The chapter's counterpart to section 11.1: a short list of properties, followed by the section where the machinery is actually used. The five facts are that the curve is skewed right rather than symmetrical, that there is a different curve for each set of degrees of freedom, that the F statistic is always at least zero, that the curve approximates the normal as both degrees of freedom grow, and that the distribution has other uses including comparing two variances. Those first four parallel chi-square's facts almost word for word, which is no coincidence — an F statistic is a ratio of quantities built from sums of squares, so it inherits non-negativity and right skew from the same source. The section then works the chapter's three complete tests: tomato yields under five mulches, which reject at 5 percent; sorority grade means, which do not reject at 1 percent; and bean plant heights, which do not reject at 3 percent and are computed by the balanced shortcut.
Subject: Statistics · 65 slides · symbolic lesson
Open the interactive version of this deck
Title
Statistics · Chapter 13 — F Distribution and One-Way ANOVA
Facts About the F Distribution
Objectives
Six outcomes, and three of them are complete tests.
OpenStax Introductory Statistics 2e, §13.3 Facts About the F Distribution §13.3, pp. 684-689 — the section these objectives are drawn from
Warm-up
Section 13.2 built the F ratio; section 11.1 listed the facts about a different skewed distribution.
Discussion prompt
Without looking, predict three properties of the F distribution.
Hint: Ask what an F statistic is made of, and compare with chi-square.
Answer:
It cannot be negative, since both mean squares are built from sums of squares and a ratio of non-negative quantities is non-negative. It is skewed right, since it is bounded below at zero and unbounded above — the same argument section 11.1 made for chi-square.
And there is a different curve for every pair of degrees of freedom, since the ratio's behaviour depends on how many quantities went into each of its two parts.
\[ F \ge 0, \quad \text{skewed right}, \quad \text{one curve per } (df_1, df_2) \]
All three are on the book's list, and the fourth — that the curve approaches the normal as both degrees of freedom grow — follows the same pattern chi-square did. What is genuinely new is having TWO degrees of freedom rather than one, and the order in which they are written.
Concept
The curve is not symmetrical but skewed to the right; there is a different curve for each set of degrees of freedom; the F statistic is greater than or equal to zero; and as the degrees of freedom for the numerator and denominator get larger, the curve approximates the normal. The one-way ANOVA hypothesis test is always right-tailed.
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 the null. Differing means can only add to the numerator, so only large values are evidence.
\[ F \ge 0; \qquad p\text{-value} = P(F > F_{\text{obs}}) \]
The fifth fact points in two directions. Comparing two variances is section 13.4, so the chapter uses the same distribution twice; two-way analysis of variance is named and set aside as beyond the scope of this chapter. Between them they explain why the F distribution is worth a chapter rather than a paragraph.
Figure (svg): The book's five facts about the F distribution
OpenStax Introductory Statistics 2e, §13.3 Facts About the F Distribution §13.3, p. 684
Section
Section 1
Concept
The F statistic is greater than or equal to zero, so the curve begins at zero; it is not symmetrical but skewed to the right; and there is a different curve for each set of degrees of freedom. As both degrees of freedom get larger, the curve approximates the normal.
two degrees of freedom — Unlike chi-square's single parameter, an F distribution is indexed by a pair — the numerator's and the denominator's — and swapping them gives a different curve.
\[ F \sim F_{df_1, df_2}, \qquad F \ge 0 \]
Non-negativity follows immediately from the construction. Both mean squares are sums of squared quantities divided by positive counts, so both are non-negative, and a ratio of non-negative numbers cannot be negative. A negative F is therefore always an arithmetic error, exactly as a negative chi-square was in section 11.1.
Figure (svg): Four F curves of increasing degrees of freedom, each starting at zero, skewed right and peaking near one
OpenStax Introductory Statistics 2e, §13.3 Facts About the F Distribution §13.3, p. 684 — the five facts
Picture it
At increasing degrees of freedom.
Figure (svg): Four F curves of increasing degrees of freedom, each starting at zero, skewed right and peaking near one
Every curve begins at zero and peaks near one, which is what the null predicts. The two-and-six curve is severely skewed and the thirty-and-sixty curve is nearly symmetric, illustrating the fourth fact across the range.
Worked example
Tracing it back to the two mean squares.
\[ \text{can } F \text{ be } -2? \]
MSbetween
Why: A sum of squares over k-1.
MSwithin
Why: A sum of squares over n-k.
