A simple random sample (SRS) of 150 is taken from a population with a 0.6 proportion of success. An independent SRS of 250 is ta

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A simple random sample (SRS) of 150 is taken from a population with a 0.6 proportion of success. An independent SRS of 250 is ta

ken from a population with a 0.3 proportion of success. What is the standard deviation of the sampling distribution for the difference in sample proportions?

Mathematics

2 answers:

Virty [35] · Answer 1 · 2022-02-02T23:33:58+03:00

Answer:

$SE_{\hat p_1 -\hat p_2}= \sqrt{\frac{0.6(1-0.6)}{150}+\frac{0.3(1-0.3)}{250}}= 0.0494$

Step-by-step explanation:

For this case we have the following data:

$n_1 =150$ represent the random sample 1 selected

$\hat p_1 =0.6$ represent the proportion of success for the sample 1 selected

$n_2 =250$ represent the random sample 2 selected

$\hat p_2 =0.3$ represent the proportion of success for the sample 2 selected

We know for this case that we can use the normal approximation since for both cases we have:

$n_1 p_1 =90 \geq 10, n_1 (1-p_1)=60 \geq 10$

$n_2 p_2 =75 \geq 10, n_2 (1-p_2)=175 \geq 10$

We have the randomization condition and we assume that the two samples are <10% of the entire population size.

So then we can use the following distribution for the proportions:

$p \sim N( \hat p, \sqrt{\frac{\hat p (1-\hat p)}{n}})$

For this case we want to find the distribution for the difference of these two proportions and we have this:

$p_1 -p_2 \sim N (\hat p_1 -\hat p_2 , \sqrt{\frac{\hat p_1 (1-\hat p_1)}{n_1} +\frac{\hat p_2 (1-\hat p_2)}{n_2}})$

So then the dtandard deviation would be given by:

$SE_{\hat p_1 -\hat p_2}= \sqrt{\frac{0.6(1-0.6)}{150}+\frac{0.3(1-0.3)}{250}}= 0.0494$

melamori03 [73] · Answer 2 · 2022-02-03T01:27:28+03:00

Answer:

0.0494

Step-by-step explanation:

Here, p1=0.6, n1=150 and p2=0.3, n2=250.

Standard deviation of the sampling distribution for the difference in sample proportions can be calculated as $\sqrt{\frac{p1q1}{n1}+\frac{p2q2}{n2} }$ .