Why Fisher’s Protected LSD Uses a Gate
An omnibus ANOVA test can indicate that not all population means are equal, but it does not identify which means differ. Fisher’s protected least significant difference (LSD) procedure follows a significant omnibus F-test with pairwise t-tests. The initial F-test is a gate: if it is not significant, the procedure makes no pairwise rejections.
The method is simple and can be useful when a follow-up comparison is needed, but its protection against false discoveries is limited. In particular, the gate does not generally provide strong familywise error control when some population means differ and others are equal.
The Omnibus Gate and Pairwise LSD Rule
Let group \(i\) have sample size \(n_i\) and sample mean \(\bar{Y}_i\), and let \(N=\sum_{i=1}^k n_i\). The omnibus null hypothesis is \(H_0:\mu_1=\cdots=\mu_k\). Let \(SSE\) be the within-group sum of squares. The pooled error mean square is \(MSE=SSE/(N-k)\), with \(\nu=N-k\) error degrees of freedom. The omnibus statistic is \(F=MS_{\text{group}}/MSE\), with \(k-1\) numerator and \(\nu\) denominator degrees of freedom.
Choose a significance level \(\alpha\). First reject the omnibus null only if \(F>F_{1-\alpha;\,k-1,\nu}\), the upper-tail critical value from the F distribution. If the omnibus test does not reject, stop: protected LSD makes no pairwise rejections. If it does reject, test each pair \(i,j\) using a two-sided t-test at level \(\alpha\), with the pooled ANOVA error mean square.
For a pair \(i,j\), the least significant difference is the critical t-value times this standard error. The pair is declared significantly different when its observed absolute mean difference exceeds that LSD.
Why the Pairwise Formula Holds
The formula uses the variability of a difference between two independent sample means. Under the model, \(\operatorname{Var}(\bar{Y}_i)=\sigma^2/n_i\) and \(\operatorname{Var}(\bar{Y}_j)=\sigma^2/n_j\). Independence makes the variance of their difference the sum of these variances.
Under normality, \((\bar{Y}_i-\bar{Y}_j)-(\mu_i-\mu_j)\), divided by its true standard deviation, has a standard normal distribution. Also, \(SSE/\sigma^2\) has a chi-square distribution with \(\nu=N-k\) degrees of freedom and is independent of the group means. Replacing \(\sigma^2\) by \(MSE\) therefore gives a t statistic with \(\nu\) degrees of freedom under \(H_0:\mu_i=\mu_j\). A two-sided level-\(\alpha\) test rejects when the absolute t statistic exceeds \(t_{1-\alpha/2,\nu}\), which gives the stated LSD rule after multiplying by the standard error.
Worked Example: Four Unequal Groups and the Gate
A hypothetical laboratory compares the mean calibration errors, in units, from four procedures. Assume independent normal observations with a common variance. The group sizes are \(5,7,8,6\), and the sample means are \(40,45,43,52\), respectively. The ANOVA summary gives \(MSE=25\), with \(\alpha=0.05\). Determine whether the gate opens and, if it does, which pairs differ by protected LSD.
Step 1: Degrees of freedom and omnibus test. The total sample size is \(N=5+7+8+6=26\), so the error degrees of freedom are \(\nu=26-4=22\). The grand mean is \[ \bar{Y}_{..}=\frac{5(40)+7(45)+8(43)+6(52)}{26} =\frac{1171}{26}=45.03846. \] The group sum of squares is
Thus \(MS_{\text{group}}=450.96154/3=150.32051\), and \(F=150.32051/25=6.01282\). The 5% upper-tail critical value for \(F_{3,22}\) is approximately \(3.049\). Since \(6.01282>3.049\), reject the omnibus null and proceed to the pairwise tests.
Step 2: Calculate all six pair-specific LSDs. The two-sided critical value is \(t_{0.975,22}\approx2.074\). For example, for groups 1 and 2, \[ SE_{12}=\sqrt{25(1/5+1/7)}=\sqrt{60/7}\approx2.92770, \] so \(LSD_{12}=2.074(2.92770)\approx6.072\). Applying the same calculation to every pair gives:
| Pair | Absolute mean difference | Standard error | LSD | Decision |
|---|---|---|---|---|
| 1 and 2 | 5 | 2.92770 | 6.072 | Not significant |
| 1 and 3 | 3 | 2.85044 | 5.912 | Not significant |
| 1 and 4 | 12 | 3.02765 | 6.279 | Significant |
| 2 and 3 | 2 | 2.58775 | 5.367 | Not significant |
| 2 and 4 | 7 | 2.78174 | 5.769 | Significant |
| 3 and 4 | 9 | 2.70031 | 5.600 | Significant |
Each LSD is computed using the unrounded standard error and rounded to three decimals. The comparisons identify differences between procedures 1 and 4, 2 and 4, and 3 and 4 at the stated pairwise level. The omnibus rejection itself only establishes that at least one population mean differs; the pairwise results provide the follow-up detail.
Balanced Groups and a Gate That Stays Closed
When all group sizes are equal, every pair has the same standard error and therefore the same LSD. The next example shows this convenient feature and the required stop rule when the omnibus test is not significant.
Worked Example: Equal Group Sizes and a Significant Gate
A hypothetical study compares mean response times under three settings. Each group has \(n_i=5\) independent normally distributed observations, with a common variance. The sample means are \(18,25,21\), and the ANOVA gives \(MSE=10\). Use \(\alpha=0.05\), and assume the omnibus gate is tested before the pairs.
