12.2 The F Distribution and the F-Ratio

Aligned outcomes:

Learning Objectives

By the end of this section, you will be able to:

In this section, you will learn to:

explain what the F distribution is and why it has two sets of degrees of freedom;
explain why one-way ANOVA is preferred over running many pairwise t-tests;
compute the sum of squares, degrees of freedom, and mean squares for a one-way ANOVA;
build and interpret a one-way ANOVA table, including the F-ratio;
explain why the ANOVA test is always right-tailed.

12.2.1 What Is the F Distribution?

The distribution used for the hypothesis test is a new one. It is called the F distribution, named after Sir Ronald Fisher, an English statistician. The F statistic is a ratio (a fraction). There are two sets of degrees of freedom; one for the numerator and one for the denominator.

Why a whole new distribution?

Every hypothesis test needs a distribution to measure its test statistic against. For comparing two means we used the t-distribution. When we compare three or more group means at once, the t-distribution no longer fits — so Fisher built one that does: the F distribution.

For example, if $F$ follows an F distribution and the number of degrees of freedom for the numerator is four, and the number of degrees of freedom for the denominator is ten, then $F \sim F_{4,10}$.

Try It Now 12.2.1

A study compares the fuel efficiency of four car brands using 8 cars from each brand ($n = 32$ total). Write the F distribution notation for this test, including both degrees of freedom.

Solution

Step 1 — Numerator df: $df_{\text{num}} = k - 1 = 4 - 1 = 3$.

Step 2 — Denominator df: $df_{\text{denom}} = n - k = 32 - 4 = 28$.

Answer: $F \sim F_{3,28}$.

Example 12.2.1: Writing F Distribution Notation

Suppose a one-way ANOVA compares $k = 3$ treatment groups with a total of $n = 24$ observations. Write the notation for the distribution the F statistic follows.

Solution

Step 1 — Find the numerator degrees of freedom: the numerator df come from the groups:

$$df_{\text{num}} = k - 1 = 3 - 1 = 2$$

Step 2 — Find the denominator degrees of freedom: the denominator df come from the data within the groups:

$$df_{\text{denom}} = n - k = 24 - 3 = 21$$

Step 3 — Write the notation: the numerator df go first, then the denominator df:

$$F \sim F_{2,21}$$

Answer: $F \sim F_{2,21}$ — read as "F follows an F distribution with 2 degrees of freedom in the numerator and 21 in the denominator."

12.2.2 Why ANOVA Instead of Multiple t-Tests?

The F distribution is derived from the Student's t-distribution. The values of the F distribution are squares of the corresponding values of the t-distribution. One-Way ANOVA expands the t-test for comparing more than two groups. The scope of that derivation is beyond the level of this course.

It is preferable to use ANOVA when there are more than two groups instead of performing pairwise t-tests because performing multiple tests introduces the likelihood of making a Type 1 error.

How extra tests pile up risk

Each t-test carries, say, a 5% chance of a Type 1 error. Compare five groups pairwise and you run ten tests — ten separate chances to reject a true null hypothesis. ANOVA rolls all of those comparisons into one test with one error rate.

To calculate the F ratio, two estimates of the variance are made.

Variance between samples: An estimate of $\sigma^{2}$ that is the variance of the sample means multiplied by $n$ (when the sample sizes are the same). If the samples are different sizes, the variance between samples is weighted to account for the different sample sizes. The variance is also called variation due to treatment or explained variation.
Variance within samples: An estimate of $\sigma^{2}$ that is the average of the sample variances (also known as a pooled variance). When the sample sizes are different, the variance within samples is weighted. The variance is also called the variation due to error or unexplained variation.

$SS_{\text{between}}$ = the sum of squares that represents the variation among the different samples
$SS_{\text{within}}$ = the sum of squares that represents the variation within samples that is due to chance.

To find a "sum of squares" means to add together squared quantities that, in some cases, may be weighted. We used sum of squares to calculate the sample variance and the sample standard deviation in Descriptive Statistics.

MS means "mean square." $MS_{\text{between}}$ is the variance between groups, and $MS_{\text{within}}$ is the variance within groups.