Their ratio
Why: Non-negative over positive.
Conclude
Why: Never negative.
Figure (svg): The solution to Worked example why F cannot be negative shown as a ladder of expressions, one row per legal move
\[ F = \frac{\text{MS}_{\text{between}}}{\text{MS}_{\text{within}}} \ge 0 \]
Verify: confirm this gives a free check on any computation
Why: A negative F in an output is always an error rather than an unusual result, exactly as a negative chi-square was in section 11.1. The likeliest cause is a sign slip in the correction term, since it is the only subtraction in the whole sum-of-squares calculation — and if SS between comes out negative, that subtraction is where to look.
OpenStax Introductory Statistics 2e, §13.3 Facts About the F Distribution §13.3, p. 684
Two truths and a lie
All three concern the facts.
Eliminate the wrong options
Two are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is false and contradicts the first fact. The curve is not symmetrical but skewed to the right — bounded below at zero and unbounded above, so symmetry is impossible.
Worked example
Section 11.1's list against this one.
\[ \text{how much is shared?} \]
Skewed right
Why: Both.
Never negative
Why: Both.
A curve per df
Why: Both.
Approaches normal
Why: Both.
Number of df
Why: One against two.
Figure (svg): The solution to Worked example comparing the facts with chi-square's shown as a ladder of expressions, one row per legal move
\[ \chi^2_{df} \quad\text{against}\quad F_{df_1, df_2} \]
Verify: confirm why the two distributions are so alike
Why: Both are built from sums of squared quantities, and squaring is what forces non-negativity and produces the right skew — a distribution bounded below at zero with no upper bound cannot be symmetric. The F is in fact a ratio of two scaled chi-square quantities, which is why it inherits both properties and why it needs two degrees of freedom, one for each.
OpenStax Introductory Statistics 2e, §13.3 Facts About the F Distribution §13.3, p. 684
Trap
\[ 5 \text{ groups of } 3 \;\Rightarrow\; F_{10,4} \]
Write the larger number first
Why: It looks more natural.
\[ \text{but the numerator's df comes first} \]
The numerator is k minus one, which is 4, and the denominator is n minus k, which is 10.
\[ F \sim F_{4,10} \]
Numerator first, always
Why: It is k minus one for a one-way ANOVA.
The consequence is real: the 5 percent critical value of F-sub-4-comma-10 is 3.48 while F-sub-10-comma-4's is 5.96, so a statistic of 4.481 would reject under the correct distribution and not under the reversed one. Example 13.2's entire conclusion turns on getting the order right.
Sorting
Each is a property of the F distribution.
Sort into buckets
Sort by whether chi-square shares it.
Four of five are shared because both distributions are built from sums of squares. The exception follows from F being a ratio of two such quantities, each carrying its own count.
Faded example
Three growing media with five plants each.
Fill in the blanks
F \sim F_2,\,12}
Why: That is Example 13.4's F-sub-2-comma-12, on which a statistic of 0.134 gives a right-tail p-value of about 0.876.
Prediction
Commit before reasoning.
Predict first
Why do the two distributions share four properties?
Correct: Both are built from sums of squares.
Why: Squaring makes every contribution non-negative, so both statistics are bounded below at zero and unbounded above — and a distribution shaped that way cannot be symmetric. The F is in fact a ratio of two chi-square-based quantities, which is why it inherits the shape and needs two parameters.
Section
Section 2
Concept
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 the null hypothesis.
why right-tailed — Differing means add a non-negative term to the numerator and leave the denominator alone, so they can only push the ratio up. A small F means the group means sit closer together than within-group variation predicts, which is never evidence against equality.
\[ p\text{-value} = P(F > F_{\text{obs}}) \]
This is the third time the course has met the same structural argument. Chapter 11's goodness-of-fit statistic squared every departure, so disagreement could only make it larger; the same held for contingency tests; and here differing means can only inflate the numerator. In all three cases the lower tail contains no alternative to test, so no left-tailed version exists.
Figure (svg): An F curve on four and ten degrees of freedom with a dashed line at one and the right tail beyond 4.481 shaded
OpenStax Introductory Statistics 2e, §13.3 Facts About the F Distribution §13.3, pp. 684-685 — the note that the ANOVA test is always right-tailed
Picture it
The book's Figure 13.4.
Figure (svg): An F curve on four and ten degrees of freedom with a dashed line at one and the right tail beyond 4.481 shaded
The dashed line at one marks what the null predicts; the statistic sits far to its right and the shaded tail beyond it is 0.0248. Nothing to the left of one would ever count as evidence, however far from one it fell.
Worked example
The bean plants give F = 0.134 on 2 and 12 degrees of freedom.
\[ F = 0.134 \]
Compare with one
Why: Far below.
The right tail
Why: Almost everything.
\[ 0.8759 \]
Decide at 3 percent
Why: 0.876 exceeds 0.03.
Note the direction
Why: Small is not evidence.
Figure (svg): The solution to Worked example Example 13.4's small F shown as a ladder of expressions, one row per legal move
\[ p = P(F > 0.134) = 0.8759 \]
Verify: confirm what taking the wrong tail would have given
Why: The left-tail area below 0.134 is only 0.124, which would look like moderate evidence and is entirely spurious — a small F means the three children's plants had unusually SIMILAR mean heights, which supports the null rather than contradicting it. The direction has to come from the argument, not from the statistic's distance from one.
OpenStax Introductory Statistics 2e, §13.3 Facts About the F Distribution §13.3, p. 689
Sorting
Each is an F statistic with its degrees of freedom.
Sort into buckets
Sort by whether it is evidence against the null.
Item (e) is Example 13.3 at 2.2303, whose p-value of 0.1241 is not small enough at any usual level — a reminder that being above one is necessary and not sufficient.
Worked example
The chapter's three F values.
\[ 4.481, \; 2.2303, \; 0.134 \]
Example 13.2
Why: Well above one.
Example 13.3
Why: Above one, modestly.
Example 13.4
Why: Well below one.
Their p-values
Why: In the same order.
\[ 0.0248, 0.1241, 0.8759 \]
Figure (svg): The solution to Worked example reading three statistics against one shown as a ladder of expressions, one row per legal move
\[ F \uparrow \;\Longleftrightarrow\; p \downarrow \]
Verify: confirm the comparison against one is a genuine shortcut
Why: The null predicts a ratio near one, so a glance at whether F is well above, near, or below one predicts the verdict before any lookup. It is not a substitute for the p-value, since how far above one counts as far depends on the degrees of freedom — but 4.481 against a prediction of one is visibly extreme, and 0.134 visibly is not.
OpenStax Introductory Statistics 2e, §13.3 Facts About the F Distribution §13.3, pp. 685-689
Error analysis
Which are correct?
Annotate
On: \( \begin{aligned} &(1)\; \text{large } F \text{ is evidence the means differ} \\ &(2)\; \text{small } F \text{ is evidence the means differ} \\ &(3)\; \text{the } p\text{-value is the area to the right} \\ &(4)\; F \text{ far from one in either direction is evidence} \end{aligned} \)
Error (4) is the tempting one, because departure in either direction from a predicted value usually is evidence. It is not here, because the alternative hypothesis can only produce departures in one direction.
Two truths and a lie
All three concern the tail.
Eliminate the wrong options
Two are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is false. There is no left-tailed version, because the alternative hypothesis cannot produce a small F — differing means add to the numerator and never subtract from it.
Prediction
Commit before reasoning.
Predict first
An F of 0.02 on 4 and 30 degrees of freedom. What does it indicate?
Correct: Unusually close agreement among the group means.
Why: It is the mirror of section 11.1's remark about a very small chi-square: the data fit the null better than chance ordinarily allows. That is not evidence against the null, though it can prompt a question about how the data were produced — and the right-tailed test never examines it.
Faded example
For an observed statistic on the appropriate F curve.
Fill in the blanks
p\text> = P(F right-tailed F____}), \text___ ___
Why: Always the upper tail for a one-way ANOVA, and section 13.4's test of two variances is the chapter's only procedure where the tail can go the other way.
Section
Section 3
Concept
Example 13.2 tests whether five mulching conditions produce the same mean tomato yield, at a 5 percent level. The ANOVA table gives an F of 4.4810 on 4 and 10 degrees of freedom, with a p-value of 0.0248.
the conclusion's wording — At the 5 percent level there is reasonably strong evidence that the differences in mean yields are unlikely to be due to chance alone, and we may conclude that at least some of the mulches led to different mean yields.
\[ F_{4,10} = 4.4810, \quad p = 0.0248 < 0.05 \]
The book's phrasing repays attention. It says at least SOME of the mulches led to different mean yields, which is exactly what the alternative hypothesis asserts and no more. It names no condition and no direction, even though the sample means run from about 3,500 grams for bare soil to about 7,800 for straw.
Figure (svg): An F curve on four and ten degrees of freedom with a dashed line at one and the right tail beyond 4.481 shaded
OpenStax Introductory Statistics 2e, §13.3 Facts About the F Distribution §13.3, pp. 685-686 — Example 13.2 in full
Picture it
Example 13.2's statistic against its distribution.
Figure (svg): An F curve on four and ten degrees of freedom with a dashed line at one and the right tail beyond 4.481 shaded
A p-value of 0.0248 is below 5 percent and above 1 percent, so the decision depends on a level chosen in advance — the same fragility Example 11.8 had in chapter 11.
Worked example
Fifteen tomato plants under five mulching conditions, three each, at a 5 percent level.
\[ \alpha = 0.05 \]
Hypotheses
Why: Five means.
Distribution
Why: k-1 and n-k.
\[ F s u b 4, 10 \]
The table
Why: From software.
\[ SS 36, 648, 561\text{ and } 20, 446, 726 \]
The statistic
Why: Mean squares divided.
\[ 4.4810 \]
Right tail, and decide
Why: Beyond 4.481.
\[ p = 0.0248 < 0.05,\text{ reject} \]
Figure (svg): The solution to Worked example Example 13.2 in full shown as a ladder of expressions, one row per legal move
\[ F = \frac{9{,}162{,}140}{2{,}044{,}673} = 4.4810 \]
Verify: confirm the table's internal consistency
Why: The two sums of squares add to 57,095,287, the reported total, and the two degrees of freedom add to 14, which is n minus one. Dividing each sum by its own degrees of freedom gives mean squares of 9,162,140 and 2,044,673, whose ratio is 4.481 — the reported statistic. Three independent checks, all from numbers already in the table.
OpenStax Introductory Statistics 2e, §13.3 Facts About the F Distribution §13.3, pp. 685-686
Faded example
Five conditions with three plants each.
Fill in the blanks
\text4 = 5 - 1 = 10, \qquad \text___ = 15 - 5 = ___
Why: Giving F-sub-4-comma-10, on which the observed 4.4810 has a right-tail area of 0.0248.
Worked example
The same p-value of 0.0248, at a stricter level.
\[ \alpha = 0.01 \]
Compare
Why: 0.0248 against 0.01.
The decision
Why: Not below alpha.
At 5 percent
Why: Below alpha.
The lesson
Why: Choose alpha first.
Figure (svg): The solution to Worked example what the decision would be at 1 percent shown as a ladder of expressions, one row per legal move
\[ 0.01 < 0.0248 < 0.05 \]
Verify: confirm this is a real risk rather than a hypothetical
Why: The chapter's own three examples use three different levels — 5, 1 and 3 percent — and Example 13.2's p-value falls between two of them. Chapter 9 established that the level has to be chosen before the data are seen precisely because a p-value near a boundary makes the temptation to adjust it afterwards concrete rather than abstract.
OpenStax Introductory Statistics 2e, §13.3 Facts About the F Distribution §13.3, pp. 685-687
Trap
\[ p = 0.0248 \;\Rightarrow\; \text{straw outperformed bare soil} \]
Read the largest and smallest sample means as the finding
Why: The difference is large.
\[ \text{but the alternative names no condition} \]
The test concluded only that at least some means differ, and picking the extreme pair after seeing the data is the multiple-comparison problem again.
\[ \text{at least some mulches led to different mean yields} \]
Report the omnibus conclusion, then describe the sample means separately
Why: Description and inference are different claims.
The sample means genuinely do run from about 3,500 to about 7,800 grams, and reporting that is useful. What the test does not license is treating the largest gap as a tested comparison — its error rate reflects all ten pairs that could have been chosen, not the one that was.
Two truths and a lie
All three concern Example 13.2.
Eliminate the wrong options
Two are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is false. The test concludes only that at least some mulches led to different mean yields; naming a pair is a separate comparison it never made, and the book's own wording carefully avoids it.
Estimation
On 4 and 10 degrees of freedom, the 5 percent critical value is 3.48.
Predict first
What does that tell you before computing the p-value?
Correct: It exceeds the critical value, so it rejects at 5 percent.
Why: Comparing the statistic against a critical value gives the decision without the p-value, exactly as section 12.4's table did for correlations. The p-value of 0.0248 adds precision — it says how far past the threshold the result lies — but the verdict at 5 percent is already settled.
Prediction
Commit before reasoning.
