Solution to Exercise 11.08.

  The data were entered into minitab and a two way ANOVA run with the option ``Fit Additive Model'' selected. Recall that with 1 observation per cell in a two way layout, this is the best that can be done, i.e. there are not enough degrees of freedom to estimate a full model with the interaction term. In the notation in the box on top of p. 449, we can estimate the $\alpha_i$ terms and the $\beta_j$ terms, but not the $\gamma_{ij}$ terms. The results of the minitab analysis are below.

Two-way Analysis of Variance


Analysis of Variance for epineph 
Source        DF        SS        MS        F        P
anesthet       2    1.1458    0.5729     6.11    0.009
subject        9    0.9872    0.1097     1.17    0.370
Error         18    1.6887    0.0938
Total         29    3.8217

                       Individual 95% CI
anesthet        Mean   ---------+---------+---------+---------+--
1              0.434    (---------*---------)
2              0.443    (---------*---------)
3              0.853                        (----------*---------)
                       ---------+---------+---------+---------+--
                            0.400     0.600     0.800     1.000

                       Individual 95% CI
subject         Mean   -+---------+---------+---------+---------+
 1              0.55         (----------*---------)
 2              0.75               (---------*----------)
 3              0.77               (----------*----------)
 4              0.35   (----------*----------)
 5              0.58          (----------*---------)
 6              0.92                    (---------*----------)
 7              0.40     (---------*----------)
 8              0.59          (----------*---------)
 9              0.50        (---------*----------)
10              0.35   (----------*----------)
                       -+---------+---------+---------+---------+
                     0.00      0.35      0.70      1.05      1.40

Here are the plots we produced:

(1)

Boxplot of the response epinephrine by Anesthetic;

(2)

Dotplots of the response by Subject (with only 3 responses per subject, a boxplot doesn't make much sense);

(3)

Residuals vs. fitted values;

(4)

Normal probability plot of the residuals.

Conclusions for part (a): There is a significant effect from Anesthetic at the 0.05 level of significance (p = 0.009).

Comments: The diagnostic plots suggest there could be some problem with the additive model, i.e. that there is a Subject-Anesthetic interaction. Five subjects (numbers 1, 2, 3, 5, and 9) show a pattern of two low values and one high value - I wish I could plot symbols ``1'', ``2'', and ``3'' so I could see it on the graph, but a check of the numerical values shows that for subjects 1, 2, 5, and 9 that the high value is for Anesthetic number 3. Four other subjects (numbers 4, 7, 8, and 10) seem to have little variation in the response, and subject 6 has a wide variation in the 3 responses. The boxplots of response by Anesthetic shows wide variation in Anesthetic 3. Thus, we see a possible pattern of ``strong response to Anesthetic 3'' about 1/2 the time, and occasional strong response to the other two anesthetics (see the ``outliers'' in the box plot). All this variation just feeds into the mean square due to error (MSE) and inflates the denominator of the F-statistic, making it harder to find an effect. Since we did find a significant effect from Anesthetic, we conclude it is probably real (but may not exist in all patients!). On the other hand, the P-value for the effect from subjects (p = 0.370) being not statistically significant should be taken with a ``grain of salt.''

(b) To compute the Tukey simultaneous intervals, we have to apply the results on p. 440 ``by hand'' as minitab doesn't compute these intervals for anything but one way ANOVA. We are interested in factor ``A'' - the anesthetic. The width of the intervals is

$$w \; = \; Q_{\alpha,I,(I-1)(J-1)} * \sqrt{MSE/J}$$

where

Our estimated means for the anesthetic effect from the minitab output are, in increasing order:

$$\hat{\mu}_1 = 0.434 \quad \hat{\mu}_2 = 0.443 \quad \hat{\mu}_3 = 0.853$$

The differences:

We see that the differences between the estimated mean response of Anesthetic 3 exceeds the critical value of 0.350, and on the positive side, so we can conclude that the mean response of Anesthetic 3 is statistically significantly larger than the other two, and there is not a significant difference between the other two.

Now we apply Bonferroni's method. We either calculate confidence intervals for the difference of means or perform t-tests and adjust the level of significance. The latter is easier to do in minitab, so let's do that. Note that we don't need to assume equal variances, etc. for this analysis, or even the additive model. In fact, the best way to go is to use the paired sample t-test, because when we test Anesthetic 1 vs. Anesthetic 2, we want to keep the measurements within subjects as there is clearly potential for dependence of the measurements performed on the same subject.

