10 One-Way ANOVA

Statistical Inference

One-Way ANOVA

Comparing Multiple Means

Pairwise Comparisons

Objectives

Upon successful completion of this lesson, you should be able to:

Explain why it is not appropriate to conduct multiple independent t-tests to compare the means of more than two independent groups.
Use Minitab to construct a probability plot for an F distribution.
Use Minitab to perform a one-way between groups ANOVA with Tukey’s pairwise comparisons.
Interpret the results of a one-way between groups ANOVA.
Interpret the results of Tukey’s pairwise comparisons.

This lesson corresponds to Section 8.1 in the Lock⁵ textbook.

In previous lessons you learned how to compare the means of two independent groups. In this lesson, we will learn how to compare the means of more than two independent groups. This procedure is known as a one-way between groups analysis of variance, or more often as a “one-way ANOVA.”

Why not multiple independent t-tests?

A frequently asked question is, “why not just perform multiple two independent samples $t$ tests?” If you were to perform multiple independent $t$ tests instead of a one-way between groups ANOVA you would need to perform more tests. For $k$ independent groups there are $\frac{k(k-1)}{2}$ possible pairs. If you had 5 independent groups, that would equal $\frac{5(5-1)}{2}=10$ independent t-tests! And, those 10 independent t-tests would not give you information about the independent variable overall. Most importantly, multiple $t$ tests would lead to a greater chance of making a Type I error. By using an ANOVA, you avoid inflating $\alpha$ and you avoid increasing the likelihood of a Type I error.

10.1 Introduction to the F Distribution

One-way ANOVAs, along with a number of other statistical tests, use the F distribution. Earlier in this course you learned about the $z$ and $t$ distributions. You computed $z$ and $t$ test statistics and used those values to look up p-values using statistical software. Similarly, in this lesson you are going to compute F test statistics. The F test statistic can be used to determine the p-value for a one-way ANOVA.

The video below gives a brief introduction to the F distribution and walks you through two examples of using Minitab to find the p-values for given F test statistics. The steps for creating a distribution plot to find the area under the F distribution are the same as the steps for finding the area under the $z$ or $t$ distribution. For the F distribution we will always be looking for a right-tailed probability. Later in this lesson we will see that this area is the p-value.

The F distribution has two different degrees of freedom: between groups and within groups. Minitab will call these the numerator and denominator degrees of freedom, respectively. Within groups is also referred to as error.

Between Groups (Numerator) Degrees of Freedom

\[ df_{\text{between}}=k-1 \]

$k$ = number of groups

Within Groups (Denominator, Error) Degrees of Freedom

\[ df_{\text{within}}=n-k \]

$n$ = total sample size with all groups combined

$k$ = number of groups

Minitab: Creating an F Distribution

Find a p-value from an $F$-test statistic in Minitab:

Select Graph > Probability Distribution Plot.
Select One Curve, then select View Probability.
Select OK.
Change the Distribution to F.
Enter the Numerator degrees of freedom and Denominator degrees of freedom.
Select Options.
Select A specified x value.
Select Right tail.
Under X value, enter the $F$-test statistic.
Select OK, then select OK again.

Scenario: An F test statistic of 2.57 is computed with 3 and 246 degrees of freedom. What is the p-value for this test?

We can create a distribution plot. Our distribution is the F distribution. The numerator df ($df_1$) is 3 and the denominator df ($df_2$) is 246. We want to shade the area in the right tail. Our “X Value” is 2.57.

Open Minitab.
Select Graph > Probability Distribution Plot > View Probability.
Change the Distribution to F.
Fill in the Numerator degrees of freedom with 3 and the Denominator degrees of freedom with 246.
Select the Options button.
Select A specified x value.
Use the default Right tail.
For the X value, enter 2.57.
Select OK and OK.

Distribution Plot of Density vs X - Fm df1=3, df2=246 — Fig 10.1

The area beyond an F-value of 2.57 with 3 and 246 degrees of freedom is 0.05487. The p-value for this F test is 0.05487.

Note: When you conduct an ANOVA in Minitab, the software will compute this p-value for you.

10.2 Hypothesis Testing (ANOVA)

A one-way between groups ANOVA is used to compare the means of more than two independent groups. A one-way between groups ANOVA comparing just two groups will give you the same results at the independent $t$ test that you conducted in Lesson 8. We will use the five-step hypothesis testing procedure again in this lesson.

Check assumptions and write hypotheses.

The assumptions for a one-way between groups ANOVA are:
1. Samples are independent
2. The response variable is approximately normally distributed for each group or all group sample sizes are at least 30
3. The population variances are equal across responses for the group levels (if the largest sample standard deviation divided by the smallest sample standard deviation is not greater than two, then assume that the population variances are equal)
Given that you are comparing $k$ independent groups, the null and alternative hypotheses are:

$H_{0}: \mu_1 = \mu_2 = \cdots = \mu_k$
$H_{a}:$ Not all $\mu_\cdot$ are equal

In other words, the null hypothesis is that at all of the groups’ population means are equal. The alternative is that they are not all equal; there are at least two population means that are not equal to one another.

Calculate the test statistic.

ANOVA uses an F test statistic. Hand calculations for ANOVAs require many steps. In this class, you will be working primarily with Minitab outputs.

Conceptually, the F statistic is a ratio: \[ F=\dfrac{\text{Between groups variability}}{\text{Within groups variability}} \]

Numerically this translates to

\[ F=\dfrac{MS_{\text{Between}}}{MS_{\text{Within}}} \]

In other words how much do individuals in different groups vary from one another over how much to individuals within groups vary from one another.

Statistical software will compute the F ratio for you and produce what is known as an ANOVA source table. The ANOVA source table will give you information about the variability between groups and within groups. The table below gives you all the formulas, but you will not be responsible for performing these calculations by hand. Minitab will do all of these calculations for you and provide you with the full ANOVA source table.

