OPTIMA Statistics

Subsection 6.3.1 Two Types of Hypotheses

In this lesson, we will actually look at two types of hypothesis tests. The only difference between these two tests is the way in which the null hypothesis is stated. Because of this, the difference may seem like a technicality. However, it is important that we recognize the difference between these two types of tests.

The difference between these two tests has to do with the number of populations. If we are looking at a single characteristic across different populations, then we are conducting a test of homogeneity. On the other hand, if we are looking at the relationship between two characteristics in the same population, we need to conduct a test of independence.

In both cases, the alternative hypothesis is simply that the distribution is not independent or homogeneous. In the following examples, we will take a look at the hypotheses for each of these tests.

Subsection 6.3.2 Observed and Expected Counts

No matter which type of hypothesis test we are doing, a test of homogeneity or a test of independence, the underlying assumption of the null hypothesis is the same. That assumption is that the rows and columns of our contingency table are independent. Put another way, the probability of being in a certain row is the same no matter what column you are in. Let's explore that idea with the following table.

Our goal is to find out what counts we would expect in each entry if we assume that the row and column probabilities are independent. Recall that we saw a test for independence in Section 2.4. It stated that two events, \(A\) and \(B\text{,}\) are independent if and only if

Suppose that \(A\) is the event of being in column \(i\) and \(B\) is the event of being in column \(j\text{.}\) Using the table above, we can figure out what these probabilities are.

If these two events are independent, then the probability of being in the single cell in both row \(i\) and column \(j\) should be

This gives us the probability of being in the cell in column \(i\) and row \(j\text{.}\) To get the expected number of counts, we multiply that probability by the total number of individuals, \(T\text{.}\) This gives the following formulas for computing the expected counts in a contingency table.

Let's practice finding the expected counts using one of our previously seen examples.

Pay special attention to the third step in the process outlined in the example above. The sum of the observed and expected counts in any row or columnn should be the same. In fact, this can be used to help reduce the number of computations you have to perform. If you find the expected count in all but one of the cells in a row or column, you can subtract those expected count from the row or column total, and what is left must be the expected count in the final cell.

Subsection 6.3.3 The Test Statistic

As in the goodness-of-fit test, a test of homogeneity or independence is based on measuring the differences between what we observed in our sample, and what we expected if the null hypothesis is really true. We just saw how to use the observed counts in a ontingency table to find the expected counts for each entry in that table. The test statistic is then computed as shown below.

Don't let the double summation sign confuse you. All that this means is that we add together the observed count minus the expected count squared divided by the expected count for every cell in the contingency table. Below we perform this computation on the table we found in Example 6.3.15.

Subsection 6.3.4 Traditional Tests

The last step in conducting a test of homogeneity or independence is to find the appropriate critical value from the \(\chi^2\)-distribution table. To do that, we need the number of degrees of freedom. Because we are dealing with contingency tables, instead of the categories we saw in the goodness-of-fit test, the degrees of freedom must be computed differently.

Thus, if we take one less than the number of columns times one less than the number of rows, we have the degrees of freedom to use when looking up our critical value. This process is shown in the following examples.

A student-body representative wishes to determine if a student's class standing affects their housing arrangements. To test this, the following data is collected.

	Freshman	Sophomore	Junior	Senior
Dorm	114	99	90	78
Off-Campus College-Owned Housing	42	83	104	124
Off-Campus Other Housing	16	19	17	21

Use a \(\chi^2\) test at the \(\alpha = 0.01\) significance level to make this determination.

Solution

Recall that our hypotheses are:

\begin{align*} H_0\amp:\ \text{Housing distributions in the four populations are homogeneous}\\ H_A\amp:\ \text{Housing distributions are not homogeneous across the four populations} \end{align*}

which means that the alternative is that they are not homogeneous, or in other words, that students in different class standings have different types of housing.

Our test statistic from Example 6.3.25 was:

\begin{equation*} \chi^2_{\text{test}} = 45.35\text{.} \end{equation*}

At the 0.01 significance level, with \((4-1)(3-1) = 6\) degrees of freedom, the critical value from the \(\chi^2\)-distribution table is \(\chi^2_{0.01} = 16.812\) as shown below.

Since our test statistic is further into the right tail than the critical value of 16.812, we must reject the null hypothesis. There is highly significant evidence that students with different class standing live in different types of housing.

An efficiency expert at a certain large company believes that the number of years experience that an employee has is independent from their efficiency rating. To test this claim, the following data is collected.

	\(\lt\) 5 Years	5-15 Years	15+ Years
Low Efficiency	21	14	11
Moderate Efficiency	45	28	24
High Efficiency	62	38	32

Conduct a test of independence at the \(\alpha = 0.05\) significance level.

Solution

You may recall that the hypotheses for this test are:

\begin{align*} H_0\amp:\ \text{Efficiency and years of experience are independent characteristics}\\ H_A\amp:\ \text{These characteristics are dependent} \end{align*}

Next, we compute the expected counts.

	\(\lt\) 5 Years		5-15 Years		15+ Years		Total
Low Efficiency	20		15		11		46
		21.41		13.38		11.20
Moderate Efficiency	43		27		27		97
		45.15		28.22		23.63
High Efficiency	65		38		29		132
		61.44		38.40		32.16
Total	128		80		67		275

And then we compute the \(\chi^2\) test statistic using this table.

\begin{align*} \chi^2_{\text{test}} \amp= \frac{(20-21.41)^2}{21.41} + \frac{(15-13.38)^2}{13.38} + \frac{(11-11.20)^2}{11.20}\\ \amp+ \frac{(43-45.15)^2}{45.15} + \frac{(27-28.22)^2}{28.22} + \frac{(27-23.63)^2}{23.63}\\ \amp+ \frac{(65-61.44)^2}{61.44} + \frac{(38.38.40)^2}{38.40} + \frac{(29-32.16)^2}{32.16}\\ \amp\approx 1.449 \end{align*}

At the \(\alpha = 0.05\) significance level with \(df = (3-1)(3-1) = 4\text{,}\) the critical value from the \(\chi^2\)-distribution is 9.488. The distribution and test statistic are shown below.

Since our test statistic is not further into the tail than the critical value of \(\chi^2_{0.05} = 9.488\text{,}\) we must fail to reject the null hypothesis. There is no statistically significant evidence that efficiency and years worked are related. These two variables appear to be independent.