Predict first
Example 13.2's p-value is 0.0248 and its decision is to reject at 5 percent. Why give both?
Correct: It shows how close to the boundary the result lies.
Why: A bare reject at 5 percent looks identical whether the p-value is 0.0248 or 0.0001. Here it is close enough to the boundary that a stricter level would reverse it, and a reader can only know that if the number is reported.
Section
Section 4
Concept
Example 13.3 compares four sororities' grade means at a 1 percent level, giving F = 2.2303 on 3 and 16 degrees of freedom with a p-value of 0.1241. Example 13.4 compares three growing media at a 3 percent level, giving F = 0.134 on 2 and 12 with a p-value of 0.8759. Neither rejects.
a balanced design — Example 13.3's note: each factor has the same number of observations. Four sororities with five sisters each, and three children with five plants each.
\[ F_{3,16} = 2.2303, \; p = 0.1241; \qquad F_{2,12} = 0.134, \; p = 0.8759 \]
The two non-rejections fail differently, which is worth noticing. The sororities' F of 2.23 is above one, so their grade means DID differ somewhat more than within-sorority variation alone predicts — just not enough to rule out chance. The bean plants' F of 0.134 is below one, so their means differed less than chance would typically produce. Both give the same decision from opposite situations.
Figure (svg): A table of the chapter's three complete tests with their statistics and decisions
OpenStax Introductory Statistics 2e, §13.3 Facts About the F Distribution §13.3, pp. 686-689 — Examples 13.3 and 13.4 in full
Picture it
Statistics, p-values, levels and decisions.
Figure (svg): A table of the chapter's three complete tests with their statistics and decisions
Each example states its significance level before computing anything, and the three levels differ. Only Example 13.2's decision would change under a different one.
Worked example
Four sororities, five sisters each, at a 1 percent level.
\[ \alpha = 0.01 \]
Note the design
Why: Five each.
Distribution
Why: 3 and 20 minus 4.
\[ F s u b 3, 16 \]
The table
Why: From the calculator.
\[ SS 2.88732\text{ and } 6.9044 \]
The statistic
Why: 0.96244 over 0.431525.
\[ 2.2303 \]
Right tail, and decide
Why: 0.1241 exceeds 0.01.
Figure (svg): The solution to Worked example Example 13.3, the sororities shown as a ladder of expressions, one row per legal move
\[ F = \frac{0.96244}{0.431525} = 2.2303 \]
Verify: confirm the decision does not depend on the level chosen
Why: The p-value of 0.1241 exceeds 5 percent and 10 percent as well as 1 percent, so no conventional level would reject. That makes this a cleaner non-rejection than Example 13.2's rejection was — the choice of 1 percent, which looks strict, turns out not to matter to the outcome at all.
OpenStax Introductory Statistics 2e, §13.3 Facts About the F Distribution §13.3, pp. 687-688
Sorting
Each pairs a p-value with a stated significance level.
Sort into buckets
Sort by the decision.
Items (a) and (d) are the same p-value at two different levels, with opposite decisions — the clearest possible illustration of why the level must be fixed before the data are examined.
Worked example
Three children's plants, five each, at a 3 percent level, computed by the balanced shortcut.
\[ \alpha = 0.03 \]
Group means
Why: First step.
\[ 24.2, 25.4, 24.4 \]
Their variance, times n
Why: 5 times 0.413.
\[ \text{MSbetween } = 2.067 \]
Mean of the group variances
Why: 11.7, 18.3, 16.3.
\[ \text{MSwithin } = 15.433 \]
The statistic
Why: The ratio.
\[ 0.134 \]
Right tail on 2 and 12
Why: Beyond 0.134.
\[ 0.8759,\text{ do not reject} \]
Figure (svg): The solution to Worked example Example 13.4, the bean plants shown as a ladder of expressions, one row per legal move
\[ F = \frac{5(0.413)}{15.433} = 0.134 \]
Verify: confirm the shortcut against the sums-of-squares route
Why: The general computation gives SS between of 4.1333 on 2 degrees of freedom, so MSbetween is 2.0667 — the shortcut's answer. SS within is 185.2 on 12, giving 15.4333. Both routes must agree for a balanced design, and checking one against the other confirms the shortcut was applied correctly.
OpenStax Introductory Statistics 2e, §13.3 Facts About the F Distribution §13.3, p. 689
Error analysis
Which are defensible?