To set up the minitab worksheet to do this, we went to Manip $\rightarrow$ Stack/Unstack $\rightarrow$ Unstack One Column. In the dialogue box, we selected ``Unstack the data in'' epineph, the response variable, ``Using Subscripts in'' anesthet, and ``Store the Unstacked Data'' in c4-c6. We subsequently relabelled c4-c6 as A1-A3, added a column with the subject numbers 1 to 10 (from Calc $\rightarrow$ Make Patterned Data $\rightarrow$ Simple Sets of Numbers), and the resulting worksheet (including our original data) is shown in Table 4.

 

anesthet subject epineph A1 A2 A3 subject2

1 1 0.28 0.28 0.30 1.07 1

2 1 0.30 0.51 0.39 1.35 2

3 1 1.07 1.00 0.63 0.69 3

1 2 0.51 0.39 0.38 0.28 4

2 2 0.39 0.29 0.21 1.24 5

3 2 1.35 0.36 0.88 1.53 6

1 3 1.00 0.32 0.39 0.49 7

2 3 0.63 0.69 0.51 0.56 8

3 3 0.69 0.17 0.32 1.02 9

1 4 0.39 0.33 0.42 0.30 10

2 4 0.38

3 4 0.28

1 5 0.29

2 5 0.21

3 5 1.24

1 6 0.36

2 6 0.88

3 6 1.53

1 7 0.32

2 7 0.39

3 7 0.49

1 8 0.69

2 8 0.51

3 8 0.56

1 9 0.17

2 9 0.32

3 9 1.02

1 10 0.33

2 10 0.42

3 10 0.30

The results of performing three paired sample t-tests:

Paired T-Test and Confidence Interval

Paired T for A1 - A2

                  N      Mean     StDev   SE Mean
A1               10    0.4340    0.2443    0.0772
A2               10    0.4430    0.1916    0.0606
Difference       10   -0.0090    0.2347    0.0742

95% CI for mean difference: (-0.1769, 0.1589)
T-Test of mean difference = 0 (vs not = 0): T-Value = -0.12  P-Value = 0.906


Paired T-Test and Confidence Interval

Paired T for A1 - A3

                  N      Mean     StDev   SE Mean
A1               10     0.434     0.244     0.077
A3               10     0.853     0.448     0.142
Difference       10    -0.419     0.550     0.174

95% CI for mean difference: (-0.812, -0.026)
T-Test of mean difference = 0 (vs not = 0): T-Value = -2.41  P-Value = 0.039


Paired T-Test and Confidence Interval

Paired T for A2 - A3

                  N      Mean     StDev   SE Mean
A2               10     0.443     0.192     0.061
A3               10     0.853     0.448     0.142
Difference       10    -0.410     0.453     0.143

95% CI for mean difference: (-0.734, -0.086)
T-Test of mean difference = 0 (vs not = 0): T-Value = -2.86  P-Value = 0.019

To correct for the multiple comparisons via Bonferroni's method, we simply divide the desired Family-wise error rate (overall level of significance) by the number of comparisons (3 in this case) and perform each test at that level. Thus, we perform the individual comparisons at the level

$$\alpha_{individual} \; = \; \alpha_{overall}/\mbox{no. of comparisons} \; = \; 0.05/3 \; = \; 0.0167 .$$

We see that none of the three P-values (0.906 for A1-A2, 0.039 for A1-A3, and 0.019 for A2-A3) is significant at this level. Interestingly, this analysis also picks up rather large differences in the standard deviations of the differences: 0.2347 for A1-A2, 0.550 for A1-A3, and 0.453 for A2-A3. I'm somewhat inclined to think that the assumption of equal standard deviations and an additive model in the previous 2-way ANOVA allowed the smaller standard deviations for A1-A2 to deflate the overall standard deviation and produce a possibly spurious statistical significance for the A1,A2 comparisons with A3. Anyway, this analysis is close to giving us the same results as before (since the p-values for A1-A3 and A2-A3 were pretty small), so any disagreement is somewhat overstated.