12.2.3 Calculation of Sum of Squares and Mean Square

Here are the symbols we will use throughout this section:

$k$ = the number of different groups
$n_j$ = the size of the $j^{th}$ group
$s_j$ = the sum of the values in the $j^{th}$ group
$n$ = total number of all the values combined (total sample size: $\sum n_j$)
$x$ = one value: $\sum x = \sum s_{j}$
Sum of squares of all values from every group combined: $\sum x^{2}$
Total sum of squares: $SS_{\text{total}} = \sum x^2 - \frac{(\sum x)^2}{n}$
Explained variation: sum of squares representing variation among the different samples:

$$SS_{\text{between}} = \sum \left[ \frac {(s _ {j}) ^ {2}}{n _ {j}} \right] - \frac {(\sum s _ {j}) ^ {2}}{n}$$

Unexplained variation: sum of squares representing variation within samples due to chance:

$$SS_{\text{within}} = SS_{\text{total}} - SS_{\text{between}}$$

Degrees of freedom for the different groups (df's for the numerator): $df_{\text{between}} = k - 1$
Equation for errors within samples (df's for the denominator): $df_{\text{within}} = n - k$
Mean square (variance estimate) explained by the different groups: $MS_{\text{between}} = \frac{SS_{\text{between}}}{df_{\text{between}}}$
Mean square (variance estimate) that is due to chance (unexplained): $MS_{\text{within}} = \frac{SS_{\text{within}}}{df_{\text{within}}}$

$MS_{\text{between}}$ and $MS_{\text{within}}$ can be written as follows:

$MS_{\text{between}} = \frac{SS_{\text{between}}}{df_{\text{between}}} = \frac{SS_{\text{between}}}{k - 1}$
$MS_{\text{within}} = \frac{SS_{\text{within}}}{df_{\text{within}}} = \frac{SS_{\text{within}}}{n - k}$

The one-way ANOVA test depends on the fact that $MS_{\text{between}}$ can be influenced by population differences among means of the several groups. Since $MS_{\text{within}}$ compares values of each group to its own group mean, the fact that group means might be different does not affect $MS_{\text{within}}$.

The null hypothesis says that all groups are samples from populations having the same normal distribution. The alternate hypothesis says that at least two of the sample groups come from populations with different normal distributions. If the null hypothesis is true, $MS_{\text{between}}$ and $MS_{\text{within}}$ should both estimate the same value.

The F distribution comes from the t-distribution

The values of the F distribution are squares of the corresponding values of the t-distribution, so one-way ANOVA is an expansion of the t-test to more than two groups. (The derivation itself is beyond this course.) Use ANOVA rather than a pile of pairwise t-tests, because running many tests raises the chance of a Type I error.

What "equal means" actually implies here

The null hypothesis says all the group population means are equal. Because we have already assumed the populations are normal with equal variances, equal means implies something stronger: the populations have the same normal distribution.

Try It Now 12.2.2

A researcher measures the reaction times (in seconds) of three groups of five drivers each: Group A (no distraction), Group B (radio playing), Group C (texting simulator). Without computing anything yet, state in words what $H_0$ and $H_a$ would say for this test, and identify which mean square could be inflated if $H_0$ is false — and why.

Solution

Step 1 — State the hypotheses: $H_0$: all three group population means are equal ($\mu_A = \mu_B = \mu_C$). $H_a$: at least two of the population means differ.

Step 2 — Identify the affected mean square: $MS_{\text{between}}$ would be inflated. It compares the group means to each other, so real differences between the populations push it up. $MS_{\text{within}}$ compares values inside each group to that group's own mean, so differences between populations do not affect it.

12.2.4 F-Ratio or F Statistic

$$F = \frac {M S _ {\text {between}}}{M S _ {\text {within}}}$$

If $MS_{\text{between}}$ and $MS_{\text{within}}$ estimate the same value (following the belief that $H_{0}$ is true), then the F-ratio should be approximately equal to one. Mostly, just sampling errors would contribute to variations away from one. As it turns out, $MS_{\text{between}}$ consists of the population variance plus a variance produced from the differences between the samples. $MS_{\text{within}}$ is an estimate of the population variance. Since variances are always positive, if the null hypothesis is false, $MS_{\text{between}}$ will generally be larger than $MS_{\text{within}}$. Then the F-ratio will be larger than one. However, if the population effect is small, it is not unlikely that $MS_{\text{within}}$ will be larger in a given sample.