Source	df	SS	MS	F	p
Between Groups (Factor)	$k-1$	$\sum_{k}n_k(\bar{x}_k-\bar{x}_\cdot)^2$	$\dfrac{SS_{\text{Between}}}{df_{\text{Between}}}$	$\dfrac{MS_{\text{Between}}}{MS_{\text{Within}}}$	Area to the right of Fk-1, n-k
Within Groups (Error)	$n-k$	$\sum_k \sum_i(x_{ik}-\bar{x}_k)^2$	$\dfrac{SS_{\text{Within}}}{df_{\text{Within}}}$
Total	$n-1$	$\sum_k \sum_i(x_{ik}-\bar{x}_\cdot)^2$

$k$: Number of groups

$n$: Total sample size (all groups combined)

$n_k$: Sample size of group $k$

$\bar{x}_k$: Sample mean of group $k$

$\bar{x}_\cdot$: Grand mean (i.e., mean for all groups combined)

$SS$: Sum of squares

$MS$: Mean square

$df$: Degrees of freedom

$F$: F-ratio (the test statistic)

Some of the notation above should look familiar, while other terms will be new to you. The sum of squares that appears in the ANOVA source table is similar to the sum of squares that you computed in Lesson 2 when computing variance and standard deviation. Recall, the sum of squares is the squared difference between each score and the mean. Here, there are three different sum of squares each measuring a different type of variability.

The ANOVA source table also has three different degrees of freedom: $df_{\text{between}}$, $df_{\text{within}}$, and $df_{\text{total}}$. If you were to look up an F value using statistical software you would need to know two of these degrees of freedom: $df_1 = df_{\text{between}}$ and $df_2=df_{\text{within}}$.

Determine the p-value.

When performing an ANOVA using statistical software, you will be given the p-value in the ANOVA source table. If performing an ANOVA by hand, you would use the F distribution. Similar to the t distribution, the F distribution varies depending on degrees of freedom.
Make a decision.

If $p \leq \alpha$ reject the null hypothesis. If $p>\alpha$ fail to reject the null hypothesis.
State a “real world” conclusion.

Based on your decision in Step 4, write a conclusion in terms of the original research question.

10.3 Pairwise Comparisons

While the results of a one-way between groups ANOVA will tell you if there is what is known as a main effect of the explanatory variable, the initial results will not tell you which groups are different from one another. In order to determine which groups are different from one another, a post-hoc test is needed. Post-hoc tests are conducted after an ANOVA to determine which groups differ from one another. There are many different post-hoc analyses that could be performed following an ANOVA. Here, we will learn about one of the most common tests known as Tukey’s Honestly Significant Differences (HSD) Test.

Most statistical software, including Minitab, will compute Tukey’s pairwise comparisons for you. This specific post-hoc test makes all possible pairwise comparisons. In this class we will be relying on statistical software to perform these analyses, if you are interested in seeing how the calculations are performed, this information is contained in the notes for STAT 502: Analysis of Variance and Design of Experiments. This analysis takes into account the fact that multiple tests are being performed and makes the necessary adjustments to ensure that Type I error is not inflated.

In the examples later in this lesson you will see a number of Tukey post-hoc tests. Next, you will also learn how to obtain these results using Minitab.

For each pairwise comparison, $H_0: \mu_i - \mu_j=0$ and $H_a: \mu_i - \mu_j \ne 0$.

10.4 Minitab: One-Way ANOVA

Conduct a one-way ANOVA with Tukey pairwise comparisons in Minitab:

Open the data file in Minitab.
Select Stat > ANOVA > One-Way.
From the dropdown menu, select the option that matches the organization of the data:
- If all responses are in one column, select Response data are in one column for all factor levels. Enter the quantitative response variable under Response and the categorical explanatory variable under Factor.
- If each group is in a separate column, select Responses are in a separate column for each factor level. Enter the group columns under Responses.
Select Comparisons and check Tukey. Under Results, also select Tests.
Select OK, then OK.

Example 10.1 (Minitab: Pig Weights by Diet) In one research study, 20 young pigs are assigned at random among 4 experimental groups. Each group is fed a different diet. (This design is a completely randomized design.) The data are the pigs’ weights in kg after being raised on these diets for 10 months. We wish to determine if there are any differences in mean pig weights for the 4 diets.

$H_0: \mu_1 = \mu_2 = \mu_3 = \mu_4$
$H_a:$ Not all $\mu$ are equal

Feed_1	Feed_2	Feed_3	Feed_4
60.8	68.3	102.6	87.9
57.1	67.7	102.2	84.7
65.0	74.0	100.5	83.2
58.7	66.3	97.5	85.8
61.8	69.9	98.9	90.3

Contained in the Minitab file: (ANOVA_ex.mpx)

Note that in this file the data were entered so that each group is in its own column. In other words, the responses are in a separate column for each factor level. In later examples, you will see that Minitab will also conduct a one-way ANOVA if the responses are all in one column with the factor codes in another column.

To perform an Analysis of Variance (ANOVA) test in Minitab:

Open the Minitab file: (ANOVA_ex.mpx)
From the menu bar, select Stat > ANOVA > One-Way.
Click the drop-down menu and select ‘Responses are in a separate column for each factor level’.
Enter the variables Feed_1, Feed_2, Feed_3, and Feed_4 to insert them into the Responses box.
Choose the Comparisons button and check the box next to Tukey. Under Results, also select Tests.
Select OK and OK.

The result should be the following output:

Method

Null hypothesis	All means are equal
Alternative hypothesis	At least one mean is different
Significance level	$\alpha=0.05$

Equal variances were assumed for the analysis

Factor Information

Factor	Levels	Values
Factor	4	Feed_1, Feed_2, Feed_3, Feed_4

Analysis of Variance

Source	DF	Adj SS	Adj MS	F-Value	P-Value
Factor	3	4703.2	1567.73	206.72	0.000
Error	16	121.3	7.58
Total	19	4824.5

Model Summary

S	R-sq	R-sq(adj)	R-sq(pred)
2.75386	97.48%	97.01%	96.07%

Means

Factor	N	Mean	StDev	95% CI
Feed_1	5	60.68	3.03	(58.07, 63.29)
Feed_2	5	69.24	2.96	(66.63, 71.85)
Feed_3	5	100.340	2.164	(97.729, 102.951)
Feed_4	5	86.38	2.78	(83.77, 88.99)

Pooled StDev = 2.75386

Grouping Information Using the Tukey Method and 95% Confidence

Factor	N	Mean	Grouping
Feed_3	5	100.34	A
Feed_4	5	86.38
Feed_2	5	69.24
Feed_1	5	60.68

Means that do not share a letter are significantly different.

Difference ofLevels	Differenceof Means	SE of Difference	95% CI	T-Value	Adjusted P-Value
Feed_2 - Feed_1	8.56	1.74	(3.57, 13.55)	4.91	0.001
Feed_3 - Feed_1	39.66	1.74	(34.67, 44.65)	22.77	0.000
Feed_4 - Feed_1	25.70	1.74	(20.71, 30.69)	14.76	0.000
Feed_3 - Feed_2	31.10	1.74	(26.11, 36.09)	17.86	0.000
Feed_4 - Feed_2	17.14	1.74	(12.15, 22.13)	9.84	0.000
Feed_4 - Feed_3	-13.96	1.74	(-18.95, -8.97)	-8.02	0.000

Tukey Simultaneous Tests for Differences of Means

Individual confidence level = 98.87%

Example 10.2 (SAT-Math Scores by Award Preference) In this example, we are comparing the SAT scores of students who said that they would prefer to win an Academy Award, a Nobel Prize, or an Olympic gold medal.

The example uses the StudentSurvey dataset provided by the Lock5 textbook.

Let’s apply the five-step hypothesis testing process to this example.

Check assumptions and write hypotheses.

The assumptions for a one-way between-groups ANOVA are:

Assumption 1: Samples are independent: Each student selected either Olympic, Academy, or Nobel. Each student is in only one group and those groups are in no way matched or paired. This assumption is met.

Assumption 2: The response variable is approximately normally distributed for each group or all group sample sizes are at least 30: To check this we can construct a histogram with groups in Minitab. These plots show that the distributions are all approximately normal.

Histograms of SAT grouped by award — Fig 10.4

Assumption 3: The population variances are equal across responses for the group levels (if the largest sample standard deviation divided by the smallest sample standard deviation is not greater than two, then assume that the population variances are equal). Again we can use Minitab to look at the standard deviations across the groups. The largest standard deviation is 151.3 and the smallest is 114.1 for a ratio of 1.33 which is less than 2. So this assumption is met.

Descriptive Statistics

Award	N	Mean	SE Mean	StDev	Minimum	Q1	Median	Q3	Maximum
Academy	31	1191.0	27.2	151.3	820.0	1070.0	1200.0	1300.0	1530.0
Nobel	149	1239.1	9.38	114.5	920.0	1160.0	1230.0	1310.0	1550.0
Olympic	182	1176.7	8.46	114.1	800.0	1100.0	1190.0	1250.0	1470.0

Use Minitab to run a one-way ANOVA with the Minitab file: (StudentSurvey.mpx)

The following will describe the output within the context of the five-step hypothesis testing process.

Given that you are comparing $k$ independent groups, the null and alternative hypotheses are:

Method

Null hypothesis: All means are equal.

Alternative hypothesis: At least one mean is different.

Significance level: $\alpha=0.05$

Equal variances were assumed for the analysis

Calculate the test statistic.

You should get the following output:

Factor Information

Factor Levels Values

Award 3 Academy, Nobel, Olympic

The factor information tells us that our factor is the Award.

Analysis of Variance

Source DF Adj SS Adj MS F-Value P-Value

Award 2 323269 162134 11.67 0.000

Error 359 4986078 13889

Total 361 5310347

We need to know two of these degrees of freedom: $df_1 = df_{\text{between}}$ and $df_2=df_{\text{within}}$.

The analysis of variance table is also known as our ANOVA source table. The source of the Award is our between-groups variation so the DF K -1 or 3-1 = 2. Error is our within-groups variation, so the degrees of freedom are n - K or 362-3 = 359. The ‘Adj’ stands for adjusted. For a one-way ANOVA, nothing is being adjusted. The F-value is our F-test statistic, which in this case is 11.67 with a p-value of 0.000. The F value could be written as F (2, 359) = 11.67.

Means

Factor N Mean StDev 95% CI

Academy 31 1191.0 151.3 (1149.3, 1232.6)

Nobel 149 1239.11 114.54 (1220.13, 1258.10)

Olympic 182 1176.73 114.13 (1159.55, 1193.91)

Pooled StDev = 117.851

The Means table is just a table of descriptive statistics. The sample size, mean, and the standard deviation is presented for each group. The last column is an unadjusted 95% confidence interval. You should not refer to these confidence intervals since these do not take into account that there are three different groups and three different confidence intervals being computed at the same time. The confidence intervals that we’re interested in come with our Tukey pairwise comparison.
Determine the p-value.

The p-value is 0.000 from the Minitab output.
Make a decision.

$p \leq \alpha$ so reject the null hypothesis.
State a “real world” conclusion.

We can conclude that the group means are not all equal.

Factor	Levels	Values
Award	3	Academy, Nobel, Olympic

Source	DF	Adj SS	Adj MS	F-Value	P-Value
Award	2	323269	162134	11.67	0.000
Error	359	4986078	13889
Total	361	5310347

Factor	N	Mean	StDev	95% CI
Academy	31	1191.0	151.3	(1149.3, 1232.6)
Nobel	149	1239.11	114.54	(1220.13, 1258.10)
Olympic	182	1176.73	114.13	(1159.55, 1193.91)

Remember that ANOVA is just an omnibus test. It only tells us there is a difference somewhere. To determine where the differences are you would have to look at the Tukey pairwise comparison output.

Grouping Information Using the Tukey Method and 95% Confidence

{.w-auto .table-sm .row-header} *Means that do not share a letter are significantly different.*
Award	N	Mean	Grouping
Nobel	149	1239.11	A
Academy	31	1190.97	A
Olympic	182	1176.73

In the grouping table, Nobel and Olympic do not share a grouping letter. The means of these two groups are significantly different. Both the Nobel and Academy belong to Group A and both Academy and Olympic belong to Group B. These pairs are not significantly different.

Tukey Simultaneous Tests for Differences of Means

Difference of Levels	Differenceof Means	SE ofDifference	95% CI	T-Value	AdjustedP-Value
Nobel - Academy	48.1	23.3	(-6.3, 102.6)	2.07	0.096
Olympic - Academy	-14.2	22.9	(-67.8, 39.4)	-0.62	0.808
Olympic - Nobel	-62.4	13.0	(-92.9, -31.9)	-4.79	0.000

Individual confidence level = 98.87%

The Tukey Simultaneous Tests table provides the adjusted 95% confidence intervals. These are the confidence intervals and p-values you should be looking at when you’re making pairwise comparisons. These are adjusted to take into account that there are three different pairwise tests being performed simultaneously. To determine which groups are statistically significant we can look at the adjusted p-value. In this case, the only p-value less than 0.05 is the Olympic - Nobel pairwise comparison. The means of the Olympic and Nobel group are significantly different.

In the Tukey Simultaneous 95% CI graph any interval not containing zero indicates a statistically significant difference. In this case, we see that the Olympic - Nobel group does not contain zero, so this pairing has a statistically significant difference.

Example 10.3 (Exam Grade by Professor) Scenario:

Three professors were each teaching one section of a course. They all gave the same final exam, and they want to know if there are any differences between their sections’ mean scores.

Write hypotheses and check assumptions.

$H_0:\mu_1=\mu_2=\mu_3$

$H_a$: Not all $\mu$ are equal.

Means

Instructor N Mean StDev 95% CI

Dr. Al 60 68.367 17.719 (63.977, 72.756)

Dr. Oh 87 71.448 16.702 (67.803, 75.094)

Dr. Pa 98 67.939 17.465 (64.504, 71.373)

Pooled StDev = 17.2609

The standard deviations for all three classes are all similar.
Compute test statistic.
1. Open the Minitab file: (ANOVA_Exam_Profs.mpx)
2. Using Minitab: Stat > ANOVA > One-Way
The result is the following ANOVA source table:

Analysis of Variance

Source DF Adj SS Adj MS F-Value P-Value

Instructor 2 635.3 317.7 1.07 0.346

Error 242 72101.1 297.9

Total 244 72736.4

F (2, 242) = 1.07
Find the p-value.

From our ANOVA source table, p = .346
Make a decision.

Because $p > \alpha$, we fail to reject the null hypothesis.
State a conclusion.

There is not enough evidence to conclude that the mean scores from the three different professors’ sections are different.

Instructor	N	Mean	StDev	95% CI
Dr. Al	60	68.367	17.719	(63.977, 72.756)
Dr. Oh	87	71.448	16.702	(67.803, 75.094)
Dr. Pa	98	67.939	17.465	(64.504, 71.373)

Source	DF	Adj SS	Adj MS	F-Value	P-Value
Instructor	2	635.3	317.7	1.07	0.346
Error	242	72101.1	297.9
Total	244	72736.4

Example 10.4 (Tukey Pairwise Comparisons) There is some debate as to whether pairwise comparisons are appropriate when the overall one-way ANOVA is not statistically significant. Some argue that if the overall ANOVA is not significant then pairwise comparisons are not necessary. Others argue that if the pairwise comparisons were planned before the ANOVA was conducted (i.e., “a priori”) then they are appropriate.

The results of our Tukey pairwise comparisons were as follows:

Grouping Information Using the Tukey Method and 95% Confidence

Instructor	N	Mean	Grouping
Dr. Oh	87	71.448	A
Dr. Al	60	68.367	A
Dr. Pa	98	67.939	A

Means that do not share a letter are significantly different.

Tukey Simultaneous Tests for Differences of Means

Difference of Levels	Difference of Means	SE of Difference	95% CI	T-Value	Adjusted P-Value
Dr. Oh-Dr. Al	3.08	2.90	(-3.70, 9.86)	1.06	0.537
Dr. Pa-Dr. Al	-0.43	2.83	(-7.05, 6.19)	-0.15	0.987
Dr. Pa-Dr. Oh	-3.51	2.54	(-9.46, 2.44)	-1.38	0.351

Individual confidence level = 97.99%

Looking at the first table, all three instructors are in group A. Means that share a letter are not significantly different from one another (i.e., they are in the same group). Because all three instructors share the letter A, there are no significantly different pairs of instructors.

We could also look at the second table which gives us the t-test statistic and adjusted p-value for each possible pairwise comparison. This p-value is adjusted to take into account that multiple tests are being conducted. You can compare these p-values to the standard alpha level of .05. All p-values are greater than .05, therefore no pairs are significantly different from one another.

10.5 Summary

In this lesson you learned how to compare the means of three or more groups using a one-way between groups ANOVA. A one-way between groups ANOVA is used instead of multiple independent $t$ tests in order to avoid increasing the likelihood of committing a Type I error.

An ANOVA provides information about the explanatory variable overall, but not about differences between the different levels of that variable. In order to compare the different pairs we need to conduct a post-hoc analysis such as Tukey’s HSD test.