Annotate
On: \( \begin{aligned} &(1)\; \text{the sorority means are all equal} \\ &(2)\; \text{there is insufficient evidence that they differ} \\ &(3)\; \text{the bean plant media produce identical mean heights} \\ &(4)\; \text{neither study detected a difference at its stated level} \end{aligned} \)
Errors (1) and (3) are chapter 9's asymmetry appearing one last time. A non-rejection says the data are consistent with equality, and with small groups a great deal is consistent with equality.
Two truths and a lie
All three concern the two non-rejections.
Eliminate the wrong options
Two are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is false. Failing to reject means the data are consistent with equality, not that equality holds — and with five observations per group in both studies, a good deal would be consistent with equality.
Faded example
Example 13.3's calculator output gives MS Factor 0.96244 and MS Error 0.431525.
Fill in the blanks
F = \frac0.4315252.2303} = ___
Why: An F of 2.23 is above one, so the sorority means did vary somewhat more than within-sorority variation predicts — but the p-value of 0.1241 says that much variation is unremarkable at this sample size.
Prediction
Commit before reasoning.
Predict first
One study has F = 2.23 and the other F = 0.134, and neither rejects. What distinguishes them?
Correct: One found more variation than predicted and one found less.
Why: An F above one means the group means varied more than within-group noise alone would produce; below one means they varied less. Both fail to reject, but from opposite situations — and a reader given only the decisions could not tell them apart, which is another argument for reporting the statistic.
Section
Section 5
Concept
Each of the book's examples reports the distribution for the test, the test statistic, a probability statement, a comparison with alpha, a decision and a conclusion stated in the problem's own terms.
the conclusion — Stated in context and no stronger than the alternative hypothesis. Example 13.2's says at least some of the mulches led to different mean yields — naming neither a condition nor a direction.
\[ F_{df_1,df_2} = \text{value}, \; p = \text{value}, \; \text{decision}, \; \text{conclusion} \]
Reporting the degrees of freedom alongside the statistic is not optional. An F of 4.481 is decisive on 4 and 10 degrees of freedom and unremarkable on 1 and 3, so neither the statistic nor the p-value can be interpreted without them — the same point section 11.1 made about a chi-square value.
Figure (svg): Running a complete one-way ANOVA
OpenStax Introductory Statistics 2e, §13.3 Facts About the F Distribution §13.3, pp. 685-689 — the structure of all three worked solutions
Picture it
Six stages, the same rhythm as every test since chapter 9.
Figure (svg): Running a complete one-way ANOVA
The note beneath is the one specific to this chapter. Every other test in the course concluded about a named parameter; this one concludes only that some unnamed pair of means differs.
Worked example
For the chapter's three tests, in the book's own terms.
\[ \text{three conclusions} \]
Tomatoes, rejected
Why: At 5 percent.
Sororities, not rejected
Why: At 1 percent.
Bean plants, not rejected
Why: At 3 percent.
What none of them says
Why: Which groups.
Figure (svg): The solution to Worked example writing the three conclusions shown as a ladder of expressions, one row per legal move
\[ \text{reject} \to \text{some differ}; \quad \text{else} \to \text{insufficient evidence} \]
Verify: confirm the two non-rejections are worded as insufficiency rather than equality
Why: Both say the evidence is not sufficient to conclude a difference, and neither says the means are equal. That distinction is chapter 9's and it survives intact into the course's final chapter — the data being consistent with a null is not the same as the null being true.
OpenStax Introductory Statistics 2e, §13.3 Facts About the F Distribution §13.3, pp. 685-689
Matching
Match each stage of Example 13.2's solution.
Match the pairs
Why: The same four stages appear in every test since chapter 9. What changed across the course is only the second and third — which statistic, and which distribution its tail is read from.
Worked example
Everything a reader needs from Example 13.2.
\[ \text{the report} \]
The design
Why: Groups and sizes.
The statistic and df
Why: Together.
\[ F(4, 10) = 4.4810 \]
The p-value and level
Why: Both.
\[ p = 0.0248, \alpha = 0.05 \]
The conclusion
Why: In context.
Figure (svg): The solution to Worked example what a complete report carries shown as a ladder of expressions, one row per legal move
\[ F(4,10) = 4.4810, \; p = 0.0248 \]
Verify: confirm why the degrees of freedom must travel with the statistic
Why: An F of 4.481 has a p-value of 0.0248 on 4 and 10 degrees of freedom and about 0.12 on 1 and 3 — a fivefold difference from the same statistic. So a reported F is uninterpretable alone, which is why the convention writes them together as F of 4 and 10 equals 4.4810.
