Respuesta :
Answer:
Option a. [52.78,64.22]
Step-by-step explanation:
We are given that a random sample of size 18 is drawn from a population that is normally distributed.
Sample mean is 58.5, and the sample standard deviation is found to be 11.5 i.e., X bar = 58.5 and s = 11.5
The pivotal quantity for calculating 95% confidence interval is;
[tex]\frac{Xbar - \mu}{\frac{s}{\sqrt{n} } }[/tex] ~ [tex]t_n_-_1[/tex]
So, 95% confidence interval about population mean is given by;
P(-2.110 < [tex]t_1_7[/tex] < 2.110) = 0.95
P(-2.110 < [tex]\frac{Xbar - \mu}{\frac{s}{\sqrt{n} } }[/tex] < 2.110) = 0.95
P(-2.110 * [tex]\frac{s}{\sqrt{n} }[/tex] < [tex]{Xbar - \mu}[/tex] < 2.110 * [tex]\frac{s}{\sqrt{n} }[/tex] ) = 0.95
P(X bar - 2.110 * [tex]\frac{s}{\sqrt{n} }[/tex] < [tex]\mu[/tex] < X bar + 2.110 * [tex]\frac{s}{\sqrt{n} }[/tex] ) = 0.95
95% confidence interval about [tex]\mu[/tex] = [ X bar - 2.110 * [tex]\frac{s}{\sqrt{n} }[/tex] , X bar + 2.110 * [tex]\frac{s}{\sqrt{n} }[/tex] ]
= [ 58.5 - 2.110 * [tex]\frac{11.5}{\sqrt{18} }[/tex] , 58.5 + 2.110 * [tex]\frac{11.5}{\sqrt{18} }[/tex] ]
= [ 52.78 , 64.22 ]
Therefore, 95% confidence interval about population mean is [52.78 , 64.22].