--- categories: [Statistical Inference, One-Way ANOVA, Comparing Multiple Means, Pairwise Comparisons] image: /assets/L10card.png # change the image path ---  # One-Way ANOVA ::: objectiveblock ## <i class="bi bi-check2-circle"></i>Objectives {.unnumbered .unlisted} Upon successful completion of this lesson, you should be able to: 1. Explain why it is not appropriate to conduct multiple independent t-tests to compare the means of more than two independent groups. 2. Use Minitab to construct a probability plot for an F distribution. 3. Use Minitab to perform a one-way between groups ANOVA with Tukey's pairwise comparisons. 4. Interpret the results of a one-way between groups ANOVA. 5. Interpret the results of Tukey's pairwise comparisons. ::: []{.bi .bi-book .fs-4} This lesson corresponds to Section 8.1 in the Lock<sup>5</sup> textbook. In previous lessons you learned how to compare the means of two independent groups. In this lesson, we will learn how to compare the means of more than two independent groups. This procedure is known as a one-way between groups analysis of variance, or more often as a "one-way ANOVA." ## Why not multiple independent t-tests? {.unnumbered .unlisted} A frequently asked question is, "why not just perform multiple two independent samples $t$ tests?" If you were to perform multiple independent $t$ tests instead of a one-way between groups ANOVA you would need to perform more tests. For $k$ independent groups there are $\frac{k(k-1)}{2}$ possible pairs. If you had 5 independent groups, that would equal $\frac{5(5-1)}{2}=10$ independent t-tests! And, those 10 independent t-tests would not give you information about the independent variable overall. Most importantly, multiple $t$ tests would lead to a greater chance of making a Type I error. By using an ANOVA, you avoid inflating $\alpha$ and you avoid increasing the likelihood of a Type I error. ## Introduction to the F Distribution One-way ANOVAs, along with a number of other statistical tests, use the F distribution. Earlier in this course you learned about the $z$ and $t$ distributions. You computed $z$ and $t$ test statistics and used those values to look up p-values using statistical software. Similarly, in this lesson you are going to compute F test statistics. The F test statistic can be used to determine the p-value for a one-way ANOVA. The video below gives a brief introduction to the F distribution and walks you through two examples of using Minitab to find the p-values for given F test statistics. The steps for creating a distribution plot to find the area under the F distribution are the same as the steps for finding the area under the $z$ or $t$ distribution. For the F distribution we will always be looking for a right-tailed probability. Later in this lesson we will see that this area is the p-value. The F distribution has two different degrees of freedom: between groups and within groups. Minitab will call these the numerator and denominator degrees of freedom, respectively. Within groups is also referred to as error. ::: {.callout-note appearance="minimal"} [**Between Groups (Numerator) Degrees of Freedom**]{.fs-4} $$ df_{\text{between}}=k-1 $$ $k$ = number of groups ::: ::: {.callout-note appearance="minimal"} [**Within Groups (Denominator, Error) Degrees of Freedom**]{.fs-4} $$ df_{\text{within}}=n-k $$ $n$ = total sample size with all groups combined $k$ = number of groups ::: ### ![ ](assets/minitab.png){.icon-image} Minitab: Creating an F Distribution {.unnumbered .unlisted} {{< include _minitab/find-a-p-value-from-an-f-test-statistic.qmd >}} ::: {.ms-3 .border-start .border-secondary-subtle .ps-3} **Scenario:** An F test statistic of 2.57 is computed with 3 and 246 degrees of freedom. What is the p-value for this test? We can create a distribution plot. Our distribution is the F distribution. The numerator df ($df_1$) is 3 and the denominator df ($df_2$) is 246. We want to shade the area in the right tail. Our “X Value” is 2.57. 1. Open Minitab. 2. Select **Graph > Probability Distribution Plot > View Probability**. 3. Change the Distribution to *F*. 4. Fill in the **Numerator degrees of freedom** with 3 and the **Denominator degrees of freedom** with 246. 5. Select the **Options** button. 6. Select **A specified x value**. 7. Use the default **Right tail**. 8. For the **X value**, enter 2.57. 9. Select **OK** and **OK**. ![](assets/200L9Fdistr.png){#fig-200L9Fdistr fig-alt="Distribution Plot of Density vs X - Fm df1=3, df2=246" .mx-auto .d-block width="60%" .lightbox} The area beyond an F-value of 2.57 with 3 and 246 degrees of freedom is 0.05487. The p-value for this F test is 0.05487. **Note**: When you conduct an ANOVA in Minitab, the software will compute this p-value for you. ::: ## Hypothesis Testing (ANOVA) A one-way between groups ANOVA is used to compare the means of more than two independent groups. A one-way between groups ANOVA comparing just two groups will give you the same results at the independent $t$ test that you conducted in [Lesson 8](Lesson08.qmd). We will use the five-step hypothesis testing procedure again in this lesson. ::: {.steps} 1. **Check assumptions and write hypotheses.** The assumptions for a one-way between groups ANOVA are: 1. Samples are independent 2. The response variable is approximately normally distributed for each group or all group sample sizes are at least 30 3. The population variances are equal across responses for the group levels (if the largest sample standard deviation divided by the smallest sample standard deviation is **not** greater than **two**, then assume that the population variances are equal) Given that you are comparing $k$ independent groups, the null and alternative hypotheses are: $H_{0}: \mu_1 = \mu_2 = \cdots = \mu_k$ $H_{a}:$ Not all $\mu_\cdot$ are equal In other words, the null hypothesis is that at all of the groups' population means are equal. The alternative is that they are not all equal; there are at least two population means that are not equal to one another. 2. **Calculate the test statistic.** ANOVA uses an F test statistic. Hand calculations for ANOVAs require many steps. In this class, you will be working primarily with Minitab outputs. Conceptually, the F statistic is a ratio: $$ F=\dfrac{\text{Between groups variability}}{\text{Within groups variability}} $$ Numerically this translates to $$ F=\dfrac{MS_{\text{Between}}}{MS_{\text{Within}}} $$ In other words how much do individuals in different groups vary from one another over how much to individuals within groups vary from one another. Statistical software will compute the F ratio for you and produce what is known as an ANOVA source table. The ANOVA source table will give you information about the variability between groups and within groups. **The table below gives you all the formulas, but you will not be responsible for performing these calculations by hand. Minitab will do all of these calculations for you and provide you with the full ANOVA source table.