Question
In: Statistics and Probability
Suppose the population variance is unknown and the sample standard deviation is 1.92. Compute the correct...
Suppose the population variance is unknown and the sample
standard deviation is 1.92. Compute the correct quantile for each
confidence interval: (Use 3 decimal places)
(a) A 95% confidence interval from a sample of size 10.
(c) A 80% confidence interval from a sample of size 6.
Solutions
Expert Solution
Solution :
Given that,
(a)
s = 1.92
Point estimate = s2 = 3.6864
2R
=
2
/2,df
= 19.023
2L
=
21 -
/2,df = 2.700
The 95% confidence interval for 
2 is,
(n - 1)s2 /
2/2
<
2 < (n - 1)s2 /
21 -
/2
(9) (3.6864) / 19.023 <
2 < (9) (3.6864) / 2.700
1.744 <
2 < 12.286
(1.744 , 12.286)
(c)
s = 1.92
Point estimate = s2 = 3.6864
2R
=
2/2,df
= 9.236
2L
=
21 -
/2,df = 1.610
The 80% confidence interval for
2 is,
(n - 1)s2 /
2/2
<
2 < (n - 1)s2 /
21 -
/2
(5) (3.6864) / 9.236 <
2 < (5) (3.6864) / 1.610
1.996 <
2 < 11.446
(1.996 , 11.446 )
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