Why Recovering Entries Requires a System
A one-way analysis of variance table summarizes variation among groups and variation within groups. In an exam problem, some entries may be given while others are omitted. The task is not simply to fill blanks by pattern: each entry is linked to specific sums of squares, degrees of freedom, and mean-square calculations. A reliable solution identifies which relationships apply, uses them in a logical order, and checks the completed table.
The table is based on a one-way fixed-effects model. For observation \(j\) in group \(i\), write \(Y_{ij}=\mu_i+\varepsilon_{ij}\), where \(\mu_i\) is the mean for group \(i\). For the usual exact F test, assume the errors are independent, normally distributed with mean zero, and have a common variance \(\sigma^2\). The arithmetic identities used to complete a table hold for the observed data even when these assumptions fail; however, the usual F-distribution reference for inference then may not apply.
These entries answer different questions. The sums of squares measure amounts of variation, their degrees of freedom count independent pieces of information associated with that variation, and each mean square is a sum of squares divided by its degrees of freedom. The F statistic is a ratio of the two mean squares, not a ratio of the two sums of squares.
The Identities That Link the Entries
The central sum-of-squares relationship is \(SS_T=SS_B+SS_W\). The degrees of freedom obey an analogous addition: \((k-1)+(N-k)=N-1\). These two partitions provide immediate checks and often allow one missing entry to be recovered. Once the sums of squares and degrees of freedom are known, calculate the mean squares; calculate \(F\) last unless the problem asks you to work backward from it.
The sum-of-squares partition follows from how each observation differs from the grand mean. Let \(\bar{Y}_{i\cdot}\) denote group \(i\)'s sample mean and \(\bar{Y}_{\cdot\cdot}\) the grand mean. Decompose each deviation as \((Y_{ij}-\bar{Y}_{\cdot\cdot})=(Y_{ij}-\bar{Y}_{i\cdot})+(\bar{Y}_{i\cdot}-\bar{Y}_{\cdot\cdot})\). Squaring and summing gives the within-group and between-group squared deviations, plus a cross-product. For each group, the cross-product contains \(\sum_j(Y_{ij}-\bar{Y}_{i\cdot})\), which is zero because deviations from a group mean sum to zero. The cross-product therefore vanishes, leaving \(SS_T=SS_W+SS_B\).
Record \(k\) and \(N\), then calculate \(df_B=k-1\), \(df_W=N-k\), and \(df_T=N-1\). If the table gives degrees of freedom, use them to check the stated group and observation counts.
Use \(SS_T=SS_B+SS_W\). A missing sum of squares can be found by addition or subtraction, but it cannot be negative.
Divide each sum of squares by its matching degrees of freedom. Do not divide \(SS_B\) by \(df_W\), or \(SS_W\) by \(df_B\).
Divide \(MS_B\) by \(MS_W\). If \(F\) is given instead, work backward using \(MS_B=FMS_W\) or \(MS_W=MS_B/F\), as appropriate.
Recovering Entries in Practice
The examples use hypothetical studies and show different directions of calculation. Keep exact fractions during intermediate steps where practical, and round only a final reported statistic. Every completed table should satisfy both partitions, as well as the definitions of its mean squares and F ratio.
Worked Example: Recovering Within-Group Variation
A hypothetical study compares the average calibration error from four measurement procedures. There are 24 independent readings, with six readings per procedure. The reported table gives \(SS_B=36\) and \(SS_T=156\), but omits \(SS_W\), the mean squares, and \(F\). Complete the table.
Method. First find the degrees of freedom from the group count and total sample size: \(df_B=4-1=3\), \(df_W=24-4=20\), and \(df_T=24-1=23\). Check that \(3+20=23\).
Use the sum-of-squares partition to find the missing within-group sum of squares: \(SS_W=SS_T-SS_B=156-36=120\). The check is \(36+120=156\).
Now calculate the mean squares using their own degrees of freedom: \(MS_B=36/3=12\), and \(MS_W=120/20=6\). Thus \(F=12/6=2\). Rounded to two decimal places, \(F=2.00\).
| Source | Sum of squares | Degrees of freedom | Mean square | F |
|---|---|---|---|---|
| Between groups | 36 | 3 | 12 | 2.00 |
| Within groups | 120 | 20 | 6 | — |
| Total | 156 | 23 | — | — |
Check. The sums of squares add to 156, the degrees of freedom add to 23, and \(12/6=2\). The common mistake this method avoids is treating the total sum of squares as the within-group sum of squares; here the within-group value is \(156-36\), not 156.
Worked Example: Finding the F Statistic From Partial Sums of Squares
A hypothetical evaluation compares three training formats using 30 participant scores in total. The table reports \(SS_T=210\) and \(SS_B=60\). Find the missing within-group sum of squares, the two mean squares, and the F statistic.
Method. There are \(k=3\) groups and \(N=30\) observations, so \(df_B=3-1=2\), \(df_W=30-3=27\), and \(df_T=30-1=29\). The degree-of-freedom check gives \(2+27=29\).
