Respuesta :
Answer:
90 percent confidence interval = [72.674 ,77.326]
Step-by-step explanation:
We are given that weight of a product is measured in pounds.
A random sample of 50 units is taken from a recent production. The sample yielded y bar = 75 lb, and we know that [tex]\sigma^{2}[/tex] = 100 lb.
The Pivotal quantity for 9% confidence interval is given by;
[tex]\frac{Ybar - \mu}{\frac{\sigma}{\sqrt{n} } }[/tex] ~ N(0,1)
where, Y bar = sample mean = 75
[tex]\sigma[/tex] = population standard deviation = 10
n = sample size = 50
So, 90% confidence interval for population mean, is given by;
P(-1.6449 < N(0,1) < 1.6449) = 0.90
P(-1.6449 < [tex]\frac{Ybar - \mu}{\frac{\sigma}{\sqrt{n} } }[/tex] < 1.6449) = 0.90
P(-1.6449 * [tex]{\frac{\sigma}{\sqrt{n} }[/tex] < [tex]{Ybar - \mu}[/tex] < 1.6449 * [tex]{\frac{\sigma}{\sqrt{n} }[/tex] ) = 0.90
P(Y bar - 1.6449 * [tex]{\frac{\sigma}{\sqrt{n} }[/tex] < [tex]\mu[/tex] < Y bar + 1.6449 * [tex]{\frac{\sigma}{\sqrt{n} } }[/tex] ) = 0.90
90% confidence interval for [tex]\mu[/tex] = [ Y bar - 1.6449 * [tex]{\frac{\sigma}{\sqrt{n} }[/tex] , Y bar + 1.6449 * [tex]{\frac{\sigma}{\sqrt{n} }[/tex] ]
= [ 75 - 1.6449 * [tex]{\frac{10}{\sqrt{50} }[/tex] , 75 + 1.6449 * [tex]{\frac{10}{\sqrt{50} }[/tex] ]
= [ 72.674 , 77.326 ]
Therefore, 90% confidence interval for population mean is [72.674 ,77.326] .
