Answer:
Explanation:
To find a 95% confidence interval for the population standard deviation, we can use the chi-square distribution with n-1 degrees of freedom, where n is the sample size. The formula for the confidence interval is:
sqrt((n-1)s^2 / χ^2α/2,n-1)) <= σ <= sqrt((n-1)s^2 / χ^21-α/2,n-1))
where s is the sample standard deviation and χ^2α/2,n-1 and χ^21-α/2,n-1 are the critical values from the chi-square distribution with n-1 degrees of freedom and α/2 and 1-α/2 as the upper and lower tail probabilities.
From the given data, we have:
n = 8
s = 10.737
α = 0.05
Using a chi-square distribution table or calculator, we can find the critical values:
χ^2α/2,n-1 = 2.1797
χ^21-α/2,n-1 = 14.0671
Substituting these values and the sample data into the formula, we get:
sqrt((n-1)s^2 / χ^2α/2,n-1)) <= σ <= sqrt((n-1)s^2 / χ^21-α/2,n-1))
sqrt((8-1)(10.737)^2 / 2.1797) <= σ <= sqrt((8-1)(10.737)^2 / 14.0671)
6.830 <= σ <= 21.114
Rounding the confidence interval limits to one more decimal place than the original data, we get:
(6.8, 21.1)
Therefore, a 95% confidence interval for the population standard deviation is (6.8, 21.1).
The critical value tα/2 corresponding to a sample size of 24 and a confidence level of 95% can be found using a t-distribution table or calculator.
Using a table, we find that the value of tα/2 is 2.064 for 24 degrees of freedom and a 95% confidence level.
Therefore, the answer is A. 2.064.