Why F Near 1 Supports the Null Hypothesis

Think of the F-ratio as a tug-of-war score. If the only thing separating the groups is random chance, both sides pull equally hard and the ratio lands near 1. A big F means the "between-groups" side is winning — something besides chance separates the groups.

Try It Now 12.2.3

An ANOVA comparing three teaching methods produces $MS_{\text{between}} = 18.4$ and $MS_{\text{within}} = 5.2$.

a) Compute the F-ratio. b) Does this value look more consistent with $H_0$ being true or false? Explain in one sentence.

Solution

Part a:

$$F = \frac{MS_{\text{between}}}{MS_{\text{within}}} = \frac{18.4}{5.2} \approx 3.54$$

Part b: The ratio is well above 1, which suggests $H_0$ may be false — the differences between the group means are much larger than the random variation within the groups. (A formal decision still requires comparing $F = 3.54$ to a critical value from the F distribution.)

12.2.5 F-Ratio Formula When the Groups Are the Same Size

The foregoing calculations were done with groups of different sizes. If the groups are the same size, the calculations simplify somewhat and the F-ratio can be written as:

$$F = \frac {n \cdot s _ {\overline{x}} ^ {2}}{s ^ {2} _ {\text{pooled}}}$$

Here is what each symbol in the simplified formula stands for. Two different counts appear, and keeping them apart is what makes the degrees of freedom come out right:

$k$ = the number of groups
$n$ = the size of each group (they are all the same size here)
$N = nk$ = the total number of data values across all groups
$df_{\text{numerator}} = k - 1$
$df_{\text{denominator}} = N - k$
$s^2_{\text{pooled}}$ = the mean of the sample variances (pooled variance)
$s_{\overline{x}}^2$ = the variance of the sample means

Try It Now 12.2.4

Five groups each contain $n = 6$ plants, so the total sample size is $N = 30$. Find $df_{\text{numerator}}$ and $df_{\text{denominator}}$ for the F test.

Solution

Step 1 — Numerator: $df_{\text{num}} = k - 1 = 5 - 1 = 4$.

Step 2 — Denominator: $df_{\text{denom}} = N - k = 30 - 5 = 25$.

Answer: $F \sim F_{4,25}$.

Data are typically put into a table for easy viewing. One-Way ANOVA results are often displayed in this manner by computer software.

**Table 12.2.1** — The general one-way ANOVA table layout.
Source of Variation	Sum of Squares (SS)	Degrees of Freedom (df)	Mean Square (MS)	F
Factor (Between)	SS(Factor)	k − 1	MS(Factor) = SS(Factor)/(k − 1)	F = MS(Factor)/MS(Error)
Error (Within)	SS(Error)	N − k	MS(Error) = SS(Error)/(N − k)
Total	SS(Total)	N − 1

12.2.6 Notation

The notation for the F distribution is $F \sim F_{df(num), df(denom)}$

where $df(num) = df_{\text{between}}$ and $df(denom) = df_{\text{within}}$

The mean for the F distribution is $\mu = \frac{df(denom)}{df(denom) - 2}$

Example 12.2.2: Three Diet Plans

Three different diet plans are tested for mean weight loss. The entries in the table are the weight losses for the different plans. The one-way ANOVA results are shown in Table 12.2.2.

**Table 12.2.2** — Weight loss (in pounds) under three diet plans.
Plan 1: $n_{1} = 4$	Plan 2: $n_{2} = 3$	Plan 3: $n_{3} = 3$
5	3.5	8
4.5	7	4
4	4.5	3.5
3

The three group totals are

$$T _ {1} = 16.5, \quad T _ {2} = 15, \quad T _ {3} = 15.5$$

where $T _ {j}$ is the sum of the values in group $j$. (The original text writes these as $s _ {j}$; we use $T$ here so they are not mistaken for standard deviations.)

Following are the calculations needed to fill in the one-way ANOVA table. The table is used to conduct a hypothesis test.

