10: OneWay ANOVA
10: OneWay ANOVAObjectives
 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 oneway between groups ANOVA with Tukey's pairwise comparisons
 Interpret the results of a oneway between groups ANOVA
 Interpret the results of Tukey's pairwise comparisons
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 oneway between groups analysis of variance, or more often as a "oneway ANOVA."
Why not multiple independent ttests?
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 oneway between groups ANOVA you would need to perform more tests. For \(k\) independent groups there are \(\frac{k(k1)}{2}\) possible pairs. If you had 5 independent groups, that would equal \(\frac{5(51)}{2}=10\) 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
10.1  Introduction to the F DistributionOneway 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 pvalues 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 pvalue for a oneway ANOVA.
The video below gives a brief introduction to the F distribution and walks you through two examples of using Minitab to find the pvalues 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 righttailed probability. Later in this lesson we will see that this area is the pvalue.
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

\(df_{between}=k1\)

\(k\) = number of groups
 Within Groups (Denominator, Error) Degrees of Freedom

\(df_{within}=nk\)

\(n\) = total sample size with all groups combined
\(k\) = number of groups
Minitab^{®} – Creating an F Distribution
Scenario: An F test statistic of 2.57 is computed with 3 and 246 degrees of freedom. What is the pvalue 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
 OK and OK
The area beyond an Fvalue of 2.57 with 3 and 246 degrees of freedom is 0.05487. The pvalue for this F test is 0.05487.
Note: When you conduct an ANOVA in Minitab, the software will compute this pvalue for you.
10.2  Hypothesis Testing
10.2  Hypothesis TestingA oneway between groups ANOVA is used to compare the means of more than two independent groups. A oneway 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.
The assumptions for a oneway between groups ANOVA are:
 Samples are independent
 The response variable is approximately normally distributed for each group or all group sample sizes are at least 30
 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.
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: \(F=\frac{Between\;groups\;variability}{Within\;groups\;variability}\). Numerically this translates to \(F=\frac{MS_{Between}}{MS_{Within}}\). 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 of 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)  \(k1\)  \(\sum_{k}n_k(\overline{x}_k\overline{x}_\cdot)^2\)  \(\dfrac{SS_{Between}}{df_{Between}}\)  \(\dfrac{MS_{Between}}{MS_{Within}}\)  Area to the right of F_{k1, nk} 
Within Groups (Error)  \(nk\)  \(\sum_k \sum_i(x_{ik}\overline{x}_k)^2\)  \(\dfrac{SS_{Within}}{df_{Within}}\)  
Total  \(n1\)  \(\sum_k \sum_i(x_{ik}\overline{x}_\cdot)^2\) 
\(k\)  Number of groups 
\(n\)  Total sample size (all groups combined) 
\(n_k\)  Sample size of group \(k\) 
\(\overline{x}_k\)  Sample mean of group \(k\) 
\(\overline{x}_\cdot\)  Grand mean (i.e., mean for all groups combined) 
SS  Sum of squares 
MS  Mean square 
df  Degrees of freedom 
F  Fratio (the test statistic) 
Some of the terms in the table above should look familiar, while others 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: \(df_{between}\), \(df_{within}\), and \(df_{total}\). If you were to look up an F value using statistical software you would need to know two of these degrees of freedom: \(df_1 = df_{between}\) and \(df_2=df_{within}\).
When performing an ANOVA using statistical software, you will be given the pvalue 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.
If \(p \leq \alpha\) reject the null hypothesis. If \(p>\alpha\) fail to reject the null hypothesis.
Based on your decision in Step 4, write a conclusion in terms of the original research question.
10.3  Pairwise Comparisons
10.3  Pairwise ComparisonsWhile the results of a oneway 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 posthoc test is needed. Posthoc tests are conducted after an ANOVA to determine which groups differ from one another. There are many different posthoc 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 posthoc 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 posthoc 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: OneWay ANOVA
10.4  Minitab: OneWay ANOVAIn 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 oneway ANOVA if the responses are all in one column with the factor codes in another column.
Minitab^{®} – OneWay ANOVA
To perform an Analysis of Variance (ANOVA) test in Minitab:
 Open the Minitab file: ANOVA_ex.mpx
 From the menu bar, select Stat > ANOVA > OneWay.
 Click the dropdown 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.
 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  FValue  PValue 

