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F on a Test (ANOVA F-Test Statistic)

In statistics, “f on a test” can refer to the F-statistic for an F-test. For a one-way ANOVA with three groups (n=5 each), sample means 72, 78, 84 and sample variances 16, 20, 18, compute the ANOVA test statistic \(F\) and decide at \(\alpha=0.05\) given \(F_{0.95}(2,12)=3.89\).

Subject: Statistics Chapter: Analysis of Variance Topic: One Way Analysis of Variance, Calculating the Test Statistic F Answer included
f on a test F-test ANOVA test statistic one-way analysis of variance mean square between mean square within MSB MSW SSB SSW
Accepted answer Answer included

Problem

In statistics, the phrase “f on a test” often means the F-statistic used in an F-test. Consider a one-way ANOVA with three groups (each with \(n=5\)). The sample means are \(72\), \(78\), \(84\) and the sample variances are \(16\), \(20\), \(18\). Compute the ANOVA test statistic \(F\). At significance level \(\alpha=0.05\), decide using the critical value \(F_{0.95}(2,12)=3.89\).


Step 1: state the one-way ANOVA F-test structure

\[ F=\frac{\text{MSB}}{\text{MSW}} \]

\[ \text{MSB}=\frac{\text{SSB}}{k-1},\qquad \text{MSW}=\frac{\text{SSW}}{N-k} \]

Here \(k\) is the number of groups and \(N\) is the total sample size. For this problem, \(k=3\) and \(N=5+5+5=15\).

Step 2: compute the overall mean

With equal group sizes, the overall mean \(\bar{y}\) is the average of the group means:

\[ \bar{y}=\frac{72+78+84}{3}=\frac{234}{3}=78 \]

Step 3: compute SSB (between-groups sum of squares)

For equal group size \(n=5\),

\[ \text{SSB}=\sum_{j=1}^{k} n\left(\bar{y}_j-\bar{y}\right)^2 \]

\[ \text{SSB}=5\left[(72-78)^2+(78-78)^2+(84-78)^2\right] \]

\[ \text{SSB}=5\left[36+0+36\right]=5\times 72=360 \]

Step 4: compute SSW (within-groups sum of squares)

Using sample variances \(s_j^2\),

\[ \text{SSW}=\sum_{j=1}^{k}(n-1)s_j^2 \]

\[ \text{SSW}=4\times 16+4\times 20+4\times 18=4\times (16+20+18)=4\times 54=216 \]

Step 5: compute degrees of freedom, mean squares, and \(F\)

\[ \text{df}_{B}=k-1=3-1=2,\qquad \text{df}_{W}=N-k=15-3=12 \]

\[ \text{MSB}=\frac{360}{2}=180,\qquad \text{MSW}=\frac{216}{12}=18 \]

\[ F=\frac{180}{18}=10 \]

The F-statistic in a one-way ANOVA is always nonnegative because it is a ratio of mean squares; a negative “F on a test” value indicates a computation or transcription error.

Step 6: decision using the critical value

The rejection rule for a right-tailed ANOVA F-test is: reject \(H_0\) if \(F \ge F_{0.95}(2,12)\).

\[ 10 \ge 3.89 \quad \Rightarrow \quad \text{Reject } H_0 \text{ at } \alpha=0.05 \]

Conclusion: the data provide sufficient evidence that at least one population mean differs among the three groups.

ANOVA summary table

Source SS df MS F
Between (Treatments) \(360\) \(2\) \(180\) \(10\)
Within (Error) \(216\) \(12\) \(18\)
Total \(576\) \(14\)

Visualization: right-tail F-test region and the observed “F on a test” value

F density \(F_{0.95}(2,12)=3.89\) \(F=10\) right-tail rejection region 0 2 3.89 10
The shaded region is the ANOVA F-test rejection region to the right of the critical value \(3.89\). The observed “F on a test” value \(F=10\) lies far into that region, leading to rejection of \(H_0\) at \(\alpha=0.05\).
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