Respuesta :
Answer:
A 95% confidence interval for the population mean is [3315.13, 22480.87] .
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 mean is given by;
P.Q. = [tex]\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }[/tex] ~ [tex]t_n_-_1[/tex]
where, [tex]\bar X[/tex] = sample proportion of defective items = 12,898
s = sample standard deviation = 7,719
n = sample size = 5
[tex]\mu[/tex] = population mean
Here for constructing a 95% confidence interval we have used a One-sample t-test statistics as we don't know about population standard deviation.
So, 95% confidence interval for the population mean, [tex]\mu[/tex] is ;
P(-2.776 < [tex]t_4[/tex] < 2.776) = 0.95 {As the critical value of t at 4 degrees of
freedom are -2.776 & 2.776 with P = 2.5%}
P(-2.776 < [tex]\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }[/tex] < 2.776) = 0.95
P( [tex]-2.776 \times {\frac{s}{\sqrt{n} } }[/tex] < [tex]{\bar X-\mu}[/tex] < [tex]2.776 \times {\frac{s}{\sqrt{n} } }[/tex] ) = 0.95
P( [tex]\bar X-2.776 \times {\frac{s}{\sqrt{n} } }[/tex] < [tex]\mu[/tex] < [tex]\bar X+2.776 \times {\frac{s}{\sqrt{n} } }[/tex] ) = 0.95
95% confidence interval for [tex]\mu[/tex] = [ [tex]\bar X-2.776 \times {\frac{s}{\sqrt{n} } }[/tex] , [tex]\bar X+2.776 \times {\frac{s}{\sqrt{n} } }[/tex] ]
= [ [tex]12,898-2.776 \times {\frac{7,719}{\sqrt{5} } }[/tex] , [tex]12,898+2.776 \times {\frac{7,719}{\sqrt{5} } }[/tex] ]
= [3315.13, 22480.87]
Therefore, a 95% confidence interval for the population mean is [3315.13, 22480.87] .
