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.

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:

For the distribution of differences, it is the square root of the sum of the standard errors of each sample squared, hence:

Option A.

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A)0.062.

Step-by-step explanation:

Given : The proportion of ales in a population are carriers of a certain infectious disease :

The proportion of all females in the population are carriers :

Sample size of men :

Sample size of women :

The standard error of the sampling distribution of is given by :

Substitute all value s, we get

Hence, the standard deviation of the sampling distribution of is 0.062.

Thus , the correct answer is A)0.062.

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step-by-step explanation: