Statistics formulas using MathML

Construct a confidence interval for the standard deviation, $σ$ of a population using sample data from that population.

The mean salary of 41 college graduates who took a statistics course in college is $67,200. The standard deviation of the salaries of those same 41 college graduates who took a statistics course in college is $18,277. Construct a 95% confidence interval for the standard deviation of all college graduates who took a statistics course in college.

Use the formula on page 347 for computing the confidence interval directly.

\sqrt{\frac{{(n - 1) s}^{2}}{χ_{R}^{2}}} < σ < \sqrt{\frac{{(n - 1) s}^{2}}{χ_{L}^{2}}}

Note: Because the chi-squared distribution is not symmetric, the margin of error in each tail of the distribution is different so the margin of error is not calculated.

Step 1: Identify the known variables.

$n = 41$ $\bar{x} = $ 67,200$ $s = $ 18,277$

Step 2: Look up the critical values of $χ^{2}$ in Table A-4 (Figure 7-10 on page 346). At a 95% confidence level for $n = 41$ the critical values of $χ^{2}$ are

χ_{R}^{2} = 59.342

χ_{L}^{2} = 24.433

Step 3 : Compute the confidence interval.

\sqrt{\frac{(41 - 1) \times {18,277}^{2}}{59.342}} < σ < \sqrt{\frac{(41 - 1) \times {18,277}^{2}}{24.433}}

$ 15,006 < σ < $ 23,385

Step 4: Be sure to properly word the statement that correctly interprets the confidence interval you have just calculated so that ordinary mortals (people who have not taken statistics) can understand.

At the 95% confidence level, the standard deviation of the salaries of all students who took a statistics course in college is between $15,006 and $23,385.