Use the given degree of confidence and sample data to find a confidence interval for the population standard deviation σ. Assume that the population has a normal distribution. Round the confidence interval limits to the same number of decimal places as the sample standard deviation. A sociologist develops a test to measure attitudes about public transportation, and 27 randomly selected subjects are given the test. Their mean score is 76.2 and their standard deviation is 21.4. Construct the 95% confidence interval for the standard deviation, σ, of the scores of all subjects.
Solution :
Given that,
c = 95% = 0.95
s = 21.4
n = 27
d.f. = n - 1 = 27 - 1 = 26
= 1 - 0.95 = 0.05
/ 2 = 0.025
1 - (
/ 2) = 0.975
Now , using chi square table ,
=
0.025,26
= 41.92
=
0.975,26
= 13.84
The 98% confidence interval for
is,
21.4 [(27 - 1 )
/ 41.92] <
<
21.4
[(27 - 1 )
/ 13.84]
16.85 < <
29.33
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