It appears from the overlapped histograms that the unmodified mortar tends to produce stronger bonds than the modified mortar. The unmodified mortar has a mean strength of 17.04 kgf/cm2 with a standard deviation of 0.25 kgf/cm2. The modified mortar has a mean strength of 16.76 kgf/cm2 with a standard deviation of 0.32 kgf/cm2. A naïve comparison of mean strength indicates that the unmodified mortar outperforms the modified mortar. However, the difference in means could simply be a result of sampling fluctuation. Using statistical theory, our goal is to incorporate the sample standard deviations (and sample sizes) to quantify how likely it is that the difference in mean strengths is due only to sampling error. If it turns out to be unlikely, we will conclude that a true difference exists between the mortar strengths.
11. Select Analyze > Fit Y by X.
12. Select Strength for Y, Response and Mortar for X, Grouping.
The Fit Y by X platform recognizes this as a one-way ANOVA since the response, Strength, is a continuous factor, and the factor Mortar is a nominal factor. When JMP is used to create experimental designs, it assigns the appropriate variable type to each column. For imported data, JMP assigns a modeling type—continuous
13. Click OK.
14. To create box plots, click the red triangle next to One-way Analysis of Strength by Mortar and select Quantiles.
The median modified mortar strength (represented by the line in the middle of the box) is lower than the median unmodified mortar strength. The similar length of the two boxes (representing the interquartile ranges) indicates that the two mortar formulations result in approximately the same variability in strength.
15. Keep the Fit Y by X platform open for the next exercise.
Section 2.4.1 Hypothesis Testing
1. Return to the Fit Y by X platform from the previous exercise.
2. Click the red triangle next to One-way Analysis of Strength by Mortar and select Means/Anova/Pooled t.
The t-test report shows the two-sample t-test assuming equal variances. Since we have a two-sided alternative hypothesis, we are concerned with the p-value labeled Prob > |t|= 0.0422. Since we have set α=0.05, we reject the null hypothesis that the mean strengths produced by the two formulations of mortar are equal and conclude that the mean strength of the modified mortar and the mean strength of the unmodified mortar are (statistically) significantly different. In practice, our next step would be to decide from a subject-matter perspective if the difference is practically significant.
Before accepting the conclusion of the t test, we should use diagnostics to check the validity of assumptions made by the model. Although this step is not shown for every example in the text, it is an essential part of every analysis. For example, a quantile plot may be used to check the assumptions of normality and identical population variances. Though not shown here, a plot of the residuals against run order could help identify potential violations of the assumed independence across runs (the most important of the three assumptions).
3. Click the red triangle next to One-way Analysis of Strength by Mortar and select Normal Quantile Plot > Plot Quantile by Actual.
The points fall reasonably close to straight lines in the plot, suggesting that the assumption of normality is reasonable. The slopes of the lines are proportional to the standard deviations in each comparison group. These slopes appear to be similar, supporting the decision to assume equal population variances.
4. Select Window > Close All.
Section 2.4.3 Choice of Sample Size
1. To determine the necessary sample size for a proposed experiment, select DOE > Sample Size and Power.
2. Click Two Sample Means.
3. Enter 0.25 for Std Dev, 0.5 for Difference to detect, and 0.95 in Power. Notice that the Difference to detect requested here is the actual difference between group means, not the scaled difference, δ, described in the textbook.
4. Click Continue. A value of 16 then appears in Sample Size. Thus, we should allocate 8 observations to each treatment (n1 = n2 = 8).
5. Suppose we use a sample size of n1 = n2 = 10. What is the power for detecting difference of 0.25 kgf/cm2? Delete the value 0.95 from the Power field, change Difference to detect to 0.25, and set Sample Size to 20.
6. Click Continue.
The 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