13.3: Test of a Single Variance

Thus far our interest has been exclusively on the population parameter \(\mu\) or it's counterpart in the binomial, p. Surely the mean of a population is the most critical piece of information to have, but in some cases we are interested in the variability of the outcomes of some distribution. In almost all clinical and athletic testing processes, quality is measured not only by how closely the individual matches the target, but also the variability of the measurements. If one were testing the range of motion in a patient’s joint, not only would there be interest in the average degrees of rotation, but also how much variation there was across repeated trials. No one wants to be assured that their average flexibility is normal when some measurements are dangerously low and others are high. Blood glucose levels may meet a healthy average level over 24 hours, but great variability, or "spikes," can cause serious damage to physiological systems and long-term metabolic health. You would not only like to have a high mean score on your fitness assessments, but also low variation about this mean to ensure your performance is consistent. In short, statistical tests concerning the variance of a distribution have great value and many applications in health and human performance.

A test of a single variance assumes that the underlying distribution is normal. The null and alternative hypotheses are stated in terms of the population variance. The test statistic is:

\[\chi_c^2=\dfrac{(n-1) s^2}{\sigma_0^2}\]

where:

• \(n=\) the total number of observations in the sample data
• \(s^2=\) sample variance
• \(\sigma_0^2=\) hypothesized value of the population variance
• \(H_0: \sigma^2=\sigma_0^2\)
• \(H_a: \sigma^2 \neq \sigma_0^2\)

You may think of \(s\) as the random variable in this test. The number of degrees of freedom is \(d f=n-1\). A test of a single variance may be right-tailed, left-tailed, or two-tailed. Example 11.1 will show you how to set up the null and alternative hypotheses. The null and alternative hypotheses contain statements about the population variance.

Exercise \(\PageIndex{1}\)

Physical education instructors are not only interested in how their students perform on fitness assessments, on average, but how those scores vary. To many instructors, the variance (or standard deviation) may be more important than the average because it indicates how uniform the class’s skill level is.

Suppose a PE instructor believes that the standard deviation for the shuttle run test is five seconds. One of the graduate assistants thinks otherwise. The assistant claims that the standard deviation is actually more than five seconds, suggesting the group's performance is more inconsistent than the instructor believes.

If the student were to conduct a hypothesis test, what would the null and alternative hypotheses be?

Answer

Even though we are given the population standard deviation, we can set up the test using the population variance as follows.

• \(H_0: \sigma^2 \leq 5^2\)
• \(H_a: \sigma^2>5^2\)

Exercise \(\PageIndex{2}\)

With individual check-in stations at its various entrances, a community health clinic finds that the standard deviation for patient waiting times on Friday afternoon is 7.2 minutes. The clinic experiments with a single, centralized triage system and finds that for a random sample of 25 patients, the waiting times have a standard deviation of 3.5 minutes on a Friday afternoon.

With a significance level of 5%, test the claim that a centralized triage system causes lower variation among patient waiting times.

Answer

Since the claim is that a centralized system causes less variation, this is a test of a single variance. The parameter is the population variance, \(\sigma^2\).

Random Variable: The sample standard deviation, \(s\), is the random variable. Let \(s=\) standard deviation for the waiting times.

• \(H_0: \sigma^2 \geq 7.2^2\)
• \(H_a: \sigma^2<7.2^2\)

The word "less" tells you this is a left-tailed test.

Distribution for the test: \(\chi_{24}^2\), where:

• \(n=\) the number of customers sampled
• \(d f=n-1=25-1=24\)

Calculate the test statistic:

\(\chi_c^2=\dfrac{(n-1) s^2}{\sigma^2}=\dfrac{(25-1)(3.5)^2}{7.2^2}=5.67\)

where \(n=25, s=3.5\), and \(\sigma=7.2\).

\Figure \(\PageIndex{1}\)

The graph of the Chi-square shows the distribution and marks the critical value with 24 degrees of freedom at \(95 \%\) level of confidence, \(\alpha=0.05,13.85\). The critical value of 13.85 came from the Chi squared table which is read very much like the students \(t\) table. The difference is that the students \(t\) distribution is symmetrical and the Chi squared distribution is not. At the top of the Chi squared table we see not only the familiar \(0.05,0.10\), etc. but also \(0.95,0.975\), etc. These are the columns used to find the left hand critical value. The graph also marks the calculated \(X^2\) test statistic of 5.67.

Comparing the test statistic with the critical value, as we have done with all other hypothesis tests, we reach the conclusion.

Make a decision: Because the calculated test statistic is in the tail we cannot accept \(H_0\). This means that you reject \(\sigma^2 \geq 7.2^2\). In other words, you do not think the variation in waiting times is 7.2 minutes or more; you think the variation in waiting times is less.

Conclusion: At a \(5 \%\) level of significance, from the data, there is sufficient evidence to conclude that a centralized triage system causes a lower variation among the waiting times, or in other words, the waiting times vary less than 7.2 minutes.

Exercise \(\PageIndex{3}\)

A clinical nutritionist is reviewing a brand of rehydration salts used for patients with chronic dehydration. For the dosage to be safe and effective, the mineral content must be consistent. A sample of 24 packets reveals a mean sodium concentration of 0.04 grams per liter above the baseline, with a sample standard deviation of 0.11 grams.

The nutritionist is interested in the average sodium level, but they are particularly concerned if one packet is radically different from another. In clinical nutrition, high variability can lead to inconsistent patient outcomes or potential electrolyte imbalances.

Problem

Test at 95% confidence (5% level of significance) the null hypothesis that the population variance of the sodium concentration is significantly different from the expected value of 0.04.

Answer

This is a problem dealing with a single population variance. We are checking if the "spread" of the sodium levels is what it should be. The null and alternative hypotheses are:

The null and alternative hypotheses are thus:

\[\begin{array}{l}
H_0: \sigma^2=0.04 \\
H_0: \sigma^2 \neq 0.04
\end{array}\]

The test is two-tailed because the nutritionist is concerned with any deviation from the standard—whether the packets are too inconsistent (high variance) or unexpectedly uniform (low variance). The test statistic should be calculated:

\[\chi^2 c=\dfrac{(n-1) s^2}{\sigma_o^2}=\dfrac{(24-1) 0.11^2}{0.04^2}=6.9575\]

The calculated \(\chi^2\) test statistic, 6.96 , is in the tail therefore at a 0.05 level of significance, we cannot accept the null hypothesis. We conclude that the variance in sodium content is significantly different from 0.04, suggesting the manufacturing process is not meeting the required consistency for clinical use.

Figure \(\PageIndex{2}\)