Solution

Step 1 — Compute SS(between):

$$SS(\text{between}) = \sum \left[ \frac {(T _ {j}) ^ {2}}{n _ {j}} \right] - \frac {\left(\sum T _ {j}\right) ^ {2}}{N} = \frac {T _ {1} ^ {2}}{4} + \frac {T _ {2} ^ {2}}{3} + \frac {T _ {3} ^ {2}}{3} - \frac {(T _ {1} + T _ {2} + T _ {3}) ^ {2}}{10}$$

where $n _ {1} = 4, n _ {2} = 3, n _ {3} = 3$ and $N = n _ {1} + n _ {2} + n _ {3} = 10$

$$= \frac {(16.5) ^ {2}}{4} + \frac {(15) ^ {2}}{3} + \frac {(15.5) ^ {2}}{3} - \frac {(47) ^ {2}}{10} = 68.0625 + 75 + 80.0833 - 220.9$$ $$SS(\text{between}) = 2.2458$$

Step 2 — Compute SS(total):

$$SS(\text{total}) = \sum x ^ {2} - \frac {\left(\sum x\right) ^ {2}}{N}$$ $$= \left(5 ^ {2} + 4.5 ^ {2} + 4 ^ {2} + 3 ^ {2} + 3.5 ^ {2} + 7 ^ {2} + 4.5 ^ {2} + 8 ^ {2} + 4 ^ {2} + 3.5 ^ {2}\right) - \frac {(5 + 4.5 + 4 + 3 + 3.5 + 7 + 4.5 + 8 + 4 + 3.5) ^ {2}}{10}$$ $$= 244 - \frac {47 ^ {2}}{10} = 244 - 220.9$$ $$SS(\text{total}) = 23.1$$

Step 3 — Compute SS(within):

$$SS(\text{within}) = SS(\text{total}) - SS(\text{between}) = 23.1 - 2.2458$$ $$SS(\text{within}) = 20.8542$$

Answer: SS(between) = 2.2458, SS(total) = 23.1, SS(within) = 20.8542.

Figure 12.2.1 — TI calculator ANOVA output for the diet plan data

Figure 12.2.1 — TI calculator ANOVA output for the diet plan data.

12.2.7 Using the TI-83, 83+, 84, 84+ Calculator

One-Way ANOVA Table: The formulas for SS(Total), SS(Factor) = SS(Between) and SS(Error) = SS(Within) are shown previously. The same information is provided by the TI calculator hypothesis test function ANOVA in STAT TESTS (syntax is ANOVA(L1, L2, L3) where L1, L2, L3 have the data from Plan 1, Plan 2, Plan 3 respectively).

**Table 12.2.3** — Completed one-way ANOVA table for the diet plan data.
Source of Variation	Sum of Squares (SS)	Degrees of Freedom (df)	Mean Square (MS)	F
Factor (Between)	SS(Factor) = SS(Between) = 2.2458	k − 1 = 3 groups − 1 = 2	MS(Factor) = SS(Factor)/(k − 1) = 2.2458/2 = 1.1229	F = MS(Factor)/MS(Error) = 1.1229/2.9792 = 0.3769
Error (Within)	SS(Error) = SS(Within) = 20.8542	N − k = 10 total data − 3 groups = 7	MS(Error) = SS(Error)/(N − k) = 20.8542/7 = 2.9792
Total	SS(Total) = 2.2458 + 20.8542 = 23.1	N − 1 = 10 total data − 1 = 9

Figure 12.2.2 — TI calculator screen showing the completed ANOVA results

Figure 12.2.2 — TI calculator screen showing the completed ANOVA results.

Notice what the completed table shows here: $F = 0.3769$, far below 1. That tells us the weight losses do not differ noticeably among the three diet plans — the between-group variation is actually smaller than the chance variation within groups.

