7.5: Claims on Population Variances - Optional Material

Learning Objectives

• Conduct hypothesis testing on claims regarding population variance using the \(p\)-value method

Section 7.5 Excel File: (contains all of the data sets for this section)

Review and Preview

Having developed hypothesis testing for claims on population means, paired variables, and proportions, we are aware that the process is supported by our understanding of the sampling distributions of particular sample statistics. This remains the case, when considering claims on population variance and standard deviation. Recall that the sample standard deviation is not an unbiased estimator of the population standard deviation but that the sample variance is an unbiased estimator of the population variance. Therefore, to test any claims on a population's standard deviation, we must first translate them into equivalent claims regarding the population's variance, test these new claims, and then translate the results back into the realm of standard deviation.

Once we formulate our hypotheses and collect our evidence, we assess the significance of the evidence using the \(p\)-value. Difficulties arise in determining the \(p\)-value when conducting a two-tailed test; they stem from determining what is equally extreme in the opposite direction when the distribution is not symmetric. We will address this difficulty in detail later in the section.

Recall that when the parent distribution is normal, we transformed the sampling distribution of sample variances into a \(\chi^2\)-distribution with \(n-1\) degrees of freedom to compute probabilities. We will need to utilize test statistics when testing claims on population variance. The test statistic is the value produced by mapping the evidence from a particular sample into the common distribution under the assumption that the null hypothesis is true. In assuming the null hypothesis is true, we will have some hypothesized value of the population variance, \(\sigma^2_0,\) leading to the following test statistic.\[\chi^2_{n-1}=\frac{(n-1)}{\sigma^2_0}\cdot s^2\nonumber\]With the test statistic in hand, we compute the \(p\)-value and make a conclusion based on the comparison between the \(\alpha\) value and the \(p\)-value. Let us begin testing claims on population variance and standard deviation.

Claims on Population Variance: One-Tailed Tests

In order to conduct hypothesis testing on claims regarding population variance, we will need to have a random sample taken from a normally distributed parent population. As with all hypothesis tests, checking that the requirements of the test are met is important! Let us consider an example situation together.

Many farmers spray their fields to prevent weeds and pests from negatively affecting their harvests. When spraying a field, it is important to get sufficient and even coverage. We need the average ratio of volume to area high enough to meet our needs and the standard deviation to be low enough to imply consistent application.

A company that manufactures sprayers conducted a test on a recently developed prototype to see if it met company standards regarding consistent, even coverage. The company will not produce a sprayer unless the standard deviation is less than a quarter of a gallon per acre. To test the consistency of the sprayer, the prototype sprayed three fields each containing \(100\) collection devices scattered sporadically throughout the field. When all was said and done, the \(300\) measurements averaged out to \(15.3\) gallons per acre with a standard deviation of \(0.235\) gallons per acre. They formulated the hypothesis test choosing a significance level of \(0.10\) and the following hypotheses.\[\begin{align*}H_0&:\sigma\ge0.25\text{ gallons per acre}\\H_1&:\sigma<0.25\text{ gallons per acre}\end{align*}\]

This formulation of the hypotheses, however, is not the formulation that the company used in testing because the hypothesis testing needs to be done in the realm of variance, which yields the following set of hypotheses.\[\begin{align*}H_0&:\sigma^2\ge0.0625\text{ gallons per acre}^2\\H_1&:\sigma^2<0.0625\text{ gallons per acre}^2\end{align*}\]To conduct the hypothesis test, we need that the sample was randomly selected from a parent distribution that is normally distributed. Given the random placement of the \(300\) collection devices, the sample was randomly chosen. The company felt confident that the distribution was normally distributed based on past history, but they conducted a test on the sample data to see if it was reasonable based on the observed data (recall that such tests exist but are outside of the scope of this course). The test affirmed the reasonableness of the assumption that the parent distribution was normally distributed. So, the hypothesis test could be conducted. Note that the sample variance is \(0.055225\) square gallons per square acre. We compute our test statistic and produce our visualization.\[\chi^2_{299}=\frac{(300-1)}{0.0625}\cdot 0.055225\approx264.1964\nonumber\]

