Inferences about One-Pop. Standard Deviation on the TI-83/84
Copyright © 2008 by Stan Brown, Oak Road Systems
Copyright © 2008 by Stan Brown, Oak Road Systems
Summary: This Web page presents a downloadable TI-83/84 program that lets you perform a hypothesis test on standard deviation or variance of a population, or compute a confidence interval about standard deviation or variance of a population.
The tests on this page require that the underlying population must be normal. They are not robust, meaning that even moderate departures from normality can invalidate your analysis. See Normality Check on the TI-83/84 or TI-89 for procedures to test whether a population is normal by testing the sample.
Outliers are also unacceptable and must be ruled out. See Make a Box-Whisker Plot for an easy way to test for outliers.
See also: Inferences about One Population Standard Deviation gives the statistical concepts with examples of calculation “by hand” and in an Excel workbook.
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. 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.
This TI-83/84 program can perform hypothesis tests and compute confidence intervals for the standard deviation of a population. Since variance is the square of standard deviation, it can also do those calculations for the variance of a population.
Please observe the cautions above, and
always check the requirements before using this program. If
the requirements are not met, the results from
SDINFER will be, in Gene Wilder’s words from
Young Frankenstein, “doo-doo”.
There are three methods to get the program into your calculator:
ENTER], and then on hers press [
3], select the program, then press [
To use the program, first check the requirements for your
sample; see Cautions above.
Then press [
SDINFER, and press [
When prompted, enter the standard deviation and size of the sample,
ENTER] after each one. If you know the variance of
the sample rather than the standard deviation, use the square root
operation since s is the square root of the variance s².
The program then presents you with a five-item menu: confidence interval for the population standard deviation σ, confidence interval for the population variance σ², and three hypothesis tests for σ or σ² less than, different from, or greater than a number. Make your selection by pressing the appropriate number.
If you select one of the confidence intervals, the program will prompt you for the confidence level and then compute the interval. Because this involves a process of successive approximations, it can take some time, so please be patient.
The program displays the endpoints of the interval on screen
and also leaves them in variables
case you want to use them in further calculations. You can include
them in any formula by [
L] and [
If you select one of the hypothesis tests, the program will prompt you for σ, the population standard deviation in the null hypothesis. If your H0 is about population variance σ² rather than σ, use the square root symbol to convert the hypothetical variance to standard deviation.
The program then displays the χ²o test statistic, the
degrees of freedom, and the p-value. These are also left in variables
P in case you wish to use
them in further calculations. You can include them in any formula with
D], and [
Example 1: A machine packs cereal into boxes, and you require a standard deviation of no more than five grams. You randomly select and weigh 45 boxes and find a sample standard deviation of 6.2 grams. Is the machine operating within specification?
You have tested the sample and find that it is normally distributed with no outliers, so you are confident that the population is also normally distributed.
Solution: n = 45, s = 6.2, σo = 5. 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.
SDINFER program. Enter s:6.2 and
n:45, and select
5:TEST σ>CONST. Enter 5
for H0 σ.
The results are shown at right. The test statistic is χ²o = 67.65 with 44 degrees of freedom, and the p-value is 0.0125.
Since p<α, you reject H0 and accept H1. At the 0.05 level of significance, the population standard deviation σ is greater than 5, and the machine is not operating within specificaton.
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, σo = 100, α = 0.05. Your hypotheses are
H0: σ = 100
H1: σ ≠ 100
This time in the
SDINFER program you select
Results are shown at right. χ²o = 29.69 with 19 degrees of freedom, and the p-value is 0.1118.
p>0; fail to reject H0. At the 0.05 significance level, you can’t say whether the population standard deviation σ is different from 100 or not.
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.)
SDINFER and select
1:σ INTERVAL with a C-level of 95 or .95.
The results screen is shown at right.
Conclusion: We’re 95% confident that the standard deviation of test scores for all students is between 13.9 and 21.8.
Example 4: Heights of U.S. males aged 18–25 are normally distributed. You take a random sample of 100 from that population and find a mean of 65.3 in and a variance of 7.3 in². (Remember that the units of variance are the square of the units of the original measurement.)
Estimate the mean and variance of the height of U.S. males aged 18–25, with 95% confidence.
Solution for mean:
Computing a confidence interval for the mean is a straightforward
TInterval. Just remember that for
calculator wants the sample standard deviation, but you have the
sample variance, which is s². Therefore you take the square root
of sample variance to get sample standard deviation, as shown in the
input screen at near right.
The output screen at far right shows the confidence interval. We’re 95% confident that the mean height of U.S. males aged 18–25 is between 64.8 and 65.8 in.
Solution for variance:
SDINFER program. Enter s:√7.3 and n:100.
2:σ² INTERVAL and enter C-Level:.95 (or
95). The program computes the confidence interval for population
variance as 5.6 ≤ σ² ≤ 9.9.
We’re 95% confident that the variance in heights of U.S. males
aged 18–25 is between 5.6 and 9.9 in².
Complete answer: We’re 95% confident that the heights of U.S. males aged 18–25 have mean 64.8–65.9 in and variance 5.6–9.9 in².