Inferences about One-Population Standard Deviation

Summary: In class you learn to estimate population means and test hypotheses about them. It can also be important to estimate or test variability — standard deviation or variance of a population. This page shows you how. All operations can be done in the accompanying Excel workbook (43 KB), or in the downloadable TI-83/84 program MATH200B part 5.

You already know how to test the mean of a population with a t test, or estimate a population mean using a t interval. Why would you want to do that for the standard deviation of a population?

The standard deviation measures variability. In many situations not just the average is important, but also the variability. Another way to look at it is that consistency is important: the variability must not be too great.

For example, suppose you are thinking about investing in one of two mutual funds. Both show an average annual growth of 3.8% in the past 20 years, but one has a standard deviation of 8.6% and the other has a standard deviation of 1.2%. Obviously you prefer the second one, because with the first one there’s quite a good chance that you’d have to take a loss if you need money suddenly.

Industrial processes, too, are monitored not only for average output but for variability within a specified tolerance. If the diameter of ball bearings produced varies too much, many of them won’t fit in their intended application. On the other hand, it costs more money to reduce variability, so you may want to make sure that the variability is not too low either.

Hypothesis Test of Standard Deviation σ

Write your hypotheses in the usual way. For H₀, you compare (=, ≤, or ≥) the population standard deviation σ to the claimed value σ_o. H₁ contains ≠, >, or < as usual.

The test statistic is

χ² = (n−1) s² / σ_o² with df = n− 1

You perform a one-tailed test by computing the cumulative probability from 0 to the χ² (left tail) or from χ² to 10^99 or ∞ (right tail). For a two-tailed test, compute the cumulative probability and double it.

Example 1: A machine packs cereal into boxes, and you don’t want too much variation from box to box. You decide that a standard deviation of no more than five grams (about 1/6 ounce) is acceptable. To determine whether the machine is operating within specification, you randomly select 45 boxes. Here are the weights of the boxes, in grams:

Solution: First, use 1-Var Stats to find the sample standard deviation, which is 6.42 g. Obviously this is greater than the target standard deviation of 5 g, but is it enough greater that you can say the machine is not operating correctly, or could it have come from a population with standard deviation no more than 5 g? Your hypotheses are

H₀: σ = 5, the machine is within spec (some books would say H₀: σ ≤ 5)

H₁: σ > 5, the machine is not working right

No α was specified, but for an industrial process with no possibility of human injury α = 0.05 seems appropriate.

Next, check the requirements: is the sample normally distributed and free of outliers? You can do it with a TI calculator or with a statistics package. (Excel users, please see Normality Check and Finding Outliers in Excel.) On my TI-84, I used MATH200A part 4 to make a quantile plot to check for normality, and MATH200A part 2 to make a box-whisker plot to check for outliers. The results are shown at right, and demonstrate that the sample of box weights is normally distributed with no outliers.

Now compute the test statistic:

χ² = (n−1) s² / σ_o² = 44×6.42²/5² = 72.54

Next, compute the p-value. Use either the accompanying Excel workbook or your TI calculator’s χ²cdf function; see keystrokes at right for your model. (If you don’t have Excel or a suitable calculator, you can probably find a table of the χ² distribution at the back of your textbook.) Use n−1 = 44 degrees of freedom.

p-value = χ²cdf(Ans, 10^99, 44) = 0.0043

Conclusion: p-value < α; reject H₀ and accept H₁. The machine’s output is too variable: at the 0.05 level of significance the standard deviation is greater than 5 g.

Example 2: You have a random sample of size 20, with a standard deviation of 125. You have good reason to believe that the underlying population is normal. Is the population standard deviation different from 100, at the 0.05 significance level?

Solution: n = 20, s = 125, σ_o = 100, α = 0.05. Your hypotheses are

H₀: σ = 100

H₁: σ ≠ 100

This is a two-tailed test.

Compute the test statistic:

χ² = (n−1) s² / σ_o² = 19×125²/100² = 29.6875

Compute the p-value. Since this is a two-tailed test, find the one-tailed p and double it. (If the one-tailed p-value is >0.5, subtract from 1 and then double.) Either use the accompanying Excel workbook or use your TI calculator’s χ²cdf function with degrees of freedom n−1 = 19:

p = 2 × χ²cdf(Ans, 10^99, 19) = 0.1118

p-value > α; you fail to reject H₀ and cannot reach a conclusion. The population standard deviation may be different from 100, or it may not.

Hypothesis Test of Variance σ²

You may have noticed that the test for σ actually uses the sample variance s² and the hypothetical population variance σ_o². Therefore, to make a test for variance you follow exactly the same procedure except that you already have the variance and you don’t square it to obtain the test statistic.

