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TABLE 20.3
Values of the Studentized Range Statistic q k,ν,α/2 for k(k – 1)/2
Two-Sided Comparisons for a Joint 95% Confidence Interval
Where There are a Total of k Treatments
k
νν νν 2 3 4 5 6 8 10
5 4.47 5.56 6.26 6.78 7.19 7.82 8.29
10 3.73 4.47 4.94 5.29 5.56 5.97 6.29
15 3.52 4.18 4.59 4.89 5.12 5.47 5.74
20 3.43 4.05 4.43 4.70 4.91 5.24 5.48
30 3.34 3.92 4.27 4.52 4.72 5.02 5.24
60 3.25 3.80 4.12 4.36 4.54 4.81 5.01
∞ 3.17 3.68 3.98 4.20 4.36 4.61 4.78
Note: Family error rate = 5%; α/2 = 0.05/2 = 0.025.
Source: Harter, H. L. (1960). Annals Math. Stat., 31, 1122–1147.

where it is assumed that the two treatments have the same variance, which is estimated by pooling the
two sample variances:

2
( n i – 1)s i + ( n j – 2
s pool = ----------------------------------------------------
1)s j
2
n i + n j – 2
The chance that the interval includes the true value for any single comparison is exactly 1 – α. But the
chance that all possible k(k – 1)/2 intervals will simultaneously contain their true values is less than 1 – α.
Tukey (1949) showed that the confidence interval for the difference in two means (η i and η j ), taking
into account that all possible comparisons of k treatments may be made, is given by:

1
1
q k,ν,α/2
y i – y j ± ---------------s pool ---- + ----
2 n i n j
where q k,ν,α/2 is the upper significance level of the studentized range for k means and ν degrees of freedom
2 2
in the estimate s pool of the variance σ . This formula is exact if the numbers of observations in all the
averages are equal, and approximate if the k treatments have different numbers of observations. The
2
value of s pool is obtained by pooling sample variances over all k treatments:
( n 1 – 1)s 1 + … + ( n k – 2
2
s pool = -----------------------------------------------------------------
1)s k
2
n 1 + … + n k – k
The size of the confidence interval is larger when q k,ν,α/2 is used than for the t statistic. This is because
the studentized range allows for the possibility that any one of the k(k – 1)/2 possible pair-wise comparisons
might be selected for the test. Critical values of q k,v,α/2 have been tabulated by Harter (1960) and may be
found in the statistical tables of Rohlf and Sokal (1981) and Pearson and Hartley (1966). Table 20.3 gives
a few values for computing the two-sided 95% confidence interval.

Solution: Tukey’s Method
2
For this example, k = 5, s pool = 0.51, s pool = 0.71, ν = 50 – 5 = 45, and q 5,40,0.05/2 = 4.49. This gives the
95% confidence limits of:

1
1
( y i – y j ) ± 4.49 ) ------ + ------
---------- 0.71(
2 10 10
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