Answer:
The 98% confidence interval for population variance is (4.13, 14.59) and for population standard deviation is (2.03, 3.82).
Explanation:
To construct the confidence interval for the population variance, we use the chi-squared distribution. The formula for the confidence interval is:
[ (n - 1) * s^2 / chi2(a/2,n-1), (n - 1) * s^2 / chi2(1 - a/2,n-1) ]
where a is the level of significance, n is the sample size, s^2 is the sample variance, and chi2 is the chi-squared distribution function.
Substituting the given values, we get:
[ (26 - 1) * 7.29 / chi2(0.01, 25), (26 - 1) * 7.29 / chi2(0.99, 25) ]
Using a chi-squared table or a calculator, we can find the critical values for the chi-squared distribution:
chi2(0.01, 25) = 9.143
chi2(0.99, 25) = 43.773
Substituting these values, we get:
[ 174.474, 33.573 ]
Therefore, with 98% confidence, the population variance lies between 174.474 and 33.573.
To construct the confidence interval for the population standard deviation, we take the square root of the endpoints of the interval for the population variance. Therefore, we get:
[ sqrt(174.474), sqrt(33.573) ]
= [ 13.202, 5.793 ]
Therefore, with 98% confidence, the population standard deviation lies between 13.202 and 5.793.
Hope this helps! I'm sorry if it doesn't. If you need more help, ask me! :]