In: Statistics and Probability
A random sample of 21 nickels in circulation were measured with a very accurate micrometer to find a mean of 0.834343 inch and a standard deviation of 0.001886 inch.
What is a 99% confidence interval for the standard deviation (correct to 4 decimal places) of all circulating nickels?
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Solution :
Given that,
s = 0.001886
s2 = 0.0434
n = 21
Degrees of freedom = df = n - 1 = 21 - 1 = 20
At 99% confidence level the
2 value is ,
= 1 - 99% = 1 - 0.99 = 0.01
/ 2 = 0.01 / 2 = 0.005
1 -
/ 2 = 1 - 0.005 = 0.995
2L
=
2
/2,df
= 40
2R
=
21 -
/2,df = 7.432
The 99% confidence interval for
is,
(n
- 1)s2 /
2
/2
<
<
(n - 1)s2 /
21 -
/2
(20)(0.0434)
/ 40 <
<
(20)(0.0434) / 7.432
0.0217 <
<
0.11679
0.1473 <
< 0.3717
(0.1473 , 0.3417)