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7.
Product and Process Comparisons
7.4. Comparisons based on data from more than two processes 7.4.7. How can we make multiple comparisons?
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| Tukey's method considers all possible pairwise differences of means at the same time | The Tukey method applies simultaneously to the set of all pairwise comparisons $$ \{ \mu_i - \mu_j \} \, . $$ The confidence coefficient for the set, when all sample sizes are equal, is exactly \(1 - \alpha\). For unequal sample sizes, the confidence coefficient is greater than \(1 - \alpha\). In other words, the Tukey method is conservative when there are unequal sample sizes. | ||
| Studentized Range Distribution | |||
| The studentized range \(q\) | The Tukey method uses the studentized range distribution. Suppose we have \(r\) independent observations \(y_1, \, \ldots, \, y_r\) from a normal distribution with mean \(\mu\) and variance \(\sigma^2\). Let \(w\) be the range for this set , i.e., the maximum minus the minimum. Now suppose that we have an estimate \(s^2\) of the variance \(\sigma^2\) which is based on \(\nu\) degrees of freedom and is independent of the \(y_i\). The studentized range is defined as $$ q_{r, \, \nu} = \frac{w}{s} \, . $$ | ||
| The distribution of \(q\) is tabulated in many textbooks |
The distribution of \(q\)
has been tabulated and appears in many textbooks on statistics.
As an example, let \(r\)
= 5 and \(\nu\)
= 10. The
95th percentile is \(q_{0.05; \, 5, \, 10}\)
= 4.65. This means:
$$ \mbox{P } \left\{ \frac{w}{s} \le 4.65 \right\} = 0.95 \, . $$
So, if we have five observations from a normal distribution, the
probability is 0.95 that their range is not more than 4.65 times as
great as an independent sample standard deviation estimate for
which the estimator has 10 degrees of freedom.
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| Tukey's Method | |||
| Confidence limits for Tukey's method |
The Tukey confidence limits for all pairwise comparisons with
confidence coefficient of at least \(1 - \alpha\)
are:
$$ \bar{y}_{i \scriptsize{\, \bullet}} - \bar{y}_{j \scriptsize{\, \bullet}} \pm
\frac{1}{\sqrt{2}} \, q_{\alpha; \, r, \, N-r} \,\, \hat{\sigma}_\epsilon \, \sqrt{\frac{2}{n}}
\,\,\,\,\, i, \, j = 1, \, \ldots, \, r; \,\, i \ne j \, . $$
Notice that the point estimator and the estimated variance are the
same as those for a
single pairwise comparison
that was illustrated previously. The only
difference between the confidence limits for simultaneous
comparisons and those for a single comparison is the multiple of
the estimated standard deviation.
Also note that the sample sizes must be equal when using the studentized range approach. |
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| Example | |||
| Data | We use the data from a previous example. | ||
| Set of all pairwise comparisons | The set of all pairwise comparisons consists of: $$ \mu_2 - \mu_1, \,\, \mu_3 - \mu_1, \,\, \mu_1 - \mu_4 \, , $$ $$ \mu_2 - \mu_3, \,\, \mu_2 - \mu_4, \,\, \mu_3 - \mu_4 \, . $$ | ||
| Confidence intervals for each pair | Assume we want a confidence coefficient of 95 percent, or 0.95. Since \(r\) = 4 and \(n_t\) = 20, the required percentile of the studentized range distribution is \(q_{0.05; \, 4, \, 16}\). Using the Tukey method for each of the six comparisons yields: $$ \begin{eqnarray} 0.29 \le & \mu_2 - \mu_1 & \le 4.47 \\ & & \\ 1.13 \le & \mu_3 - \mu_1 & \le 5.31 \\ & & \\ -2.25 \le & \mu_1 - \mu_4 & \le 1.93 \\ & & \\ -2.93 \le & \mu_2 - \mu_3 & \le 1.25 \\ & & \\ 0.13 \le & \mu_2 - \mu_4 & \le 4.31 \\ & & \\ 0.97 \le & \mu_3 - \mu_4 & \le 5.15 \\ \end{eqnarray} $$ | ||
| Conclusions | The simultaneous pairwise comparisons indicate that the differences \(\mu_1 - \mu_4\) and \(\mu_2 - \mu_3\) are not significantly different from 0 (their confidence intervals include 0), and all the other pairs are significantly different. | ||
| Unequal sample sizes | It is possible to work with unequal sample sizes. In this case, one has to calculate the estimated standard deviation for each pairwise comparison. The Tukey procedure for unequal sample sizes is sometimes referred to as the Tukey-Kramer Method. | ||
