Respuesta :
Answer:
A 95% confidence for the population proportion of defective items in the whole shipment is [0.075, 0.165] .
Step-by-step explanation:
We are given that for quality control purposes, we collect a sample of 200 items and find 24 defective items.
Firstly, the pivotal quantity for finding the confidence interval for the population proportion is given by;
P.Q. = [tex]\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] ~ N(0,1)
where, [tex]\hat p[/tex] = sample proportion of defective items = [tex]\frac{24}{200}[/tex] = 0.12
n = sample of items = 200
p = population proportion of defective items
Here for constructing a 95% confidence interval we have used a One-sample z-test statistics for proportions.
So, 95% confidence interval for the population proportion, p is ;
P(-1.96 < N(0,1) < 1.96) = 0.95 {As the critical value of z at 2.5% level
of significance are -1.96 & 1.96}
P(-1.96 < [tex]\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] < 1.96) = 0.95
P( [tex]-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] < [tex]{\hat p-p}[/tex] < [tex]1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] ) = 0.95
P( [tex]\hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] < p < [tex]\hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] ) = 0.95
95% confidence interval for p = [ [tex]\hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] , [tex]\hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] ]
= [ [tex]0.12-1.96 \times {\sqrt{\frac{0.12(1-0.12)}{200} } }[/tex] , [tex]0.12+1.96 \times {\sqrt{\frac{0.12(1-0.12)}{200} } }[/tex] ]
= [0.075, 0.165]
Therefore, a 95% confidence for the population proportion of defective items in the whole shipment is [0.075, 0.165] .
