Look at the attached image to find the problem. Please explain to me how you got both of the answers.
1 answer:
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Answer:
99% confidence interval for the population standard deviation = (74.97 , 635.20).
Step-by-step explanation:
We are given that the ages of a random sample of five university professors are 39, 54, 61, 72 and 59. Also, it is provided that the ages of university professors are normally distributed.
So, firstly the pivotal quantity for 99% confidence interval for the population standard deviation is given by;
P.Q. = ~
where, s = sample standard deviation
= population standard deviation
n = sample of university professors = 5
Also, =
= 144.5
So, 99% confidence interval for population standard deviation, is;
P(0.2070 < < 14.86) = 0.99 {As the table of
at 4 degree of freedom
gives critical values of 0.2070 & 14.86}
P(0.2070 < < 14.86) = 0.99
P( <
<
) = 0.99
P( <
<
) = 0.99
99% confidence interval for = (
,
)
= ( ,
)
= (5620.525 , 403483.092)
99% confidence interval for = (
,
)
= (74.97 , 635.20)
Therefore, 99% confidence interval for the population standard deviation of the ages of all professors at the university is (74.97 , 635.20).