Confidence Interval Estimate of Standard Deviation σ

To estimate the standard deviation σ of a population at confidence level 1−α, the bounds are

√(n−1)s² / χ²_crit(df,α/2) < σ < √(n−1)s² / χ²_crit(df,1−α/2)

In the formula, χ²_crit(df,rtail) is the critical χ² value that divides the curve with area (probability) rtail to the right and 1−rtail to the left. It’s an inverse χ² function, analogous to inverse t or inverse normal.

As usual when you’re dealing with standard deviations, df = n−1.

Caution: For standard deviation of a population, the confidence interval is not symmetric and the point estimate is not in the middle of the confidence interval. Therefore the confidence interval can be expressed only in endpoint form, not in s ± E form.

Example 3: Of several thousand students who took the same exam, 40 papers were selected randomly and statistics were computed. The standard deviation of the sample was 17 points. Estimate the standard deviation of the population, with 95% confidence. (Recall that test scores are normally distributed.)

Solution: 1−α = 0.95, so α = 0.05, α/2 = 0.025, and 1−α/2 = 0.975. df = n−1 =39. The confidence interval is

√37 × 17² / χ²_crit(39,0.025) < σ < √37 × 17² / χ²_crit(39,0.975)

How do we find the two required χ²_crit values? There are several methods, laid out in Finding χ²_crit(df,rtail), below. For now let’s just use the values: χ²(39,0.025) = 58.12006 and χ²(39,0.975) = 23.65432. Continuing with the calculation,

√39×17²/58.12006 < σ < √39×17²/23.65432

13.9 < σ < 21.8 with 95% confidence

Remark: The middle of the confidence interval is (13.9+21.8)/2 ≈ 17.9, but the point estimate was 17. The confidence interval is not symmetric: it extends 3.1 below and 4.8 above the point estimate.

Finding χ²_crit(df,rtail), Inverse Chi Squared

χ²_crit, also known as inverse χ², is not easy to compute, but you’re not necessarily reduced to looking up tables in a book. Here are several methods using various forms of technology:

• The TI-89 has a function to compute inverse χ². There’s also a TI-83/84 program; please see MATH200B Program part 5. That program also does all the calculations for hypothesis tests and confidence intervals.
• Excel has the CHISQ.INV.RT or CHIINV(df,rtail) function. rtail in the CHIINV() function is the area of the right-hand tail. The accompanying workbook uses this function to compute χ²_crit and the confidence intervals.
• A normal approximation for χ²_crit when df ≥ 30 is found in Spiegel 1999 [full citation in “References”, below], “Confidence Intervals for χ²”, page 245. When df ≥ 30, “√2χ² − √2df−1 is very nearly normally distributed with mean 0 and standard deviation 1”. Therefore,

  χ²_crit(df,rtail) ≈ ½ [z_crit(rtail) + √2·df−1]², for df ≥ 30

  Example: Using the normal approximation,

  χ²(39,0.025) ≈ ½ [z(0.025)+√2×39−1]²

  z(rtail) is easily found on your calculator by invNorm(1−rtail). In this case, z(0.025) = invNorm(1−0.025) = 1.9600, so

  χ²(39,0.025) ≈ (1.9600+√77)² / 2

  χ²(39,0.025) ≈ 57.61934

  That agrees quite well with the true value of 58.12006 given above. Using the same technique, χ²(39,0.975) ≈ 23.22212 and the 95% confidence interval on σ is 14.0 to 22.0, close to the interval we found above.

• The National Institute of Standards and Technology’s statistics Web site offers Critical Values of the Chi-Square Distribution for 1 through 100 degrees of freedom. This is part of the useful NIST/SEMATECH e-Handbook of Statistical Methods, located here.

Confidence Interval Estimate of Variance σ²

Variance is the square of standard deviation, so the confidence interval procedure is the same except that you don’t take square roots:

(n−1)s² / χ²_crit(df,α/2)  <  σ²  <  (n−1)s² / χ²_crit(df,1−α/2)

Example 4: Heights of US males aged 18–25 are normally distributed. You take a random sample of 100 from that population and find a variance of 7.3 in². (Remember that the units of variance are the square of the units of the original measurement.) Estimate the variance of the height of US males aged 18–25, with 95% confidence.

Solution: For a 95% confidence interval, 1−α = 0.95 and α/2 = 0.025. From the accompanying workbook we find

χ²_crit(df,α/2) = χ²_crit(99,0.025) = 128.4219887

χ²_crit(df,1−α/2) = χ²_crit(99,0.975) = 73.36108022

The endpoints of the interval are therefore

99 × 7.3 / 128.42199 < σ² < 99 × 7.3 / 73.36108

5.6275... < σ² < 9.8512...

We’re 95% confident that the variance in heights of US males aged 18–25 is between 5.6 and 9.9 in².

References

Spiegel, Murray R., and Larry J. Stephens. 1999.

Theory and Problems of Statistics. 3d ed. McGraw-Hill.