Try It Now 12.2.5

As part of an experiment to see how different types of soil cover would affect slicing tomato production, Marist College students grew tomato plants under different soil cover conditions. Groups of three plants each had one of the following treatments:

bare soil
a commercial ground cover
black plastic
straw
compost

All plants grew under the same conditions and were the same variety. Students recorded the weight (in grams) of tomatoes produced by each of the $n = 15$ plants:

**Table 12.2.4** — Tomato yield (in grams) under five soil cover treatments.
Bare: $n_{1} = 3$	Ground Cover: $n_{2} = 3$	Plastic: $n_{3} = 3$	Straw: $n_{4} = 3$	Compost: $n_{5} = 3$
2,625	5,348	6,583	7,285	6,277
2,997	5,682	8,560	6,897	7,818
4,915	5,482	3,830	9,230	8,677

Create the one-way ANOVA table.

Solution

Step 1 — Find the group sums: $s_1 = 2625 + 2997 + 4915 = 10{,}537$; $s_2 = 5348 + 5682 + 5482 = 16{,}512$; $s_3 = 6583 + 8560 + 3830 = 18{,}973$; $s_4 = 7285 + 6897 + 9230 = 23{,}412$; $s_5 = 6277 + 7818 + 8677 = 22{,}772$. Here $k = 5$, $n = 15$, and $\sum s_j = 92{,}206$.

Step 2 — Compute SS(between):

$$SS(\text{between}) = \sum \left[ \frac{s_j^2}{n_j} \right] - \frac{\left(\sum s_j\right)^2}{n} = \frac{10537^2 + 16512^2 + 18973^2 + 23412^2 + 22772^2}{3} - \frac{92206^2}{15}$$

Each squared group sum divided by 3 gives $37{,}009{,}456.33 + 90{,}882{,}048 + 119{,}991{,}576.33 + 182{,}707{,}248 + 172{,}854{,}661.33 = 603{,}444{,}989.99$, and $\frac{92206^2}{15} = 566{,}796{,}429.07$. Therefore

$$SS(\text{between}) = 603{,}444{,}989.99 - 566{,}796{,}429.07 = 36{,}648{,}560.92$$

Step 3 — Compute SS(total): $\sum x^2 = 623{,}891{,}716$, so

$$SS(\text{total}) = 623{,}891{,}716 - 566{,}796{,}429.07 = 57{,}095{,}286.93$$

Step 4 — Compute SS(within):

$$SS(\text{within}) = SS(\text{total}) - SS(\text{between}) = 57{,}095{,}286.93 - 36{,}648{,}560.92 = 20{,}446{,}726.01$$

Step 5 — Fill in the ANOVA table:

Source of Variation	Sum of Squares	Degrees of Freedom	Mean Square	F
Factor (Between)	36,648,560.92	k − 1 = 4	9,162,140.23	F = 9,162,140.23 / 2,044,672.60 ≈ 4.48
Error (Within)	20,446,726.01	n − k = 10	2,044,672.60
Total	57,095,286.93	n − 1 = 14

Answer: The completed one-way ANOVA table gives $F \approx 4.48$ with $df(4, 10)$.

The one-way ANOVA hypothesis test is always right-tailed because larger F-values are way out in the right tail of the F-distribution curve and tend to make us reject $H_{0}$.

Problem Set 12.2

Problem 1. Explain in one sentence what the F distribution is used for.

Solution

Step 1 — Identify the purpose: The F distribution is the probability distribution used to measure the F-ratio in a one-way ANOVA test, which compares three or more group means.

Answer: The F distribution is used for hypothesis tests (one-way ANOVA) that compare the means of three or more groups.

Problem 2. Write the notation for an F distribution with 3 numerator degrees of freedom and 22 denominator degrees of freedom.

Solution

Step 1 — Apply the notation convention: The numerator degrees of freedom are written first, then the denominator degrees of freedom, in the form $F \sim F_{df(num),\, df(denom)}$.

Answer: $F \sim F_{3,22}$

Problem 3. State what $MS_{\text{between}}$ measures and what $MS_{\text{within}}$ measures, in your own words.

Solution

Step 1 — Interpret $MS_{\text{between}}$: It is the variance estimate that measures how far apart the group means are from each other — the variation due to treatment (explained variation).

Step 2 — Interpret $MS_{\text{within}}$: It is the variance estimate that measures how much individual values vary inside their own groups due to chance — the variation due to error (unexplained variation).

Answer: $MS_{\text{between}}$ measures variation among the group means (treatment/explained variation); $MS_{\text{within}}$ measures chance variation within the groups (error/unexplained variation).

