Respuesta :
Answer:
[tex] SE_{\hat p_1 -\hat p_2}= \sqrt{\frac{0.6(1-0.6)}{150}+\frac{0.3(1-0.3)}{250}}= 0.0494[/tex]
Step-by-step explanation:
For this case we have the following data:
[tex] n_1 =150[/tex] represent the random sample 1 selected
[tex] \hat p_1 =0.6[/tex] represent the proportion of success for the sample 1 selected
[tex] n_2 =250[/tex] represent the random sample 2 selected
[tex] \hat p_2 =0.3[/tex] 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:
[tex] n_1 p_1 =90 \geq 10, n_1 (1-p_1)=60 \geq 10[/tex]
[tex] n_2 p_2 =75 \geq 10, n_2 (1-p_2)=175 \geq 10[/tex]
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:
[tex] p \sim N( \hat p, \sqrt{\frac{\hat p (1-\hat p)}{n}})[/tex]
For this case we want to find the distribution for the difference of these two proportions and we have this:
[tex] 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}}) [/tex]
So then the dtandard deviation would be given by:
[tex] SE_{\hat p_1 -\hat p_2}= \sqrt{\frac{0.6(1-0.6)}{150}+\frac{0.3(1-0.3)}{250}}= 0.0494[/tex]
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 [tex]\sqrt{\frac{p1q1}{n1}+\frac{p2q2}{n2} }[/tex].
q1=1-p1=1-0.6=0.4
q2=1-p2=1-0.3=0.7
[tex]S.D(p1-p2)=\sqrt{\frac{(0.6)(0.4)}{150}+\frac{(0.3)(0.7)}{250} }[/tex]
[tex]S.D(p1-p2)=\sqrt{\frac{0.24}{150}+\frac{0.21}{250} }[/tex]
[tex]S.D(p1-p2)=\sqrt{0.0016+0.00084 }[/tex]
[tex]S.D(p1-p2)=\sqrt{0.00244 }[/tex]
S.D(p1-p2)=0.049396
Rounding to four decimal places
S.D(p1-p2)=0.0494
The Standard deviation of the sampling distribution for the difference in sample proportions is 0.0494