OpenStax Introductory Statistics 2e, §13.3 Facts About the F Distribution §13.3, pp. 685-686
Trap
\[ F = 4.48, \text{ significant} \]
Give the statistic and the verdict
Why: Both numbers are reported.
\[ \text{but } 4.48 \text{ means nothing alone} \]
On 4 and 10 degrees of freedom its p-value is 0.0248; on 1 and 3 it is about 0.12.
\[ F(4, 10) = 4.481, \; p = 0.0248 \]
Always attach both degrees of freedom
Why: They identify which curve was used.
The convention of writing F with its two degrees of freedom in parentheses exists precisely so that a reader can reconstruct the p-value independently — and can spot an inconsistency between the reported statistic and the reported p-value, which is the commonest error in published ANOVA results.
Two truths and a lie
All three concern reporting.
Eliminate the wrong options
Two are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is false. The alternative hypothesis names no groups, so a conclusion that does goes beyond what was tested. The book's wording — at least some of the mulches — is careful about exactly this.
Faded example
Three groups of five, with a statistic of 0.134.
Fill in the blanks
F(2, 12) = 0.134, \; p = 0.8759
Why: That is Example 13.4's result in the standard convention, and a reader can recompute the p-value from those three numbers alone.
Explain it
A classmate reports that Example 13.2 shows straw mulch produces the highest yields.
Discussion prompt
In two sentences or fewer, correct them.
Hint: Ask what the alternative hypothesis claims.
Answer:
The test concluded only that at least some of the mulches led to different mean yields, and it named no condition — the alternative hypothesis says nothing about which groups differ or in which direction.
Straw did have the highest sample mean, and reporting that as a description of these fifteen plants is fine; presenting it as the test's finding is not.
Comparison
Fill the blanks. Four properties are shared, and one is not.
Comparison matrix
| Chi-square | F | |
|---|---|---|
| Shape | skewed right | skewed right |
| Lower bound | zero | zero |
| Degrees of freedom | one parameter | two, and ordered |
| Tail used for its tests | right, except for a single variance | right for ANOVA; any for two variances |
The last row's exceptions are the same shape in both chapters. When a statistic measures disagreement the tail is fixed; when it measures a ratio of variances, both directions carry meaning — which is section 13.4's subject.
Pattern
Six steps, and the level is fixed at the second.
Compare the statistic against one before looking anything up: well above suggests rejection, at or below suggests none.
OpenStax Introductory Business Statistics 2e, §12.4 Facts About the F Distribution §12.4 Facts About the F Distribution
Check
The facts.
Check your understanding
Which of these is NOT true of the F distribution?
Answer: A
Why: The first fact says the curve is not symmetrical but skewed to the right, since it is bounded below at zero and unbounded above.
Check
The tail.
Check your understanding
An ANOVA gives F = 0.4. What follows?
Answer: A
Why: The test is right-tailed, so an F below one leaves almost all the distribution above it and the p-value is large.
Check
The conclusion.
Check your understanding
A one-way ANOVA on five groups rejects at 5 percent. What may be reported?
Answer: A
Why: That is the alternative hypothesis, and it is the book's own wording for Example 13.2's conclusion.
Real world
A nutrition study compares mean cholesterol across six diets with twelve subjects each, reports F = 3.10 with p = 0.012, and concludes that the Mediterranean diet — which had the lowest sample mean — significantly lowers cholesterol compared with the others.
Discussion prompt
Assess the statistical claim and the conclusion separately.
Hint: Check the degrees of freedom, then ask what the alternative hypothesis says.
Answer:
The omnibus result is sound. Six groups of twelve give 5 and 66 degrees of freedom, and on that curve an F of 3.10 has a right-tail p-value near 0.014 — close to the reported 0.012 and comfortably below 5 percent. So there is real evidence that the six diets do not all produce the same mean cholesterol.
\[ \text{df} = (6-1, \; 72-6) = (5, 66) \]
The conclusion goes well beyond it. The alternative hypothesis says only that at least two of the six means differ; it names no diet and no direction. Singling out the Mediterranean diet because it had the lowest sample mean is a comparison chosen after seeing the data, which is the multiple-comparison problem section 13.1 warned about — with six groups there are fifteen pairs, and the most extreme of fifteen looks extreme even when nothing is happening.
