Answer :
Answer: (0.186, 0.298)
Step-by-step explanation:
The formula to find the confidence interval for the difference of the population proportion is given by :-
[tex]\hat{p}_1-\hat{p}_2\pm z^*\sqrt{\dfrac{\hat{p}_1(1-\hat{p}_1)}{n_1}+\dfrac{\hat{p}_2(1-\hat{p}_2)}{n_2}}[/tex] ... (i)
, where [tex]n_1=[/tex] Sample size of population 1.
[tex]n_2=[/tex] Sample size of population 2.
[tex]\hat{p_1}=[/tex] Sample proportion of population 1.
[tex]\hat{p}_2[/tex] = Sample proportion of population 2.
z* = Critical z-value corresponding to confidence interval
As per given , we have
[tex]n_1=n_2=500[/tex]
[tex]\hat{p}_1=\dfrac{378}{500}=0.756\\\hat{p}_2=\dfrac{256}{500}=0.512[/tex]
Critical value corresponds to 95% confidence interval = 1.96
Put all these values , in (i) , we get
[tex]0.756-0.512\pm 1.96\sqrt{\dfrac{0.756(1-0.756)}{500}+\dfrac{0.512(1-0.512)}{500}}\\\\=0.244\pm1.96(\sqrt{0.00086864})\\\\=0.244\pm1.96(0.0294727)\\\\=0.244\pm0.0577665\\\\=(0.244-0.0577664,\ 0.24+0.0577664)\\\\=(0.1862336,\ 0.2977664)\approx(0.186,\ 0.298)[/tex]
Hence, the 95% confidence interval for the difference of the population proportions= (0.186, 0.298)