Figure \(\PageIndex{1}\): \(\chi^2\)-distribution

We note that the \(\chi^2\)-distribution appears to be symmetric as opposed to the asymmetrical appearance we have come to recognize. This is because the sample size is so large. The amount of skew present in \(\chi^2\)-distributions decreases as the degrees of freedom increase. From our visualization we compute our \(p\)-value in order to conclude the hypothesis test.\[p\text{-value}\approx\text{CHISQ.DIST}(264.1964,299,1)\approx0.0728\nonumber\]Given that the level of significance for this test was \(0.10,\) there is sufficient evidence to reject the null hypothesis. The company can begin to produce the first generation of this prototype sprayer.

Text Exercise \(\PageIndex{1}\)

An amateur game developer is designing a game with AI generated open worlds in hopes of building a game that is essentially endless. The developer does not, however, want the game to become monotonous and has tried to incorporate a great variability between worlds. One of the metrics the developer decided to use to test if the AI is producing enough variability is the distance the first significant encounter occurs from the starting position. The developer does not want the distance to be too long and does not want it to be too consistent. The developer designed the AI to produce worlds so that the average distance is about \(550\) game paces with a standard deviation of more than \(170\) game paces.

To make sure the AI was working properly, the developer randomly chose \(50\) game backers to play randomly chosen AI generated worlds in order to find the distances to the first significant encounter. The sample data was analyzed and was found to have an average distance of \(497\) game paces and standard deviation of \(200\) game paces. Test the hypothesis at the \(0.05\) significance level under the assumption that the distribution of the number of game paces to the first significant encounter is normally distributed.

Answer

We can conduct the hypothesis test because the sample was randomly selected and we were told to assume the parent distribution is normally distributed. The problem is framed within the context of standard deviation; so, we must translate the problem to variance. If the standard deviation is supposed to be more than \(170\) game paces, the variance would need to be more than \(170^2\) \(=28,900\) square game paces. Since this game is just being developed and tested to see if it is working correctly, we do not want to assume that the population variance is greater than \(28,900\) square game paces. This helps us to set our hypotheses as follows.\[\begin{align*}H_0&:\sigma\leq 28,900\text{ game paces}^2\\H_1&:\sigma>28,900\text{ game paces}^2\end{align*}\]We have a right-tailed test. We compute our test statistic using the sample variance and then produce our visualization to help compute the \(p\)-value.\[\chi^2_{49}=\frac{(50-1)}{40,000}\cdot 28,900\approx67.8201\nonumber\]

Figure \(\PageIndex{2}\): \(\chi^2\)-distribution

\[p\text{-value}\approx1-\text{CHISQ.DIST}(67.8201,49,1)\approx0.0387\nonumber\]The \(p\)-value is smaller than the level of significance; therefore we reject the null hypothesis. This provides sufficient evidence for the developer to assert that the AI is working for the variation in the game. It looks like it may not be meeting specifications regarding the average distance though. That would require a test on means. An interested reader is encouraged to consider how to conduct such a test.

Claims on Population Variance: Two-Tailed Tests

As hinted in the Review and Preview section, we will need to add nuance to our approach in order to conduct two-tailed tests on population variances. Let us examine why an issue arises with two-tailed tests on population variances. Recall two main ideas: the standard normal distribution and the \(t\)-distributions are symmetric about \(0\) (the expected value of each distribution) and the \(p\)-value is the probability of obtaining something at least as extreme as what was observed under the assumption that the null hypothesis is true. In the two-tailed case, we needed to consider the value of a test statistic equally extreme as the test statistic computed from the observed sample statistic but in the opposite direction. Thus far, we chose the value that was the same distance away from the mean on the other side of the mean; this will no longer suffice. Given the symmetry of the previous distributions, three facts about the two values coincide: equidistant from the mean, equal probability in the tails, and equal probability densities (the heights of the density function at those values match). As it turns out, these serve as three different possibilities for determining the value that would be considered equally extreme in the opposite direction. Since the \(\chi^2\)-distribution is not symmetric, these three facts do not coincide in the \(\chi^2\)-distribution, and a decision regarding how to define equally extreme must be made.