Factor  3  4703.2  1567.73  206.72  0.000 
Error  16  121.3  7.58  
Total  19  4824.5 
Model Summary
S  Rsq  Rsq(adj)  Rsq(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  B  
Feed_2  5  69.24  C  
Feed_1  5  60.68  D 
Means that do not share a letter are significantly different.
Difference of Levels 
Difference of Means 
SE of Difference 
95% CI  TValue  Adjusted PValue 

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%
10.5  Example: SATMath Scores by Award Preference
10.5  Example: SATMath Scores by Award PreferenceIn 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 fivestep hypothesis testing process to this example.
The assumptions for a oneway betweengroups ANOVA are:
Assumption: 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: 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.
Assumption: 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.
Statistics
Award  N  N*  Mean  SE Mean  StDev  Minimum  Q1  Median  Q3  Maximum 

Academy  31  0  1191.0  27.2  151.3  820.0  1070.0  1200.0  1300.0  1530.0 
Nobel  149  0  1239.1  9.38  114.5  920.0  1160.0  1230.0  1310.0  1550.0 
Olympic  182  0  1176.7  8.46  114.1  800.0  1100.0  1190.0  1250.0  1470.0 
Use Minitab to run a oneway ANOVA with the Minitab file: StudentSurvey.mpx
The following will describe the output within the context of the fivestep 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
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  FValue  PValue 

Award  2  323269  162134  11.67  0.000 
Error  359  4986078  13889  
Total  361  5310347 
The analysis of variance table is also known as our ANOVA source table. The source of the Award is our betweengroups variation so the DF K 1 or 31 = 2. Error is our withingroups variation so the degrees of freedom are n  K or 3623 = 359. The 'Adj' stands for adjusted. For a oneway ANOVA, nothing is being adjusted. The Fvalue is our Ftest statistic, which in this case is 11.67 with a pvalue 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.
need to know two of these degrees of freedom: \(df_1 = df_{between}\) and \(df_2=df_{within}\).
The pvalue is 0.000 from the Minitab output.
\(p \leq \alpha\) so reject the null hypothesis.
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.
Grouping Information Using the Tukey Method and 95% Confidence
Award  N  Mean  Grouping  

Nobel  149  1239.11  A  
Academy  31  1190.97  A  B 
Olympic  182  1176.73  B 
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
Difference of Levels  Difference  SE of  95% CI  TValue  Adjusted 

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 pvalues 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 pvalue. In this case, the only pvalue 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.
10.6  Example: Exam Grade by Professor
10.6  Example: Exam Grade by ProfessorScenario
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.
\(H_0:\mu_1=\mu_2=\mu_3\)
\(H_a: Not\;all\;\mu\;are\;equal\)
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.
 Open the Minitab file: ANOVA_Exam_Profs.mpx
 Using Minitab: Stat > ANOVA > OneWay
The result is the following ANOVA source table:
Analysis of Variance
Source  DF  Adj SS  Adj MS  FValue  PValue 

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
From our ANOVA source table, p = .346
Because \(p > \alpha\), we fail to reject the null hypothesis.
There is not enough evidence to conclude that the mean scores from the three different professors’ sections are different.
Tukey Pairwise Comparisons
There is some debate as to whether pairwise comparisons are appropriate when the overall oneway 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  TValue  Adjusted PValue 

Dr. OhDr. Al  3.08  2.90  (3.70, 9.86)  1.06  0.537 
Dr. PaDr. Al  0.43  2.83  (7.05, 6.19)  0.15  0.987 
Dr. PaDr. 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 ttest statistic and adjusted pvalue for each possible pairwise comparison. This pvalue is adjusted to take into account that multiple tests are being conducted. You can compare these pvalues to the standard alpha level of .05. All pvalues are greater than .05, therefore no pairs are significantly different from one another.
10.7  Lesson 10 Summary
10.7  Lesson 10 SummaryObjectives
 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 oneway between groups ANOVA with Tukey's pairwise comparisons
 Interpret the results of a oneway between groups ANOVA
 Interpret the results of Tukey's pairwise comparisons
In this lesson you learned how to compare the means of three or more groups using a oneway between groups ANOVA. A oneway 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 posthoc analysis such as Tukey's HSD test.