Answer: A)0.062.
Step-by-step explanation:
Given : The proportion of ales in a population are carriers of a certain infectious disease : [tex]p_1= 0.25[/tex]
The proportion of all females in the population are carriers : [tex]p_2= 0.28[/tex]
Sample size of men : [tex]n_1=100[/tex]
Sample size of women : [tex]n_2=100[/tex]
The standard error of the sampling distribution of [tex]\hat{p_1}-\hat{p_2}[/tex] is given by :
[tex]\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}}[/tex]
Substitute all value s, we get
[tex]\sigma_{\hat{p_1}-\hat{p_2}}=\sqrt{\dfrac{0.25(1-0.25)}{100}+\dfrac{0.28(1-0.28)}{100}}[/tex]
[tex]\sigma_{\hat{p_1}-\hat{p_2}}=\sqrt{\dfrac{0.25(0.75)}{100}+\dfrac{0.28(0.72)}{100}}[/tex]
[tex]=\sqrt{0.001875+0.002016}[/tex]
[tex]=\sqrt{0.003891}=0.062377880695\apprrox0.062[/tex]
Hence, the standard deviation of the sampling distribution of [tex]\hat{p_1}-\hat{p_2}[/tex] is 0.062.
Thus , the correct answer is A)0.062.
According to the proportions given, the standard error of the sampling distribution of the differences of the proportions is given by:
A) 0.062.
For each sample, the standard errors are:
[tex]s_1 = \sqrt{\frac{0.25(0.75)}{100}} = 0.0433[/tex]
[tex]s_2 = \sqrt{\frac{0.28(0.72)}{100}} = 0.0449[/tex]
For the distribution of differences, it is the square root of the sum of the standard errors of each sample squared, hence:
[tex]s = \sqrt{s_1^2 + s_2^2}[/tex]
[tex]s = \sqrt{0.0433^2 + 0.0449^2}[/tex]
[tex]s = 0.062[/tex]
Option A.
A similar problem is given at https://brainly.com/question/16695444