Omnibus calculation. Here \(N=15\), \(\nu=15-3=12\), and the grand mean is \((18+25+21)/3=64/3\). Since the groups have equal sizes,
Therefore \(MS_{\text{group}}=(370/3)/2=185/3\), and \(F=(185/3)/10=37/6\approx6.1667\). Since \(F_{0.95;2,12}\approx3.885\) and \(6.1667>3.885\), the gate opens.
Pairwise calculations. The critical value is \(t_{0.975,12}\approx2.179\). For every pair, \[ SE=\sqrt{10(1/5+1/5)}=\sqrt{4}=2, \qquad LSD=2.179(2)=4.358. \] The absolute differences are \(|18-25|=7\), \(|18-21|=3\), and \(|25-21|=4\). Only \(7>4.358\), so only the first and second settings are significantly different by protected LSD.
Worked Example: A Nonsignificant Gate Means Stop
A hypothetical experiment compares three independent groups, each with four normally distributed observations and a common variance. The sample means are \(10,11,12\), and \(MSE=9\). At \(\alpha=0.05\), the residual sum of squares is \(SSE=81\), so \(\nu=12-3=9\).
The grand mean is \((10+11+12)/3=11\), and \[ SS_{\text{group}}=4[(10-11)^2+(11-11)^2+(12-11)^2]=8. \] Thus \(MS_{\text{group}}=8/2=4\), and \(F=4/9\approx0.4444\). The critical value is \(F_{0.95;2,9}\approx4.256\). Since \(0.4444<4.256\), do not reject the omnibus null. Protected LSD stops here: it makes no pairwise rejections, even though the observed sample means are not identical.
What “Protected” Does and Does Not Mean
The omnibus gate gives Fisher’s procedure a limited form of familywise protection. Under the complete null, all population means are equal. A false pairwise rejection can then occur only if the omnibus test rejects, because the procedure does not conduct pairwise tests unless that gate opens. Therefore the probability of at least one false pairwise rejection under the complete null is no greater than the level \(\alpha\) of the omnibus test.
That guarantee is called weak familywise error control: it applies when every pairwise null hypothesis is true. It is not a general guarantee that the probability of one or more false rejections is at most \(\alpha\) for every configuration of means. If some means differ while another pair of means is equal, the omnibus gate may open because of the real differences; the remaining equal-mean pairs are then tested at the unadjusted level \(\alpha\). Thus protected LSD does not generally provide strong familywise error control across all pairwise comparisons.
This distinction matters when many pairs are examined. “Protected” does not mean that every collection of pairwise tests has been adjusted to maintain a specified familywise error rate under every possible pattern of true and false pairwise null hypotheses. Use the procedure with that limitation in mind, and report that the pairwise tests followed a significant omnibus F-test.
Common Exam Trap: Equal-Size Shortcuts With Unequal Groups
For unequal group sizes, the pairwise standard error must use both group sizes. A tempting shortcut is to use \(2MSE/n_i\), which is valid for a pair only when both groups have the same size \(n_i\).
Wrong calculation: Suppose two groups have sizes 5 and 6, \(MSE=25\), and the error degrees of freedom are 22. Using the size-5 shortcut gives \(SE=\sqrt{2(25)/5}=\sqrt{10}\approx3.162\), and an apparent LSD of \(2.074(3.162)\approx6.559\).
Correction: Use both sample sizes: \[ SE=\sqrt{25(1/5+1/6)} =\sqrt{275/30} \approx3.02765, \qquad LSD=2.074(3.02765)\approx6.279. \] For an absolute difference of \(6.4\), the incorrect shortcut says “not significant,” while the correct calculation says “significant,” because \(6.4>6.279\). The shortcut is not justified for unequal group sizes.
Check Your Understanding
For each problem, apply the gate where information is provided and use the pooled ANOVA error variance for any pairwise LSD calculation.
- A pair has sample sizes 4 and 9, \(MSE=16\), and \(\nu=11\). Use \(t_{0.975,11}=2.201\). If the omnibus gate has opened, calculate the LSD to three decimals and decide whether an observed absolute difference of 5.5 is significant. (Answer: \(SE=\sqrt{16(1/4+1/9)}=\sqrt{52/9}=2.403701\); \(LSD=2.201(2.403701)=5.290547\), or 5.291 to three decimals; significant because \(5.5>5.291\).)
- Two groups each have \(n=6\), \(MSE=4\), and \(\nu=15\). Use \(t_{0.975,15}=2.131\). If the gate has opened, find the LSD to three decimals. (Answer: \(SE=\sqrt{4(1/6+1/6)}=\sqrt{4/3}=1.154701\); \(LSD=2.131(1.154701)=2.460667\), or 2.461 to three decimals.)
- An ANOVA has \(k=3\), \(N=18\), \(MS_{\text{group}}=12\), and \(MSE=4\). The 5% critical value for \(F_{2,15}\) is 3.682. Should protected LSD proceed? (Answer: \(F=12/4=3\), and \(3<3.682\); the gate stays closed, so make no pairwise rejections.)
- Explain what the omnibus gate guarantees about familywise error under the complete null, and what it does not generally guarantee when some population means differ. (Answer: Under the complete null, any false pairwise rejection requires rejection of the omnibus test, so the familywise error probability is at most \(\alpha\). This is weak control; strong control for every configuration of means is not generally guaranteed.)