11.4: F-Tests in One-Way ANOVA (2024)

In Chapter 9, we saw how to compare two population means \(\mu _1\) and \(\mu _2\).

In this section we will learn to compare three or more population means at the same time, which is often of interest in practical applications. For example, an administrator at a university may be interested in knowing whether student grade point averages are the same for different majors. In another example, an oncologist may be interested in knowing whether patients with the same type of cancer have the same average survival times under several different competing cancer treatments.

In general, suppose there are \(K\) normal populations with possibly different means, \(μ_1 , μ_2 , \ldots, μ_K\), but all with the same variance \(σ^2\). The study question is whether all the \(K\) population means are the same. We formulate this question as the test of hypotheses

\[H_0: \mu _1=\mu _2=\cdots =\mu _K\\ vs.\\ H_a: \text{not all K population means are equal} \nonumber \]

To perform the test \(K\) independent random samples are taken from the \(K\) normal populations. The \(K\) sample means, the \(K\) sample variances, and the \(K\) sample sizes are summarized in the table:

Define the following quantities:

The mean of the combined sample of all \(n\) observations:

\[x= \dfrac{\displaystyle Σx}{n} = \dfrac{n_1 \overline{x} + n_2 \overline{x}_2 + \ldots + n_K \overline{x}_K}{n} \nonumber \]

The mean square for treatment:

\[MST=\dfrac{n_1(\overline{x}_1−\overline{x})^2 + n_2(\overline{x}_2−\overline{x})^2 + \ldots  + n_K (\overline{x}K−\overline{x})^2}{K−1} \nonumber \]

The mean square for error:

\[MSE= \dfrac{(n_1−1)s^2_1 + (n_2−1)s^2_2 + \ldots + (n_{K−1})s^2_K}{n−K} \nonumber \]

\(MST\) can be thought of as the variance between the \(K\) individual independent random samples and \(MSE\) as the variance within the samples. This is the reason for the name “analysis of variance,” universally abbreviated ANOVA. The adjective “one-way” has to do with the fact that the sampling scheme is the simplest possible, that of taking one random sample from each population under consideration. If the means of the \(K\) populations are all the same then the two quantities \(MST\) and \(MSE\) should be close to the same, so the null hypothesis will be rejected if the ratio of these two quantities is significantly greater than \(1\). This yields the following test statistic and methods and conditions for its use.

As always the test is performed using the usual five-step procedure.

The average of all \(44\) observations is (after rounding for simplicity) \(\overline{x}=3.15\). We compute (rounding for simplicity)

\[\begin{align} MST &= \dfrac{11(2.90−3.15)^2+11(3.34−3.15)^2+11(3.36−3.15)^2+11(3.02−3.15)^2}{4−1} \nonumber \\[6pt] &=\dfrac{1.7556}{3} \nonumber \\[6pt] &=0.585 \nonumber \end{align} \nonumber \]

and

\[\begin{align} MSE &= \dfrac{(11−1)(0.188)+(11−1)(0.148)+(11−1)(0.229)+(11−1)(0.157)}{44−4} \nonumber \\[6pt] &=\dfrac{7.22}{40} \nonumber \\[6pt] &=0.181 \nonumber \end{align} \nonumber \]

so that

\[F=\dfrac{MST}{MSE}=\dfrac{0.585}{0.181}=3.232 \nonumber \]

• Step 4. The test is right-tailed. The single critical value is (since \(df_1=3\) and \(df_2=40\)) \(F_\alpha =F_{0.05}=2.84\). Thus the rejection region is \([2.84,\infty )\), as illustrated in Figure \(\PageIndex{1}\).

• Step 5. Since \(F=3.232>2.84\), we reject \(H_0\). The data provide sufficient evidence, at the \(5\%\) level of significance, to conclude that the averages of major GPAs for the four majors considered are not all equal.

The average of all \(33\) observations is (after rounding for simplicity) \(\overline{x}=73.42\). We compute (rounding for simplicity)

\[\begin{align*} MST &= \frac{16(69.75-73.42)^2+9(77.78-73.42)^2+8(75.88-73.42)^2}{31}\\ &= \frac{434.63}{2}\\ &= 217.50 \end{align*} \nonumber \]

and

\[\begin{align*} MSE &= \frac{(16-1)(34.47)+(9-1)(52.69)+(8-1)(30.69)}{33-3}\\ &= \frac{1153.4}{30}\\ &= 38.45\end{align*} \nonumber \]

so that

\[F=\dfrac{MST}{MSE}=\dfrac{217.50}{38.45}=5.65 \nonumber \]

• Step 4. The test is right-tailed. The single critical value is \(F_\alpha =F_{0.01}=5.39\). Thus the rejection region is \([5.39,\infty )\), as illustrated in Figure \(\PageIndex{2}\).

• Step 5. Since \(F=5.65>5.39\), we reject \(H_0\). The data provide sufficient evidence, at the \(1\%\) level of significance, to conclude that a treatment effect exists at least for one of the two treatments in increasing the mean survival time of mice with thymic leukemia.

It is important to to note that, if the null hypothesis of equal population means is rejected, the statistical implication is that not all population means are equal. It does not however tell which population mean is different from which. The inference about where the suggested difference lies is most frequently made by a follow-up study.