Why ANOVA Tables Can Be Reconstructed
A one-way ANOVA table organizes several related calculations into three sources of variation: between groups, within groups, and total. Because the entries are connected by fixed arithmetic identities, an omitted value can often be recovered from the remaining entries. The task is not to guess what a blank might be; it is to identify which relationship determines it and then check that the result agrees with the rest of the table.
This tutorial focuses on recovering missing sums of squares, degrees of freedom, mean squares, F statistics, and total-row entries. The table calculations are bookkeeping identities. Their use in an inferential ANOVA test rests on the usual one-way ANOVA model: observations in group \(i\) can be represented as \(Y_{ij}=\mu_i+\epsilon_{ij}\), where errors are independent and normally distributed with mean 0 and a common variance \(\sigma^2\). The arithmetic identities themselves do not require normally distributed errors; those assumptions are relevant to the usual F-test interpretation.
The Relationships That Fill the Blanks
Write the table columns in a consistent order: source, sum of squares (SS), degrees of freedom (\(df\)), mean square (MS), and, where applicable, \(F\). The between-groups and within-groups mean squares are used to form the F statistic. A total-row mean square and F statistic are not part of the standard one-way ANOVA table, so those cells are usually left blank.
| Source | SS | \(df\) | MS | \(F\) |
|---|---|---|---|---|
| Between groups | \(SS_{\text{Between}}\) | \(k-1\) | \(SS_{\text{Between}}/(k-1)\) | \(MS_{\text{Between}}/MS_{\text{Within}}\) |
| Within groups | \(SS_{\text{Within}}\) | \(N-k\) | \(SS_{\text{Within}}/(N-k)\) | — |
| Total | \(SS_{\text{Total}}\) | \(N-1\) | — | — |
The main identities can be rearranged to solve for a blank. For example, if a within-groups mean square is missing, multiply the within-groups sum of squares by the reciprocal of its degrees of freedom. If its sum of squares is missing, multiply the mean square by the degrees of freedom. If the between-groups sum of squares is missing but the total and within-groups sums are known, subtract the within-groups value from the total.
These rules give a useful order of work. First establish the group count and total sample size, if available, so the degrees of freedom are known. Then use sums-of-squares addition or subtraction. Calculate mean squares from their own rows, and calculate \(F\) only after both mean squares are available. Finally, check all the relationships again. This last check matters because one blank can be calculated correctly from one pair of entries while another supplied entry elsewhere in the table is inconsistent.
Record \(k\), the number of groups, and \(N\), the total number of observations, when provided. Use them to determine the three degrees of freedom.
Use \(SS_{\text{Total}}=SS_{\text{Between}}+SS_{\text{Within}}\), rearranging by addition or subtraction as needed.
Divide each between- or within-groups sum of squares by its own degrees of freedom. Then divide the between-groups mean square by the within-groups mean square.
Verify both degrees-of-freedom addition and sums-of-squares addition. Recalculate the mean squares and \(F\) from the displayed values, allowing for rounding.
Worked Reconstructions
Worked Example: Recovering a Within-Groups Mean Square and \(F\)
A hypothetical experiment compares the means of three groups, with 18 observations altogether. Its ANOVA table includes \(SS_{\text{Between}}=48\), \(SS_{\text{Within}}=90\), and leaves the within-groups mean square and \(F\) blank. Reconstruct the omitted values and complete the table.
There are \(k=3\) groups and \(N=18\) observations. Thus, \(df_{\text{Between}}=k-1=3-1=2\), \(df_{\text{Within}}=N-k=18-3=15\), and \(df_{\text{Total}}=N-1=18-1=17\). The degrees of freedom check because \(2+15=17\).
The total sum of squares is \(SS_{\text{Total}}=48+90=138\). The missing within-groups mean square is calculated using the within-groups row: \(MS_{\text{Within}}=90/15=6\). For the F statistic, first calculate the between-groups mean square: \(MS_{\text{Between}}=48/2=24\). Therefore, \(F=24/6=4\).
The completed entries are: between groups, SS \(=48\), \(df=2\), MS \(=24\), \(F=4\); within groups, SS \(=90\), \(df=15\), MS \(=6\); total, SS \(=138\), \(df=17\). The sums-of-squares check is \(48+90=138\), as required.
Worked Example: Finding an Omitted Sum of Squares
In a hypothetical one-way ANOVA with four groups and 24 observations, the total sum of squares is 250 and the between-groups sum of squares is 70. The within-groups sum of squares, the between-groups mean square, and \(F\) are omitted. Find them.
