Practice problems for confidence interval for sigma_x, the standard deviation for the population

We have learned the methods for finding confidence intervals for proportions and averages of populations given similar statistics from samples. There is also a method for estimating the standard deviation of a population and giving a confidence level to that interval.

Let's say we took a sample of 28 scores and got a standard deviation of sx = 16.689, rounded to three places after the decimal. The degrees of freedom is n-1, which in this case is 27. Let's look at the Chi square table at the line that corresponds to d.f. = 27.

____0.995__0.99___0.975__0.95___0.90___||_0.10___0.05___0.025__0.01___0.005
27__11.808 12.879 14.573 16.151 18.114 || 36.741 40.113 43.194 46.963 49.645

The denominators in the formulas shown above are taken from the following columns.

90% confidence: Chi square Big comes from the 0.05 column, Chi square Small comes from the 0.95 column.

95% confidence: Chi square Big comes from the 0.025 column, Chi square Small comes from the 0.975 column.

99% confidence: Chi square Big comes from the 0.005 column, Chi square Small comes from the 0.995 column.

In this example, the formulas would look as follows.

90% confidence interval: sqrt(16.689^2*27/40.113) < sigmax < sqrt(16.689^2*27/16.151)

95% confidence interval: sqrt(16.689^2*27/43.194) < sigmax < sqrt(16.689^2*27/14.573)

99% confidence interval: sqrt(16.689^2*27/49.645) < sigmax < sqrt(16.689^2*27/11.808)

If n-1 is not one of the values in the degrees of freedom chart, use the next lowest number on the list.

Exercise #1: Find the values from the equations listed above, rounded to the nearest thousandth.

Exercise #2: Find the confidence intervals for 90%, 95% and 99% if n = 102 and sigmax = 0.62. Round the answers to two places after the decimal.

Answers in the comments.