Let us examine these three possibilities. As we generalize our approach beyond test statistics in the standard normal distribution and \(t\)-distributions, we will also consider the possibility of distributions other than the \(\chi^2\)-distribution. It is possible to have a distribution that has more than two values with equal probability densities. If this is the case, which of the values do we take as equally extreme? As such, we do not use probability density values as we formulate our approach. When dealing with a skewed distribution, one of the tails extends farther in one of the directions. If the test statistic lies far along the tail in the direction of skew, it is possible, even probable, that the value that is equidistant from the mean just on the opposite side has a probability density of 0. Think about the \(\chi^2\)-distribution which takes on no negative numbers. If our test statistic lies far to the right of the mean (far along the tail in the direction of the skew), the value equidistant from the mean on the opposite side may be negative. If this is the case, that would mean there is not an equally extreme value on the other side of the mean, which we intuitively know to be false. As such, this leaves one path forward; the value that is equally extreme on the opposite side of the mean is the value such that the area in its tail is equal to the tail on the other side.

Such a determination really eases the computational process. If the value that is equally extreme is the value on the opposite side of the mean with the same area in its tail, we only need to compute the area in the tail from our actual test statistic and double it. We need not care about what specific value of test statistic is equally extreme; it is not necessary for the computation of the \(p\)-value. Note the similarities between our previous two-tailed tests.

We conclude with the logic. When conducting a two-tailed test on population variances, we compute the test statistic. We then compare the test statistic to the mean of the distribution (recall that the mean of a \(\chi^2\)-distributions is the number of degrees of freedom defining the distribution. If the test statistic is greater than the degrees of freedom, then we compute the area to the right of the test statistic and then double it to calculate the \(p\)-value. If the test statistic is less than the degrees of freedom, then we compute the area to the left of the test statistic and double it to calculate the \(p\)-value. Once we have the \(p\)-value, we proceed as in other tests.

Text Exercise \(\PageIndex{2}\)

In Text Exercise \(5.2.3\), we claimed that the heights of adult females followed a normal distribution with an average height of \(64\) inches and a standard deviation of \(2.5\) inches. A researcher thinks that the variation of adult female heights changes with time due to a combination of genetics, nutrition, and lifestyle. The researcher decides to test this claim at a level of significance of \(0.01\) by randomly sampling \(15\) adult females. Their heights are reported below.\[59,59,61,62,63,63,64,64,65,66,68,69,69,69,70\nonumber\]

Answer

The heights of adult females are known to be normally distributed and the sample was randomly selected. We can, therefore, conduct the hypothesis test. Since the researcher is interested in any difference in the variability, we will have a two-tailed test. We do not want to assume that the researcher is correct without evidence. We settle on the following hypotheses.\[\begin{align*}H_0&:\sigma^2=6.25\text{ inches}^2\\H_1&:\sigma^2\ne6.25\text{ inches}^2\end{align*}\]We now compute the sample variance from the collected data and arrive at \(s^2\approx13.4952\) square inches. We compute our test statistic.\[\chi^2_{14}\approx\frac{15-1}{6.25}\cdot13.4962\approx30.2293\nonumber\]The test statistic is larger than the degrees of freedom. We, therefore, compute the area to the right of the test statistic.\[1-\text{CHISQ.DIST(30.2293,14,1)}\approx0.007096 \nonumber\]

Figure \(\PageIndex{9}\): \(\chi^2\)-distribution

The \(p\)-value is twice the computed area which is approximately \(0.01419\). The \(p\)-value is larger than the level of significance; we conclude that there is not sufficient evidence to reject the null hypothesis. We cannot affirm the researchers' claims that the variability present in adult female heights is different than it once was with a standard deviation of \(2.5\) inches.