Problem 4. Explain why a difference among the group means inflates $MS_{\text{between}}$ but leaves $MS_{\text{within}}$ alone.

Solution

Step 1 — Look at what each mean square compares: $MS_{\text{between}}$ is built from how far each group's mean sits from the overall mean, so if the group means differ, those differences directly inflate it. $MS_{\text{within}}$ compares each value only to its own group's mean, so it never "sees" differences between groups.

Answer: Real differences between population means push the group means apart and inflate $MS_{\text{within}}$ not at all, because within-group deviations are measured against each group's own mean, which shifts along with its population.

Problem 5. An ANOVA gives $MS_{\text{between}} = 42.6$ and $MS_{\text{within}} = 14.2$. Compute F.

Solution

Step 1 — Apply the F-ratio formula:

$$F = \frac{MS_{\text{between}}}{MS_{\text{within}}} = \frac{42.6}{14.2} = 3$$

Answer: $F = 3$

Problem 6. An ANOVA gives $MS_{\text{between}} = 3.1$ and $MS_{\text{within}} = 9.4$. Compute F, and say whether the result leans toward or against $H_0$.

Solution

Step 1 — Compute the ratio:

$$F = \frac{3.1}{9.4} \approx 0.33$$

Step 2 — Interpret: An F well below 1 means the between-group variation is smaller than the chance variation within groups, which leans toward $H_0$ being true.

Answer: $F \approx 0.33$; this result leans toward $H_0$ (no difference among the group means).

Problem 7. Explain why an F-ratio near 1 is what we expect when $H_0$ is true.

Solution

Step 1 — Recall what each MS estimates under $H_0$: If all populations share the same normal distribution, both $MS_{\text{between}}$ and $MS_{\text{within}}$ estimate the same population variance $\sigma^2$, so their ratio should be about $\sigma^2/\sigma^2 = 1$.

Step 2 — Note the role of sampling error: Only random sampling error makes F deviate from 1, so small departures above or below 1 are expected by chance.

Answer: When $H_0$ is true, both mean squares estimate the same $\sigma^2$, so their ratio should be approximately 1, with only sampling error causing deviation.

Problem 8. Four groups of seven subjects each take part in a study. Find $k$, $N$, $df_{\text{numerator}}$ and $df_{\text{denominator}}$.

Solution

Step 1 — Identify $k$ and $N$: There are $k = 4$ groups, each of size 7, so $N = nk = 7 \times 4 = 28$.

Step 2 — Compute the degrees of freedom:

$$df_{\text{numerator}} = k - 1 = 4 - 1 = 3$$ $$df_{\text{denominator}} = N - k = 28 - 4 = 24$$

Answer: $k = 4$, $N = 28$, $df_{\text{numerator}} = 3$, $df_{\text{denominator}} = 24$ (so $F \sim F_{3,24}$).

Problem 9. A study has groups of sizes 6, 8, and 5. Find $df_{\text{numerator}}$ and $df_{\text{denominator}}$.

Solution

Step 1 — Count groups and total size: $k = 3$ groups with sizes $n_1 = 6$, $n_2 = 8$, $n_3 = 5$, so $N = 6 + 8 + 5 = 19$.

Step 2 — Compute the degrees of freedom:

$$df_{\text{numerator}} = k - 1 = 3 - 1 = 2$$ $$df_{\text{denominator}} = N - k = 19 - 3 = 16$$

Answer: $df_{\text{numerator}} = 2$, $df_{\text{denominator}} = 16$ (so $F \sim F_{2,16}$).

Problem 10. A classmate computes $df_{\text{denominator}}$ for five groups of six as $6 - 5 = 1$. Explain the mistake and give the correct value.

Solution

Step 1 — Identify the mistake: The classmate used the wrong quantities in the formula. The denominator degrees of freedom are $N - k$, where $N$ is the total number of data values across all groups, not the size of one group.

Step 2 — Compute correctly: With five groups of six, $N = 5 \times 6 = 30$ and $k = 5$:

$$df_{\text{denominator}} = N - k = 30 - 5 = 25$$

Answer: The mistake was using one group's size (6) instead of the total sample size $N = 30$; the correct value is $df_{\text{denominator}} = 25$.