The missing sum of squares is \(SS_W=SS_T-SS_B=210-60=150\), and \(60+150=210\) verifies the partition. The mean squares are \(MS_B=60/2=30\) and \(MS_W=150/27=50/9\), or approximately \(5.5556\).
The F ratio is \(F=30/(50/9)=30(9/50)=270/50=5.4\). Rounded to two decimal places, \(F=5.40\). Dividing by the exact fraction avoids unnecessary intermediate rounding.
Check. Multiplying the within-group mean square by its degrees of freedom gives \((50/9)(27)=150\), and \(30/(50/9)=5.4\). A tempting but incorrect calculation is \(SS_B/SS_W=60/150=0.4\). That is not the F statistic: each sum of squares must first be divided by its own degrees of freedom.
Worked Example: Working Backward From a Reported F Ratio
A hypothetical materials study compares five manufacturing settings using 40 test pieces. A partial table gives \(SS_T=360\) and \(MS_W=8\), and reports \(F=2.5\). Recover \(SS_B\), \(SS_W\), both degrees of freedom, and \(MS_B\).
Method. With \(k=5\) and \(N=40\), the degrees of freedom are \(df_B=5-1=4\), \(df_W=40-5=35\), and \(df_T=40-1=39\). The check is \(4+35=39\).
Since \(MS_W=SS_W/df_W\), recover the within-group sum of squares as \(SS_W=MS_Wdf_W=8(35)=280\). The sum-of-squares partition then gives \(SS_B=SS_T-SS_W=360-280=80\).
Work backward from the F ratio: \(F=MS_B/MS_W\), so \(MS_B=FMS_W=2.5(8)=20\). Check this against the between-group sum of squares: \(SS_B/df_B=80/4=20\), as required.
Check. The sums of squares give \(80+280=360\), and the mean-square ratio gives \(20/8=2.5\). The common mistake avoided here is multiplying the reported F ratio by \(SS_W\) instead of by \(MS_W\). The F ratio relates mean squares, not raw sums of squares.
Consistency Checks and Exam Traps
A missing-entry problem can also test whether the supplied information is mutually compatible. Before accepting a completed table, verify every identity that can be checked from the entries. Small discrepancies can arise when a problem reports rounded values; substantial discrepancies usually indicate a calculation error, a mismatched degree of freedom, or inconsistent inputs.
- Check the degrees of freedom: \(df_B+df_W=df_T\), with \(df_B=k-1\), \(df_W=N-k\), and \(df_T=N-1\).
- Check the sums of squares: \(SS_B+SS_W=SS_T\). Neither component should be negative.
- Check each mean square: divide by the degrees of freedom from the same row.
- Check the F ratio: \(F=MS_B/MS_W\), not \(MS_W/MS_B\).
- Check that the calculation is defined: the ordinary one-way F ratio requires positive within-group degrees of freedom and \(MS_W>0\). If \(MS_W=0\), the ratio is undefined or may require special treatment; do not divide by zero.
For instance, suppose a table states that \(df_B=4\), \(df_W=15\), and \(df_T=20\). The first two entries sum to 19, not 20, so the degrees-of-freedom partition fails. Do not proceed as though the table is valid; revisit the group count, sample size, or reported entry. Similarly, if \(SS_T=90\) and \(SS_B=100\), subtraction implies \(SS_W=-10\), which is impossible for a sum of squared deviations. The given values cannot describe a valid ANOVA partition as stated.
Check Your Understanding
Complete each table or calculation. Treat all study settings as hypothetical, and round only the final F statistic to two decimal places when needed.
- A study compares four inspection methods using 28 readings. Given \(SS_B=54\) and \(SS_W=216\), find \(SS_T\), the three degrees of freedom, both mean squares, and \(F\). (Answer: \(SS_T=270\); \(df_B=3\), \(df_W=24\), \(df_T=27\); \(MS_B=18\), \(MS_W=9\); \(F=2.00\).)
- A study compares three procedures using 18 observations. Given \(SS_T=170\) and \(SS_B=50\), find \(SS_W\), all degrees of freedom, both mean squares, and \(F\). (Answer: \(SS_W=120\); \(df_B=2\), \(df_W=15\), \(df_T=17\); \(MS_B=25\), \(MS_W=8\); \(F=3.125\).)
- A study has six groups and 31 observations. Given \(MS_W=4\), \(F=3.5\), and \(SS_T=170\), recover \(SS_W\), \(SS_B\), both mean squares, and all degrees of freedom. (Answer: \(SS_W=100\), \(SS_B=70\); \(MS_W=4\), \(MS_B=14\); \(df_B=5\), \(df_W=25\), \(df_T=30\).)
- A study compares four settings. The table reports \(df_W=22\), \(SS_W=110\), \(MS_B=18\), and \(F=3.6\). Recover \(N\), all degrees of freedom, \(MS_W\), \(SS_B\), and \(SS_T\). (Answer: \(N=26\); \(df_B=3\), \(df_W=22\), \(df_T=25\); \(MS_W=5\); \(SS_B=54\); \(SS_T=164\).)