In: Statistics and Probability
A commercial farm uses a machine that packages blueberries in ten pound portions. A sample of 95 packages of blueberries has a standard deviation of 0.22. Construct the 90% confidence interval to estimate the standard deviation of the weights of the packages prepared by the machine. Round your answers to two decimal places.
Solution :
Given that,
s = 0.22
s2 = 0.0484
n = 95
Degrees of freedom = df = n - 1 = 94
At 90% confidence level the
2 value is ,
2L
=
2
/2,df
= 117.632
2R
=
21 -
/2,df = 72.640
The 90% confidence interval for
is,
(n
- 1)s2 /
2
/2
<
<
(n - 1)s2 /
21 -
/2
94
* 0.0484 / 117.632 <
<
94 * 0.0484 / 72.640
0.20 <
< 0.25
(0.20 , 0.25)