Answer: (-0.620, 0.220)
Step-by-step explanation:
The formula to find 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.
Let first group be "group of prototypes by first process " and second group be "group of prototypes by second process".
[tex]p_1[/tex] = Proportion of defectives in the first batch.
[tex]p_2[/tex] = Proportion of defectives in the second batch.
From question , we have
[tex]n_1=10[/tex] , [tex]n_2=10[/tex] , [tex]\hat{p}_1=\dfrac{3}{10}=0.3[/tex] , [tex]\hat{p}_2=\dfrac{5}{10}=0.5[/tex]
By z-table , Critical value for 95% confidence interval is z* =1.96.
Substitute all values in formula , we get
[tex]0.3-0.5\pm (1.96)\sqrt{\dfrac{0.3(1-0.3)}{10}+\dfrac{0.5(1-0.5)}{10}}[/tex]
[tex]-0.2\pm (1.96)\sqrt{0.021+0.025}[/tex]
[tex]-0.2\pm (1.96)\sqrt{0.046}[/tex]
[tex]-0.2\pm (1.96)(0.21448)[/tex]
[tex]-0.2\pm 0.4203808[/tex]
[tex]=(-0.2-0.4203808,\ -0.2+0.4203808)\\\\=(-0.6203808,\ 0.2203808)\\\\\approx(-0.620,\ 0.220)[/tex]
Hence, a 95% confidence interval for the difference in the proportion of defectives is (-0.620, 0.220).