** | Source | df | SS | MS | F | p | | --- |--- |--- |--- |--- |--- | | Between Groups (Factor) | $k-1$ | $\sum_{k}n_k(\bar{x}_k-\bar{x}_\cdot)^2$ | $\dfrac{SS_{\text{Between}}}{df_{\text{Between}}}$ | $\dfrac{MS_{\text{Between}}}{MS_{\text{Within}}}$ | Area to the right of Fk-1, n-k | | Within Groups (Error) | $n-k$ | $\sum_k \sum_i(x_{ik}-\bar{x}_k)^2$ | $\dfrac{SS_{\text{Within}}}{df_{\text{Within}}}$ | | | | Total | $n-1$ | $\sum_k \sum_i(x_{ik}-\bar{x}_\cdot)^2$ | | | | : {.w-auto .table-sm .mx-auto .row-header} $k$: Number of groups $n$: Total sample size (all groups combined) $n_k$: Sample size of group $k$ $\bar{x}_k$: Sample mean of group $k$ $\bar{x}_\cdot$: Grand mean (i.e., mean for all groups combined) $SS$: Sum of squares $MS$: Mean square $df$: Degrees of freedom $F$: F-ratio (the test statistic) Some of the notation above should look familiar, while other terms will be new to you. The sum of squares that appears in the ANOVA source table is similar to the sum of squares that you computed in [Lesson 2](Lesson02.qmd) when computing variance and standard deviation. Recall, the sum of squares is the squared difference between each score and the mean. Here, there are three different sum of squares each measuring a different type of variability. The ANOVA source table also has three different degrees of freedom: $df_{\text{between}}$, $df_{\text{within}}$, and $df_{\text{total}}$. If you were to look up an F value using statistical software you would need to know two of these degrees of freedom: $df_1 = df_{\text{between}}$ and $df_2=df_{\text{within}}$. 3. **Determine the p-value.** When performing an ANOVA using statistical software, you will be given the p-value in the ANOVA source table. If performing an ANOVA by hand, you would use the F distribution. Similar to the t distribution, the F distribution varies depending on degrees of freedom. 4. **Make a decision.** If $p \leq \alpha$ reject the null hypothesis. If $p>\alpha$ fail to reject the null hypothesis. 5. **State a "real world" conclusion.** Based on your decision in **Step 4**, write a conclusion in terms of the original research question. ::: ## Pairwise Comparisons While the results of a one-way between groups ANOVA will tell you if there is what is known as a main effect of the explanatory variable, the initial results will not tell you which groups are different from one another. In order to determine which groups are different from one another, a post-hoc test is needed. Post-hoc tests are conducted after an ANOVA to determine which groups differ from one another. There are many different post-hoc analyses that could be performed following an ANOVA. Here, we will learn about one of the most common tests known as Tukey's Honestly Significant Differences (HSD) Test. Most statistical software, including Minitab, will compute Tukey's pairwise comparisons for you. This specific post-hoc test makes all possible pairwise comparisons. In this class we will be relying on statistical software to perform these analyses, if you are interested in seeing how the calculations are performed, this information is contained in the notes for STAT 502: Analysis of Variance and Design of Experiments. This analysis takes into account the fact that multiple tests are being performed and makes the necessary adjustments to ensure that Type I error is not inflated. In the examples later in this lesson you will see a number of Tukey post-hoc tests. Next, you will also learn how to obtain these results using Minitab. For each pairwise comparison, $H_0: \mu_i - \mu_j=0$ and $H_a: \mu_i - \mu_j \ne 0$. ## ![ ](assets/minitab.png){.icon-image} Minitab: One-Way ANOVA {{< include _minitab/conduct-a-one-way-anova.qmd >}} ::: {#exm-pig-weights-by-diet-in-minitab .example-minitab} ### Minitab: Pig Weights by Diet In one research study, 20 young pigs are assigned at random among 4 experimental groups. Each group is fed a different diet. (This design is a completely randomized design.) The data are the pigs' weights in kg after being raised on these diets for 10 months. We wish to determine if there are any differences in mean pig weights for the 4 diets. - $H_0: \mu_1 = \mu_2 = \mu_3 = \mu_4$ - $H_a:$ Not all $\mu$ are equal | Feed_1 | Feed_2 | Feed_3 | Feed_4 | | --- |--- |--- |--- | | 60.8 | 68.3 | 102.6 | 87.9 | | 57.1 | 67.7 | 102.2 | 84.7 | | 65.0 | 74.0 | 100.5 | 83.2 | | 58.7 | 66.3 | 97.5 | 85.8 | | 61.8 | 69.9 | 98.9 | 90.3 | : {.w-auto .table-sm .mx-auto .row-header} Contained in the Minitab file: ([ANOVA_ex.mpx](data_files/ANOVA_ex.mpx){download="" target="_blank"}) Note that in this file the data were entered so that each group is in its own column. In other words, the responses are in a separate column for each factor level. In later examples, you will see that Minitab will also conduct a one-way ANOVA if the responses are all in one column with the factor codes in another column. To perform an Analysis of Variance (ANOVA) test in Minitab: 1. Open the Minitab file: ([ANOVA_ex.mpx](data_files/ANOVA_ex.mpx){download="" target="_blank"}) 2. From the menu bar, select **Stat > ANOVA > One-Way**. 3. Click the drop-down menu and select **'Responses are in a separate column for each factor level'**. 4. Enter the variables *Feed\_1*, *Feed\_2*, *Feed\_3*, and *Feed\_4* to insert them into the **Responses** box. 5. Choose the **Comparisons** button and check the box next to **Tukey**. Under **Results**, also select **Tests**. 6. Select **OK** and **OK**. The result should be the following output: ::: {.minitab_output} ### Method {.unnumbered .unlisted} | Null hypothesis | All means are equal | | --- |--- | | Alternative hypothesis | At least one mean is different | | Significance level | $\alpha=0.05$ | : {.w-auto .table-sm .row-header} *Equal variances were assumed for the analysis* ### Factor Information {.unnumbered .unlisted} | Factor | Levels | Values | | --- |--- |--- | | Factor | 4 | Feed_1, Feed_2, Feed_3, Feed_4 | : {.w-auto .table-sm .row-header} ### Analysis of Variance {.unnumbered .unlisted} | Source | DF | Adj SS | Adj MS | F-Value | P-Value | | --- |--- |--- |--- |--- |--- | | Factor | 3 | 4703.2 | 1567.73 | 206.72 | 0.000 | | Error | 16 | 121.3 | 7.58 | | | | Total | 19 | 4824.5 | | | | : {.w-auto .table-sm .row-header} ### Model Summary {.unnumbered .unlisted} | S | R-sq | R-sq(adj) | R-sq(pred) | | --- |--- |--- |--- | | 2.75386 | 97.48% | 97.01% | 96.07% | : {.w-auto .table-sm .row-header} ### Means {.unnumbered .unlisted} | Factor | N | Mean | StDev | 95% CI | | --- |--- |--- |--- |--- | | Feed_1 | 5 | 60.68 | 3.03 | (58.07, 63.29) | | Feed_2 | 5 | 69.24 | 2.96 | (66.63, 71.85) | | Feed_3 | 5 | 100.340 | 2.164 | (97.729, 102.951) | | Feed_4 | 5 | 86.38 | 2.78 | (83.77, 88.99) | : {.w-auto .table-sm .row-header} *Pooled StDev = 2.75386* ### Grouping Information Using the Tukey Method and 95% Confidence {.unnumbered .unlisted} | Factor | N | Mean | Grouping | | --- |--- |--- |--- | | Feed_3 | 5 | 100.34 | A | | | | | Feed_4 | 5 | 86.38 | | B | | | | Feed_2 | 5 | 69.24 | | | C | | | Feed_1 | 5 | 60.68 | | | | D | : {.w-auto .table-sm .row-header} *Means that do not share a letter are significantly different.