Respuesta :
Answer: b) [-0.080, 0.060]
Step-by-step explanation:
The confidence interval for the difference in true proportion of the two groups. 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]
, where [tex]n_1[/tex] = Sample size for first group.
[tex]n_2[/tex] = Sample size for second group.
[tex]\hat{p}_1[/tex] = Sample proportion for first group.
[tex]\hat{p}_2[/tex] = Sample proportion for second group.
z* = critical z-value.
As per given , we have
[tex]n_1=118[/tex] , [tex]n_2=127[/tex] , [tex]\hat{p}_1=0.12[/tex] , [tex]\hat{p}_2=0.13[/tex]
Critical value for 90% confidence interval is 1.645. (By z-table)
Substitute all values in formula , we get
[tex]0.12-0.13\pm (1.645) \sqrt{\dfrac{0.12(1-0.12)}{118}+\dfrac{0.13(1-0.13)}{127}}[/tex]
[tex]=-0.01\pm (1.645) \sqrt{0.000895+0.000890}[/tex]
[tex]=-0.01\pm (1.645) \sqrt{0.001785}[/tex]
[tex]=-0.01\pm (1.645) (0.04225)[/tex]
[tex]\approx-0.01\pm 0.070[/tex]
[tex]=(-0.01-0.070,\ -0.01+0.070)=(-0.080,\ 0.060)[/tex]
Hence, the 90% confidence interval for the difference in true proportion of the two groups. is (-0.080, 0.060)
Hence, the correct answer is b) [-0.080, 0.060]