In: Statistics and Probability
A service station advertises that customers will have to wait no more than 30 miniutes for an oil change. A sample of 25 oil changes has a standard deviation of 5.2 minutes.
a) Find the 95% confidence interval of the population variance of the time spent waiting for oil change.
b) Find the 95% confidence interval of the population standard deviation of the time spent waiting for oil change.
At 95% confidence interval the critical values are
=
= 39.364
=
= 12.401
a) THe 95% confidence interval for
is
(n - 1)s2/
<
< (n - 1)s2/
= 24 * (5.2)^2/39.364 <
< 24 * (5.2)^2/12.401
= 16.486 <
< 52.331
b) The 95% confidence interval for
is
sqrt(16.486) <
< sqrt(52.331)
= 4.060 <
< 7.234