Respuesta :
Answer:
95% confidence interval for the mean amount of the active chemical in the drug = [ 90.088 , 94.312 ]
Step-by-step explanation:
We are given that a lab is testing the amount of a certain active chemical compound in a particular drug that has been recently developed. It is known that the standard deviation in the amount of the chemical is 6 mg.
A random sample of 31 batches of the new drug is tested and found to have a sample mean concentration of 92.2 mg of the active chemical i.e.;
Population standard deviation, [tex]\sigma[/tex] = 6 mg
Sample mean, [tex]Xbar[/tex] = 92.2 mg
Sample size, n = 31
Now, the pivotal quantity for 95% confidence interval is given by;
[tex]\frac{Xbar -\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] ~ N(0,1)
So, 95% confidence interval for the mean amount of the active chemical in the drug is given by;
P(-1.96 < N(0,1) < 1.96) = 0.95
P(-1.96 < [tex]\frac{Xbar -\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] < 1.96) = 0.95
P(-1.96 * [tex]\frac{\sigma}{\sqrt{n} }[/tex] < [tex]Xbar - \mu[/tex] < 1.96 * [tex]\frac{\sigma}{\sqrt{n} }[/tex] ) = 0.95
P(Xbar - 1.96 * [tex]\frac{\sigma}{\sqrt{n} }[/tex] < [tex]\mu[/tex] < Xbar + 1.96 * [tex]\frac{\sigma}{\sqrt{n} }[/tex] ) = 0.95
95% confidence interval for [tex]\mu[/tex] = [Xbar - 1.96 * [tex]\frac{\sigma}{\sqrt{n} }[/tex] , Xbar + 1.96 * [tex]\frac{\sigma}{\sqrt{n} }[/tex] ]
= [ 92.2 - 1.96 * [tex]\frac{6}{\sqrt{31} }[/tex] , 92.2 + 1.96 * [tex]\frac{6}{\sqrt{31} }[/tex] ]
= [ 90.088 , 94.312 ]