Problem 11. Three groups have totals $T_1 = 20$, $T_2 = 24$, $T_3 = 16$ with $n_1 = n_2 = n_3 = 4$. Compute $SS(\text{between})$.

Solution

Step 1 — Set up the formula: With equal group sizes,

$$SS(\text{between}) = \sum \left[ \frac{T_j^2}{n_j} \right] - \frac{\left(\sum T_j\right)^2}{N}$$

where $N = 4 + 4 + 4 = 12$ and $\sum T_j = 20 + 24 + 16 = 60$.

Step 2 — Substitute and compute:

$$SS(\text{between}) = \frac{20^2}{4} + \frac{24^2}{4} + \frac{16^2}{4} - \frac{60^2}{12} = 100 + 144 + 64 - 300$$ $$SS(\text{between}) = 308 - 300 = 8$$

Answer: $SS(\text{between}) = 8$

Problem 12. For the data in Problem 12.2.11, $SS(\text{total}) = 38$. Find $SS(\text{within})$, then both mean squares and F.

Solution

Step 1 — Find SS(within):

$$SS(\text{within}) = SS(\text{total}) - SS(\text{between}) = 38 - 8 = 30$$

Step 2 — Find the degrees of freedom: From Problem 12.2.11, $k = 3$ and $N = 12$:

$$df_{\text{between}} = k - 1 = 2, \qquad df_{\text{within}} = N - k = 9$$

Step 3 — Compute the mean squares:

$$MS_{\text{between}} = \frac{8}{2} = 4, \qquad MS_{\text{within}} = \frac{30}{9} \approx 3.33$$

Step 4 — Compute F:

$$F = \frac{4}{3.33} \approx 1.20$$

Answer: $SS(\text{within}) = 30$, $MS_{\text{between}} = 4$, $MS_{\text{within}} \approx 3.33$, and $F \approx 1.20$.

Problem 13. Explain why the one-way ANOVA test is always right-tailed.

Solution

Step 1 — Recall how F behaves under each hypothesis: If $H_0$ is false, $MS_{\text{between}}$ picks up extra variation from real differences between the population means, making it larger than $MS_{\text{within}}$ and pushing F above 1.

Step 2 — Connect to the tail of the distribution: Large F-values lie far out in the right tail of the F distribution, and evidence against $H_0$ always comes in the form of a large F — never a small one.

Answer: The ANOVA test is always right-tailed because evidence against $H_0$ appears as an unusually large F-ratio, and large F-values sit in the right tail of the F distribution.

Problem 14. The mean of an F distribution is $\frac{df(denom)}{df(denom) - 2}$. Compute the mean of $F_{4, 12}$, and explain why the formula needs $df(denom) > 2$.

Solution

Step 1 — Apply the mean formula:

$$\mu = \frac{df(denom)}{df(denom) - 2} = \frac{12}{12 - 2} = \frac{12}{10} = 1.2$$

Step 2 — Explain the restriction: If $df(denom) = 2$, the denominator becomes zero and the mean is undefined; if $df(denom) < 2$, it would be negative, which is impossible for a mean. So the formula requires $df(denom) > 2$.

Answer: The mean of $F_{4,12}$ is $\mu = 1.2$. The formula needs $df(denom) > 2$ because otherwise the denominator would be zero or negative, making the mean undefined or meaningless.

Key Terms

F distribution — the probability distribution used for one-way ANOVA; named after Sir Ronald Fisher, with two sets of degrees of freedom.

variance between samples — an estimate of $\sigma^2$ based on how far apart the sample means are; also called explained variation or variation due to treatment.

variance within samples — an estimate of $\sigma^2$ equal to the average of the sample variances (pooled variance); also called unexplained variation or variation due to error.

sum of squares (SS) — the total obtained by adding squared quantities, measuring variation.

mean square (MS) — a sum of squares divided by its degrees of freedom; a variance estimate.

F-ratio (F statistic) — the ratio $F = MS_{\text{between}} / MS_{\text{within}}$; approximately 1 when $H_0$ is true, larger when it is false.