* | Difference ofLevels | Differenceof Means | SE of Difference | 95% CI | T-Value | Adjusted P-Value | | --- |--- |--- |--- |--- |--- | | Feed_2 - Feed_1 | 8.56 | 1.74 | (3.57, 13.55) | 4.91 | 0.001 | | Feed_3 - Feed_1 | 39.66 | 1.74 | (34.67, 44.65) | 22.77 | 0.000 | | Feed_4 - Feed_1 | 25.70 | 1.74 | (20.71, 30.69) | 14.76 | 0.000 | | Feed_3 - Feed_2 | 31.10 | 1.74 | (26.11, 36.09) | 17.86 | 0.000 | | Feed_4 - Feed_2 | 17.14 | 1.74 | (12.15, 22.13) | 9.84 | 0.000 | | Feed_4 - Feed_3 | -13.96 | 1.74 | (-18.95, -8.97) | -8.02 | 0.000 | : {.w-auto .table-sm .row-header} ### Tukey Simultaneous Tests for Differences of Means {.unnumbered .unlisted} *Individual confidence level = 98.87%* ![](assets/200L10Anovatukey95CIgraph.png){#fig-200L10Anovatukey95CIgraph fig-alt="ANOVA Output - Example 3" .mx-auto .d-block width="60%" .lightbox} ![](assets/200L10anovatukeyintervalplot.png){#fig-200L10anovatukeyintervalplot fig-alt="ANOVA Output - Example 4" .mx-auto .d-block width="60%" .lightbox} ::: ::: ::: {#exm-sat-math-scores-by-award-preference} ### SAT-Math Scores by Award Preference In this example, we are comparing the SAT scores of students who said that they would prefer to win an Academy Award, a Nobel Prize, or an Olympic gold medal. The example uses the StudentSurvey dataset provided by the Lock5 textbook. Let's apply the five-step hypothesis testing process to this example. ::: {.steps} 1. **Check assumptions and write hypotheses.** The assumptions for a one-way between-groups ANOVA are: **Assumption 1:** Samples are independent: Each student selected either Olympic, Academy, or Nobel. Each student is in only one group and those groups are in no way matched or paired. This assumption is met. **Assumption 2:** The response variable is approximately normally distributed for each group or all group sample sizes are at least 30: To check this we can construct a histogram with groups in Minitab. These plots show that the distributions are all approximately normal. ![](assets/200L10-5exhistogram.png){#fig-200L10-5exhistogram fig-alt="Histograms of SAT grouped by award" .mx-auto .d-block width="60%" .lightbox} **Assumption 3:** The population variances are equal across responses for the group levels (if the largest sample standard deviation divided by the smallest sample standard deviation is **not** greater than **two**, then assume that the population variances are equal). Again we can use Minitab to look at the standard deviations across the groups. The largest standard deviation is 151.3 and the smallest is 114.1 for a ratio of 1.33 which is less than 2. So this assumption is met. ::: {.minitab_output} #### Descriptive Statistics | Award | N | N* | Mean | SE Mean | StDev | Minimum | Q1 | Median | Q3 | Maximum | | ------- | --- | --- | ------ | ------- | -------------- | ------- | ------ | ------ | ------ | ------- | | Academy | 31 | 0 | 1191.0 | 27.2 | [151.3]{.mark} | 820.0 | 1070.0 | 1200.0 | 1300.0 | 1530.0 | | Nobel | 149 | 0 | 1239.1 | 9.38 | [114.5]{.mark} | 920.0 | 1160.0 | 1230.0 | 1310.0 | 1550.0 | | Olympic | 182 | 0 | 1176.7 | 8.46 | [114.1]{.mark} | 800.0 | 1100.0 | 1190.0 | 1250.0 | 1470.0 | : {.w-auto .table-sm .row-header} ::: Use Minitab to run a one-way ANOVA with the Minitab file: ([StudentSurvey.mpx](data_files/StudentSurvey.mpx){download="" target="_blank"}) The following will describe the output within the context of the five-step hypothesis testing process. Given that you are comparing $k$ independent groups, the null and alternative hypotheses are: ::: {.minitab_output} ### Method {.unnumbered .unlisted} **Null hypothesis:** All means are equal. **Alternative hypothesis:** At least one mean is different. **Significance level:** $\alpha=0.05$ *Equal variances were assumed for the analysis* ::: 2. **Calculate the test statistic.** You should get the following output: ::: {.minitab_output} ### Factor Information {.unnumbered .unlisted} | Factor | Levels | Values | | --- |--- |--- | | Award | 3 | Academy, Nobel, Olympic | : {.w-auto .table-sm .row-header} The factor information tells us that our factor is the Award. ### Analysis of Variance {.unnumbered .unlisted} | Source | DF | Adj SS | Adj MS | F-Value | P-Value | | --- |--- |--- |--- |--- |--- | | Award | 2 | 323269 | 162134 | 11.67 | 0.000 | | Error | 359 | 4986078 | 13889 | | | | Total | 361 | 5310347 | | | | : {.w-auto .table-sm .row-header} ::: We need to know two of these degrees of freedom: $df_1 = df_{\text{between}}$ and $df_2=df_{\text{within}}$. The analysis of variance table is also known as our ANOVA source table. The source of the Award is our between-groups variation so the DF K -1 or 3-1 = 2. Error is our within-groups variation, so the degrees of freedom are n - K or 362-3 = 359. The 'Adj' stands for adjusted. For a one-way ANOVA, nothing is being adjusted. The F-value is our F-test statistic, which in this case is 11.67 with a p-value of 0.000. The F value could be written as F (2, 359) = 11.67. ::: {.minitab_output} ### Means {.unnumbered .unlisted} | Factor | N | Mean | StDev | 95% CI | | --- |--- |--- |--- |--- | | Academy | 31 | 1191.0 | 151.3 | (1149.3, 1232.6) | | Nobel | 149 | 1239.11 | 114.54 | (1220.13, 1258.10) | | Olympic | 182 | 1176.73 | 114.13 | (1159.55, 1193.91) | : {.w-auto .table-sm .row-header} *Pooled StDev = 117.851* ::: The Means table is just a table of descriptive statistics. The sample size, mean, and the standard deviation is presented for each group. The last column is an unadjusted 95% confidence interval. You should not refer to these confidence intervals since these do not take into account that there are three different groups and three different confidence intervals being computed at the same time. The confidence intervals that we're interested in come with our Tukey pairwise comparison. 3. **Determine the p-value.** The p-value is 0.000 from the Minitab output. 4. **Make a decision.** $p \leq \alpha$ so reject the null hypothesis. 5. **State a "real world" conclusion.