Chapter 13 How it Works

13.1 Conducting A One-Way Within-Groups ANOVA

Researchers followed the progress of 42 people undergoing inpatient rehabilitation following a spinal cord injury (White, Driver, & Warren, 2010). They assessed the patients on a variety of measures on three separate occasions—when they were admitted to the rehabilitation facility, 3 weeks later, and at discharge. Below are data that reflect the patients’ symptoms of depression on the Patient Health Questionnaire-9 (PHQ-9). (The data for these three fictional patients have the same means as the actual larger data set, as well as the same outcome in terms of the decision in step 6 of the ANOVA below.)

Admission Three Weeks Discharge
Patient 1 6.1 5.5 5.3
Patient 2 6.9 5.7 4.2
Patient 3 7.4 6.5 4.9

How can we use one-way within-groups ANOVA to determine whether depression levels changed as patients went through rehabilitation for spinal cord injury? We’ll walk through all six steps of hypothesis testing for a one-way within-groups ANOVA.

  1. Population 1: People just admitted to an inpatient rehabilitation facility following a spinal cord injury. Population 2: People 3 weeks after they were admitted to an inpatient rehabilitation facility following a spinal cord injury. Population 3: People being discharged from an inpatient rehabilitation facility following spinal cord injury.
    The comparison distribution will be an F distribution. The hypothesis test will be a one-way within-groups ANOVA. Regarding the assumptions: (1) The patients were not selected randomly (all were from the same hospital), so we must generalize with caution. (2) We do not know if the underlying population distributions are normal, but the sample data do not indicate severe skew. (3) To see if we meet the homoscedasticity assumption, we will check to see if the variances are similar (typically, when the largest variance is not more than twice the smallest) when we calculate the test statistic. (4) The experimenter could not counterbalance, so order effects might be present. With different levels of a time-related variable, it is not possible to assign someone to be measured at, for example, the final time point before the first time point.
  2. Null hypothesis: People in an inpatient rehabilitation hospital for a spinal cord injury have the same levels of depression, on average, at admission, 3 weeks later, and at discharge—H0: μ1 = μ2 = μ3. Research hypothesis: People in an inpatient rehabilitation hospital for a spinal cord injury do not have the same levels of depression, on average, at admission, 3 weeks later, and at discharge—H1 is that at least one μ is different from another μ.
  3. We use an F distribution with 2 and 4 degrees of freedom.
    dfbetween = Ngroups − 1 = 3 − 1 = 2
    dfsubjects = n − 1 = 3 − 1 = 2
    dfwithin = (dfbetween)(dfsubjects) = (2)(2) = 4
    dftotal = dfbetween + dfsubjects + dfwithin = 2 + 2 + 4 = 8 (or dftotal = Ntotal − 1 = 9 − 1 = 8)
  4. The critical value for the F statistic for a p level of 0.05 and 2 and 4 degrees of freedom is 6.95.
  5. SStotal = Σ(X − GM)2 = 8.059.
    Time X X − GM (X − GM)2
    Admission 6.1       0.267       0.071      
    Admission 6.9       1.067       1.138      
    Admission 7.4       1.567       2.455      
    Three weeks 5.5       −0.333       0.111      
    Three weeks 5.7       −0.133       0.018      
    Three weeks 6.5       0.667       0.445      
    Discharge 5.3       −0.533       0.284      
    Discharge 4.2       −1.633       2.667      
    Discharge 4.9       −0.933       0.87      
    GM = 5.833            Σ(X − GM)2 = 8.059

    SSbetween = Σ(M − GM)2 = 6.018

    Time X Group Mean (M) M − GM (M − GM)2
    Admission 6.1       6.8       0.967       0.935      
    Admission 6.9       6.8       0.967       0.935      
    Admission 7.4       6.8       0.967       0.935      
    Three weeks 5.5       5.9       0.067       0.004      
    Three weeks 5.7       5.9       0.067       0.004      
    Three weeks 6.5       5.9       0.067       0.004      
    Discharge 5.3       4.8       −1.033       1.067      
    Discharge 4.2       4.8       −1.033       1.067      
    Discharge 4.9       4.8       −1.033       1.067      
    GM = 5.833             Σ(M − GM)2 = 6.018

    SSsubjects = Σ(Mparticipant − GM)2 = 0.846

    Participant Time X Participant Mean (Mparticipant) Mparticipant − GM (Mparticipant − GM)2
    1 Admission 6.1 5.633 −0.2 0.040
    2 Admission 6.9 5.6 −0.233 0.054
    3 Admission 7.4 6.267   0.434 0.188
    1 Three weeks 5.5 5.633 −0.2 0.040
    2 Three weeks 5.7 5.6 −0.233 0.054
    3 Three weeks 6.5 6.267   0.434 0.188
    1 Discharge 5.3 5.633 −0.2 0.040
    2 Discharge 4.2 5.6 −0.233 0.054
    3 Discharge 4.9 6.267   0.434 0.188
    GM = 5.833                    Σ(Mparticipant − GM)2 = 0.846

    SSwithin = SStotal − SSbetween − SSsubjects = 8.059 − 6.018 − 0.846 = 1.195

    We now have enough information to fill in the first three columns of the source table—the source, SS, and df columns—and to divide each sum of squares by the degrees of freedom to get variance, MS.

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    We then calculate two F statistics—one for between-groups and one for subjects—by dividing each MS by the within-groups MS.

    The completed source table is:

    Source SS df MS F
    Between 6.018 2 3.009 10.06
    Subjects 0.846 2 0.423   1.41
    Within 1.195 4 0.299
    Total 8.059 8

    We want to know if there’s a statistically significant difference between groups, so we’ll look at the between-groups F statistic, 10.06.

  6. The F statistic, 10.06, is beyond the critical value, 6.95. We can reject the null hypothesis. It appears that depression scores differ based on the time point during rehabilitation. A post hoc test is necessary to know exactly which pairs of means are significantly different.