know these actual numbers are interested in estimating the difference by using random samples of 100 males and 100 females to find \\( \hat p_1−\hat p_2 \! \\) .
The standard deviation of the sampling distribution of \\( \hat p_1−\hat p_2 \\) is:

A)0.062.

B)0.004.

C)0.073.

Answer 1

Answer: A)0.062.

Step-by-step explanation:

Given : The proportion of ales in a population are carriers of a certain infectious disease : $p_1= 0.25$

The proportion of all females in the population are carriers : $p_2= 0.28$

Sample size of men : $n_1=100$

Sample size of women : $n_2=100$

The standard error of the sampling distribution of $\hat{p_1}-\hat{p_2}$ is given by :

$\sigma_{\hat{p_1}-\hat{p_2}}=\sqrt{\dfrac{\hat{p_1}(1-\hat{p_1})}{n_1}+\dfrac{p_2(1-p_2)}{n_2}}$

Substitute all value s, we get

$\sigma_{\hat{p_1}-\hat{p_2}}=\sqrt{\dfrac{0.25(1-0.25)}{100}+\dfrac{0.28(1-0.28)}{100}}$

$\sigma_{\hat{p_1}-\hat{p_2}}=\sqrt{\dfrac{0.25(0.75)}{100}+\dfrac{0.28(0.72)}{100}}$

$=\sqrt{0.001875+0.002016}$

$=\sqrt{0.003891}=0.062377880695\apprrox0.062$

Hence, the standard deviation of the sampling distribution of $\hat{p_1}-\hat{p_2}$ is 0.062.

Thus , the correct answer is A)0.062.