** We can conclude that the group means are not all equal. ::: Remember that ANOVA is just an omnibus test. It only tells us there is a difference somewhere. To determine where the differences are you would have to look at the Tukey pairwise comparison output. ::: {.minitab_output} ### Grouping Information Using the Tukey Method and 95% Confidence {.unnumbered .unlisted} | Award | N | Mean | Grouping | | --- |--- |--- |--- | | Nobel | 149 | 1239.11 | A | | | Academy | 31 | 1190.97 | A | B | | Olympic | 182 | 1176.73 | | B | : {.w-auto .table-sm .row-header} *Means that do not share a letter are significantly different.* ::: In the grouping table, Nobel and Olympic do not share a grouping letter. The means of these two groups are significantly different. Both the Nobel and Academy belong to Group A and both Academy and Olympic belong to Group B. These pairs are not significantly different. ### Tukey Simultaneous Tests for Differences of Means {.unnumbered .unlisted} ::: {.minitab_output} | Difference of Levels | Differenceof Means | SE ofDifference | 95% CI | T-Value | AdjustedP-Value | | --- |--- |--- |--- |--- |--- | | Nobel - Academy | 48.1 | 23.3 | (-6.3, 102.6) | 2.07 | 0.096 | | Olympic - Academy | -14.2 | 22.9 | (-67.8, 39.4) | -0.62 | 0.808 | | Olympic - Nobel | -62.4 | 13.0 | (-92.9, -31.9) | -4.79 | 0.000 | : {.w-auto .table-sm .row-header} *Individual confidence level = 98.87%* ::: The Tukey Simultaneous Tests table provides the adjusted 95% confidence intervals. These are the confidence intervals and p-values you should be looking at when you're making pairwise comparisons. These are adjusted to take into account that there are three different pairwise tests being performed simultaneously. To determine which groups are statistically significant we can look at the adjusted p-value. In this case, the only p-value less than 0.05 is the Olympic - Nobel pairwise comparison. The means of the Olympic and Nobel group are significantly different. ![](assets/200L10-5extukey95.png){#fig-200L10-5extukey95 fig-alt="ANOVA Output - Example 3" .mx-auto .d-block width="60%" .lightbox} In the Tukey Simultaneous 95% CI graph any interval not containing zero indicates a statistically significant difference. In this case, we see that the Olympic - Nobel group does not contain zero, so this pairing has a statistically significant difference. ::: ::: {#exm-exam-grade-by-professor} #### Exam Grade by Professor [**Scenario:**]{.fs-4} Three professors were each teaching one section of a course. They all gave the same final exam, and they want to know if there are any differences between their sections’ mean scores. ::: {.steps} 1. **Write hypotheses and check assumptions.** $H_0:\mu_1=\mu_2=\mu_3$ $H_a$: Not all $\mu$ are equal. ::: {.minitab_output} ### Means {.unnumbered .unlisted} | Instructor | N | Mean | StDev | 95% CI | | --- |--- |--- |--- |--- | | Dr. Al | 60 | 68.367 | 17.719 | (63.977, 72.756) | | Dr. Oh | 87 | 71.448 | 16.702 | (67.803, 75.094) | | Dr. Pa | 98 | 67.939 | 17.465 | (64.504, 71.373) | : {.w-auto .table-sm .row-header} *Pooled StDev = 17.2609* ::: The standard deviations for all three classes are all similar. 2. **Compute test statistic.** 1. Open the Minitab file: ([ANOVA_Exam_Profs.mpx](data_files/ANOVA_Exam_Profs.mpx){download="" target="_blank"}) 2. Using Minitab: **Stat > ANOVA > One-Way** The result is the following ANOVA source table: ::: {.minitab_output} ### Analysis of Variance {.unnumbered .unlisted} | Source | DF | Adj SS | Adj MS | F-Value | P-Value | | --- |--- |--- |--- |--- |--- | | Instructor | 2 | 635.3 | 317.7 | 1.07 | 0.346 | | Error | 242 | 72101.1 | 297.9 | | | | Total | 244 | 72736.4 | | | | : {.w-auto .table-sm .row-header} F (2, 242) = 1.07 ::: 3. **Find the p-value.** From our ANOVA source table, p = .346 4. **Make a decision.** Because $p > \alpha$, we fail to reject the null hypothesis. 5. **State a conclusion.** There is not enough evidence to conclude that the mean scores from the three different professors’ sections are different. ::: ::: ::: {#exm-tukey-pairwise-comparisons} #### Tukey Pairwise Comparisons There is some debate as to whether pairwise comparisons are appropriate when the overall one-way ANOVA is not statistically significant. Some argue that if the overall ANOVA is not significant then pairwise comparisons are not necessary. Others argue that if the pairwise comparisons were planned before the ANOVA was conducted (i.e., "a priori") then they are appropriate. The results of our Tukey pairwise comparisons were as follows: ::: {.minitab_output} ### Grouping Information Using the Tukey Method and 95% Confidence {.unnumbered .unlisted} | Instructor | N | Mean | Grouping | | --- |--- |--- |--- | | Dr. Oh | 87 | 71.448 | A | | Dr. Al | 60 | 68.367 | A | | Dr. Pa | 98 | 67.939 | A | : {.w-auto .table-sm .row-header} *Means that do not share a letter are significantly different.* ### Tukey Simultaneous Tests for Differences of Means {.unnumbered .unlisted} | Difference of Levels | Difference of Means | SE of Difference | 95% CI | T-Value | Adjusted P-Value | | --- |--- |--- |--- |--- |--- | | Dr. Oh-Dr. Al | 3.08 | 2.90 | (-3.70, 9.86) | 1.06 | 0.537 | | Dr. Pa-Dr. Al | -0.43 | 2.83 | (-7.05, 6.19) | -0.15 | 0.987 | | Dr. Pa-Dr. Oh | -3.51 | 2.54 | (-9.46, 2.44) | -1.38 | 0.351 | : {.w-auto .table-sm .row-header} *Individual confidence level = 97.99%* ::: Looking at the first table, all three instructors are in group A. Means that share a letter are not significantly different from one another (i.e., they are in the same group). Because all three instructors share the letter A, there are no significantly different pairs of instructors. We could also look at the second table which gives us the t-test statistic and adjusted p-value for each possible pairwise comparison. This p-value is adjusted to take into account that multiple tests are being conducted. You can compare these p-values to the standard alpha level of .05. All p-values are greater than .05, therefore no pairs are significantly different from one another. ::: ## Summary In this lesson you learned how to compare the means of three or more groups using a one-way between groups ANOVA. A one-way between groups ANOVA is used instead of multiple independent $t$ tests in order to avoid increasing the likelihood of committing a Type I error. An ANOVA provides information about the explanatory variable overall, but not about differences between the different levels of that variable. In order to compare the different pairs we need to conduct a post-hoc analysis such as Tukey's HSD test.

Source	df	SS	MS	F	p
Between Groups (Factor)	\(k-1\)	\(\sum_{k}n_k(\bar{x}_k-\bar{x}_\cdot)^2\)	\(\dfrac{SS_{\text{Between}}}{df_{\text{Between}}}\)	\(\dfrac{MS_{\text{Between}}}{MS_{\text{Within}}}\)	Area to the right of Fk-1, n-k
Within Groups (Error)	\(n-k\)	\(\sum_k \sum_i(x_{ik}-\bar{x}_k)^2\)	\(\dfrac{SS_{\text{Within}}}{df_{\text{Within}}}\)
Total	\(n-1\)	\(\sum_k \sum_i(x_{ik}-\bar{x}_\cdot)^2\)