What the study supports: the six diets do not all produce the same mean cholesterol, at a 5 percent level. What it does not: that any particular diet differs from any other. Those are genuinely different claims, and the second needs a follow-up procedure that controls the error rate across all the comparisons — the methods the book says are studied in much greater detail in future statistics courses.
Two further points belong in the report. The assumptions should be checked, since cholesterol distributions are often skewed and six groups of twelve give enough data to look at that; and the effect's SIZE should be given in the outcome's own units. A p-value of 0.012 says the differences are unlikely to be chance, and says nothing at all about whether they are large enough to matter clinically — which is chapter 12's distinction between significance and importance, arriving one last time.
Commit first
Answer, then rate your confidence honestly.
Predict first
Why is a one-way ANOVA always right-tailed?
Correct: Differing means can only inflate the numerator.
\[ \text{MS}_{\text{between}} = \sigma^2 + (\ge 0) \;\Longrightarrow\; F \text{ pushed upward only} \]
Why: MSbetween consists of the population variance plus a non-negative term produced by the differences among the means, while MSwithin estimates the population variance alone. So the alternative hypothesis can only push the ratio up, and a small F — which two of the chapter's three examples have — means the group means agree more closely than within-group variation predicts. The distribution has a perfectly good lower tail; no alternative lives in it.
Explain it
They got F = 0.42 and concluded that the group means are very different, since 0.42 is far from one.
Discussion prompt
In two sentences or fewer, correct them.
Hint: Ask which direction the alternative hypothesis pushes the statistic.
Answer:
Differing means add to the numerator, so they can only push F above one — an F of 0.42 means the group means sit closer together than within-group variation would normally produce.
The test is right-tailed, so the p-value here is large and the result is no evidence at all against equality.
Exit ticket
Name the weakest spot before you close the deck.
Predict first
Which of these would you least want handed to you cold?
Correct: Whichever you picked is tonight's ten minutes, and each has a one-line fix.
Why: For the first, both distributions are built from sums of squares. For the second, differing means can only add to the numerator. For the third, mean squares first, then their ratio, then the right tail. For the fourth, at least some means differ, with no pair named. Do five problems of your chosen kind rather than twenty mixed ones.
Connect it up
Paper. Fifteen minutes.
Draw it
At the top, list the five facts and mark beside each whether chi-square shares it — four do, and the exception is having two degrees of freedom rather than one. Beneath, draw three or four F curves on one axis for increasing degrees of freedom, all starting at zero and skewed right, with a dashed vertical line at F equal to one labelled what the null predicts. Shade the right tail of one curve and write beside it that the test is always right-tailed because differing means can only add to the numerator. In the middle, make a table of the chapter's three tests with columns for the example, k and n, F, the two degrees of freedom, the p-value, the significance level and the decision: tomatoes at 5, 15 with F 4.4810 on 4 and 10, p 0.0248, alpha 0.05, reject; sororities at 4, 20 with F 2.2303 on 3 and 16, p 0.1241, alpha 0.01, do not reject; bean plants at 3, 15 with F 0.134 on 2 and 12, p 0.8759, alpha 0.03, do not reject. Circle the two non-rejections and write beside them that one has F above one and one below — the same decision from opposite situations. At the bottom, write the conclusion for the rejected test in full, underlining the words at least some, and note that no condition is named.
Check your table by confirming the p-values run in the reverse order of the F statistics, which they must since all three use a right tail. Check your curves by confirming every one starts at zero — an F curve that crosses into negative territory has been drawn wrong.
Recap
Six things, and three of them are complete tests.
| If you see | Then |
|---|---|
| A negative F | An arithmetic error: it cannot happen |
| F well above one | The right tail is small; likely a rejection |
| F at or below one | The right tail is large; no evidence |
| A p-value below the stated alpha | Reject: at least two means differ |
| A p-value above it | Insufficient evidence, never proof of equality |
| An F reported without its degrees of freedom | Uninterpretable; ask for both |
| A rejection with a named group | An over-claim: the test names none |
| Very large degrees of freedom | The curve approaches the normal |
Section 13.4 puts the same distribution to its other use. Instead of comparing several means through their variances, it compares two variances directly — and because a variance can be larger or smaller than another, that test can be left-tailed, right-tailed or two-tailed.
OpenStax Introductory Statistics 2e, §13.3 Facts About the F Distribution §13.3, pp. 684-689 — everything on these slides traces back here
Want this taught 1-on-1? Alexander tutors Statistics — $55/session, free consultation.