The degrees of freedom are \(df_{\text{Between}}=4-1=3\), \(df_{\text{Within}}=24-4=20\), and \(df_{\text{Total}}=24-1=23\). These add correctly: \(3+20=23\).
Rearrange the total sum-of-squares identity to obtain \(SS_{\text{Within}}=SS_{\text{Total}}-SS_{\text{Between}}\). Substitution gives \(SS_{\text{Within}}=250-70=180\). The between-groups mean square is \(MS_{\text{Between}}=70/3=23.333\ldots\), or approximately \(23.33\). The within-groups mean square is \(MS_{\text{Within}}=180/20=9\). Thus, \(F=23.333\ldots/9=2.59259\ldots\), or approximately \(2.593\).
The reconstructed table has between-groups SS \(=70\), \(df=3\), MS \(=23.33\), \(F=2.593\); within-groups SS \(=180\), \(df=20\), MS \(=9\); and total SS \(=250\), \(df=23\). The sum check is \(70+180=250\). The unrounded values give \(F=(70/3)/9\); using the rounded mean square may produce a tiny display difference.
Worked Example: Reconstructing Several Rows From Mean Squares
A hypothetical table for three groups reports \(N=15\), \(MS_{\text{Within}}=4.5\), and \(F=3.2\). The between-groups and within-groups sums of squares and the total row are missing. Find all recoverable entries.
The degrees of freedom are \(df_{\text{Between}}=3-1=2\), \(df_{\text{Within}}=15-3=12\), and \(df_{\text{Total}}=15-1=14\). The given within-groups mean square determines its sum of squares: \(SS_{\text{Within}}=MS_{\text{Within}}\times df_{\text{Within}}=4.5\times12=54\).
Since \(F=MS_{\text{Between}}/MS_{\text{Within}}\), multiply both sides by \(MS_{\text{Within}}\) to solve for the missing between-groups mean square: \(MS_{\text{Between}}=F\times MS_{\text{Within}}=3.2\times4.5=14.4\). Then \(SS_{\text{Between}}=MS_{\text{Between}}\times df_{\text{Between}}=14.4\times2=28.8\). The total sum of squares is \(SS_{\text{Total}}=28.8+54=82.8\).
The reconstructed table is: between groups, SS \(=28.8\), \(df=2\), MS \(=14.4\), \(F=3.2\); within groups, SS \(=54\), \(df=12\), MS \(=4.5\); total, SS \(=82.8\), \(df=14\). Checks confirm the arithmetic: \(2+12=14\), \(28.8+54=82.8\), and \(14.4/4.5=3.2\).
Why the Checks Matter
Each relationship is a separate check on the table. Degrees of freedom depend on the number of groups and observations, not on the observed sizes of the sums of squares. Sums of squares must add across the between- and within-groups rows. Mean squares use the degrees of freedom from their own rows, and the F statistic uses the between-groups mean square as its numerator and the within-groups mean square as its denominator.
The identities also make it possible to diagnose a table that cannot be reconstructed consistently. Suppose a table gives \(SS_{\text{Total}}=100\), \(SS_{\text{Between}}=35\), and \(SS_{\text{Within}}=70\). The two component sums add to \(105\), not \(100\). At least one reported value is inconsistent with the others. Do not silently alter a supplied value to force agreement; state that the entries fail the sum-of-squares check and identify the discrepancy.
Displayed entries may be rounded. For instance, a reported mean square of \(23.33\) may represent an unrounded value of \(23.333\ldots\), so a recomputed F statistic can differ slightly in its final decimal place. Prefer calculations from unrounded entries when they are available. If only rounded values are shown, distinguish a small rounding difference from a substantial arithmetic mismatch.
A table reconstruction is complete only when every requested entry has been calculated and the independent identities agree. In a standard table, the total row provides total SS and total degrees of freedom; it does not provide a mean square for calculating \(F\). Keeping that structure in mind prevents filling a blank simply because another column appears in the table.
Check Your Understanding
Use the ANOVA table relationships to answer each question. Show the arithmetic and include a check where appropriate.
- There are five groups and 30 observations. What are the between-groups, within-groups, and total degrees of freedom?
- If \(SS_{\text{Total}}=96\) and \(SS_{\text{Between}}=31\), find \(SS_{\text{Within}}\).
- If \(SS_{\text{Within}}=84\) and \(df_{\text{Within}}=14\), find \(MS_{\text{Within}}\).
- If \(MS_{\text{Between}}=18\) and \(MS_{\text{Within}}=6\), find \(F\). Which mean square is in the numerator?
- A table reports total SS \(=120\), between-groups SS \(=45\), and within-groups SS \(=80\). Do these entries pass the sums-of-squares check? Explain.