Yeehah again! The theoretical work for developing a hypothesis test for a population variance \(\sigma^2\) is already behind us. Recall that if you have a random sample of size *n* from a **normal population** with (unknown) mean \(\mu\) and variance \(\sigma^2\), then:

\(\chi^2=\dfrac{(n-1)S^2}{\sigma^2}\)

follows a chi-square distribution with *n*−1 degrees of freedom. Therefore, if we're interested in testing the null hypothesis:

\(H_0 \colon \sigma^2=\sigma^2_0\)

against any of the alternative hypotheses:

\(H_A \colon\sigma^2 \neq \sigma^2_0,\quad H_A \colon\sigma^2<\sigma^2_0,\text{ or }H_A \colon\sigma^2>\sigma^2_0\)

we can use the test statistic:

\(\chi^2=\dfrac{(n-1)S^2}{\sigma^2_0}\)

and follow the standard hypothesis testing procedures. Let's take a look at an example.

##
Example 12-1
Section* *

A manufacturer of hard safety hats for construction workers is concerned about the mean and the variation of the forces its helmets transmits to wearers when subjected to an external force. The manufacturer has designed the helmets so that the mean force transmitted by the helmets to the workers is 800 pounds (or less) with a standard deviation to be less than 40 pounds. Tests were run on a random sample of *n* = 40 helmets, and the sample mean and sample standard deviation were found to be 825 pounds and 48.5 pounds, respectively.

Do the data provide sufficient evidence, at the \(\alpha = 0.05\) level, to conclude that the population standard deviation **exceeds** 40 pounds?

### Answer

We're interested in testing the null hypothesis:

\(H_0 \colon \sigma^2=40^2=1600\)

against the alternative hypothesis:

\(H_A \colon\sigma^2>1600\)

Therefore, the value of the test statistic is:

\(\chi^2=\dfrac{(40-1)48.5^2}{40^2}=57.336\)

Is the test statistic too large for the null hypothesis to be true? Well, the **critical value approach** would have us finding the threshold value such that the probability of rejecting the null hypothesis if it were true, that is, of committing a Type I error, is small... 0.05, in this case. Using Minitab (or a chi-square probability table), we see that the cutoff value is 54.572:

That is, we reject the null hypothesis in favor of the alternative hypothesis if the test statistic \(\chi^2\) is greater than 54.572. It is. That is, the test statistic falls in the rejection region:

Therefore, we conclude that there is sufficient evidence, at the 0.05 level, to conclude that the population standard deviation exceeds 40.

Of course, the ** P-value approach** yields the same conclusion. In this case, the

*P*-value is the probablity that we would observe a chi-square(39) random variable more extreme than 57.336:

As the drawing illustrates, the *P*-value is 0.029 (as determined using the chi-square probability calculator in Minitab). Because \(P = 0.029 ≤ 0.05\), we reject the null hypothesis in favor of the alternative hypothesis.

Do the data provide sufficient evidence, at the \(\alpha = 0.05\) level, to conclude that the population standard deviation **differs from** 40 pounds?

### Answer

In this case, we're interested in testing the null hypothesis:

\(H_0 \colon \sigma^2=40^2=1600\)

against the alternative hypothesis:

\(H_A \colon\sigma^2 \neq 1600\)

The value of the test statistic remains the same. It is again:

\(\chi^2=\dfrac{(40-1)48.5^2}{40^2}=57.336\)

Now, is the test statistic either too large or too small for the null hypothesis to be true? Well, the **critical value approach** would have us dividing the significance level \(\alpha = 0.05\) into 2, to get 0.025, and putting one of the halves in the left tail, and the other half in the other tail. Doing so (and using Minitab to get the cutoff values), we get that the lower cutoff value is 23.654 and the upper cutoff value is 58.120:

That is, we reject the null hypothesis in favor of the two-sided alternative hypothesis if the test statistic \(\chi^2\) is either smaller than 23.654 or greater than 58.120. It is not. That is, the test statistic does not fall in the rejection region:

Therefore, we fail to reject the null hypothesis. There is insufficient evidence, at the 0.05 level, to conclude that the population standard deviation differs from 40.

Of course, the ** P-value approach** again yields the same conclusion. In this case, we simply double the

*P*-value we obtained for the one-tailed test yielding a

*P*-value of 0.058:

\(P=2\times P\left(\chi^2_{39}>57.336\right)=2\times 0.029=0.058\)

Because \(P = 0.058 > 0.05\), we fail to reject the null hypothesis in favor of the two-sided alternative hypothesis.

The above example illustrates an important fact, namely, that the conclusion for the one-sided test does not always agree with the conclusion for the two-sided test. If you have reason to believe that the parameter will differ from the null value in a particular direction, then you should conduct the one-sided test.