power has dropped to 0.56. That is, if the model assumptions hold and the true pooled standard deviation is 0.25, only 56% of the experiments (assuming that we repeat this experiment several times) with 10 measurements from each group would successfully detect the difference of 0.25 kgf/cm2. What sample size would be necessary to achieve a power of 0.9 for this specific difference to detect?
7. Clear the Sample Size field and enter 0.9 for Power.
8. Click Continue.
The required total sample size is 45. This means that we need at least 22.5 observations per group. Rounding up, we see that we need at least 23 observations from each group to achieve a power of at least 0.9. We could have left the Power field blank, specifying only that the Difference to detect is 0.25. The Sample Size and Power platform would then have produced a power curve, displaying Power as a function of Sample Size.
9. Select Window > Close All.
Example 2.1 Hypothesis Testing
1. Open Fluorescence.jmp.
2. Click Analyze > Fit Y by X.
3. Select Fluorescence for Y, Response and Tissue for X, Factor.
4. Click OK.
5. Click the red triangle next to One-way Analysis of Fluorescence by Tissue and select Normal Quantile Plot > Plot Quantile by Actual.
Since the slopes of the lines in the normal quantile plots are proportional to the sample standard deviations of the treatments, the difference between the slopes of the lines for Muscle and Nerve indicates that the variances may be different between the groups. As a result, we will use a form of the t-test that does not assume that the population variances are equal. Formal testing for the equality of the treatment variances is illustrated in Example 2.3 at the end of this chapter.
6. Click the red triangle next to One-way Analysis of Fluorescence by Tissue and select t-Test.
The p-value for the one-sided hypothesis test is 0.0073, which is less than the set α of 0.05. We therefore reject the null hypothesis and conclude that the mean normalized fluorescence for nerve tissue is greater than the mean normalized fluorescence for muscle tissue. Subject matter knowledge would need to determine if there is a practical difference; confidence intervals for the differences (reported in JMP) can be beneficial for this assessment.
7. Select Window > Close All.
Section 2.5.1 The Paired Comparison Problem
1. Open Hardness-Testing.jmp
2. Select Analyze > Matched Pairs.
3. Select Tip 1 and Tip 2 for Y, Paired Response.
4. Click OK.
The p-value, Prob > |t| = 0.7976, indicates that there is no evidence of a difference in the performance of the two tips. This p-value is larger than the standard significance level of α = 0.05.
5. Leave Hardness-Testing.jmp open for the next exercise.
Section 2.5.2 Advantages of the Paired Comparison Design
1. Return to the Hardness-Testing table opened in the previous example.
2. Select Tables > Stack. This will create a file in long format with one observation per row. Most JMP platforms expect data to appear in long format.
3. Select Tip 1 and Tip 2 for Stack Columns.
4. Type “Depth” in the Stacked Data Column field.
5. Type “Tip” in the Source Label Column field.
6. Type “Hardness-Stacked” in the Output table name field.
7. Click OK.
8. Hardness-Stacked is now the current data table. Select Analyze > Fit Y by X.
9. Select Depth for Y, Response and Tip for X, Grouping.
10. Click OK.
11. Click the red triangle next to One-way Analysis of Depth by Tip and select Means/Anova/Pooled t.
The root mean square error of 2.315407 is the pooled standard deviation estimate from the t-test. Compared to the standard deviation estimate of 1.20 from the paired difference test, we see that blocking has reduced the estimate of variability considerably. Though we do not work through the details here, it would be possible to perform this same comparison for the Fluorescence data from Example 2.1.
12. Leave Hardness-Stacked.jmp and the Fit Y by X output window open for the next exercise.
Example 2.3 Testing for the Equality of Variances
This example demonstrates how to test for the equality of two population variances. Section 2.6 of the textbook also discusses hypothesis testing for whether the variance of a single population is equal to a given constant. Though not shown here, the testing for a single variance may be performed in the Distribution platform.
1. Return to the Fit Y by X platform from the previous example.
2. Click the red triangle next to One-way Analysis of Depth by Tip and select Unequal Variances.
3. Save Hardness-Stacked.jmp.
The p-value for the F test (described in the textbook) for the null hypothesis of equal variances (with a two-sided alternative hypothesis) is 0.8393. The data do not indicate a difference with respect to the variances of depth produced from Tip 1 versus Tip 2. Due to the use of a slightly different data set, the F Ratio of 1.1492 reported here is different from the ratio of 1.34 that appears in the book. Furthermore, the textbook uses a one-sided test with an alternative hypothesis. That hypothesis is that the variance of the depth produced by Tip 1 is greater than that produced by Tip 2. Since the sample standard deviation from Tip 1 is greater than that from Tip 2, the F Ratios for the one- and two-sided tests are both equal to 1.1492, but the p-value for the one-sided test would be 0.4197.
It is important to remember that the F test is extremely sensitive to the assumption of normality. If the population has heavier tails than a normal distribution, this test will reject the null hypothesis (that the population variances are equal) more often than it should. By contrast, the Levene test is robust to departures from normality.
4. Select Window > Close All.
3
Experiments with a Single Factor: The Analysis of Variance
