3.2.1.1 Step 1 – Perform a Descriptive Analysis of the Variable “Ingredient”

For the quantitative variable “Ingredient,” use the dotplot (recommended for small samples with less than 50 observations) or the histogram (for moderate or large datasets with sample size greater than 20) to show its distribution. Add the boxplot and calculate means and measures of variability to complete its descriptive analysis.

To display the histogram, go to: Graph > Histogram

To change the width and number of bar intervals, double‐click on any bar and select the Binning tab. A dialog box will open with several options to define. Specify 5 in Number of intervals and click OK.

Use the tips in Chapter 2, Section 2.2.2.1 to change the appearance of the histogram.

To display the boxplot, go to: Graph > Boxplot

Check the graphical option Simple and in the next screen, select “Ingredient” in Graph variables. Then click OK in the main dialog box.

To change the appearance of a boxplot, use the tips previously referred to histograms, as well as the following:

• To display the boxplot horizontally: double‐click on any scale value on the horizontal axis. A dialog box will open with several options to define. Select from these options: Transpose value and category scales.
• To add the mean value to the boxplot: right‐click on the area inside the border of the boxplot, then select Add > Data display > Mean symbol.

To display the descriptive measures (means, measures of variability, etc.), go to:

Stat > Basic Statistics > Display Descriptive Statistics

Select “Ingredient” in Variables; then click on Statistics to open a dialog box displaying a range of possible statistics to choose from. In addition to the default options, select Interquartile range and Range.

3.2.1.1.1 Interpret the Results of Step 1 Let us describe the shape, central and non‐central tendency and variability of the content of the anesthetic ingredient.

Shape: the distribution is fairly symmetric: middle values are more frequent than low and high values (Stat Tools 1.4–1.5, 1.11).

Central tendency: the mean is equal to 2.4058 mg. Mean and median are close in values (Stat Tools 1.6, 1.11).

Non‐central tendency: 25% of the values are less than 2.3706 mg (first quartile Q₁) and 25% are greater than 2.4402 mg (third quartile Q₃), (Stat Tools 1.7, 1.11).

Variability: observing the length of the boxplot (range) and the width of the box (interquartile range, IQR), the distribution shows moderate variability. The lozenges' content varies from 2.2953 mg (minimum) to 2.4976 mg (maximum). 50% of middle values vary from 2.3706 to 2.4402 mg (IQR = maximum observed difference between middle values equal to 0.0696 mg) (Stat Tools 1.8, 1.11).

The average distance of the values from the mean (standard deviation) is 0.0486 mg: the lozenges' content varies on average from (2.4058–0.0486) = 2.3572 mg to (2.4058 + 0.0486) = 2.4544 mg (Stat Tools 1.9).

From a descriptive point of view, the mean content of the anesthetic agent seems not to be different from the required value of 2.4.

Is the difference between the sample mean of 2.4058 and the target value 2.4 statistically significant or NOT? Do the sample results lead to rejection of the null hypothesis of equality of the population mean content of the anesthetic ingredient to the required value? Apply the one‐sample t‐test to answer these questions.

3.2.1.2 Step 2 – Assess the Null and the Alternative Hypotheses and Apply the One‐Sample t‐Test

The appropriate procedure to determine whether the population mean content does not differ from the required value of 2.4 mg is the one‐sample t‐test (Stat Tool 3.1). What are the null and alternative hypotheses?

Look at the alternative hypothesis: the mean content can be greater or less than 2.4. This is a bidirectional alternative hypothesis. In this case, the test is called a two‐tailed (or two‐sided) test. Next, we'll look at the test results and decide whether to reject or fail to reject the null hypothesis.

To apply the one‐sample t‐test, go to:

In the dialog box, leave the option One or more samples, each in a column, and select the variable “Ingredient.” Check the option Perform hypothesis test and specify the target value “2.4” in Hypothesized mean. Click Options if you want to modify the significance level of the test or specify a directional alternative. Click on Graphs to display the boxplot with the null hypothesis and the confidence interval for the mean.

3.2.1.2.1 Interpret the Results of Step 2 The sample mean 2.4058 mg is a point estimate of the population mean content of the anesthetic agent. The confidence interval provides a range of likely values for the population mean. We can be 95% confident that on average the true mean content is between 2.3876 and 2.4239. Notice that the CI includes the value 2.4. Because the p‐value (0.522) is greater than the significance level α = 0.05, we fail to reject the null hypothesis. There is not enough evidence that the mean content is different from the target value 2.4 mg. The difference between the sample mean 2.4058 mg and the required value 2.4 mg is not statistically significant.

3.2.2.1.1 Interpret the Results of Step 1 The table stored in the Session window and the pie chart show the frequency distribution of the variable “Recoded weight.” In 7 out of 30 lozenges (sample percentage = 23.3%; sample proportion = 0.233), the total weight is greater than 2.85 mg.

From a descriptive point of view, the new formulated lozenges seem not to have exceeded the critical value of 40%, as the sample percentage is equal to 23.3%.

Is the difference between the sample proportion of 0.233 and the critical proportion of 0.40 statistically significant or NOT? Do the sample results lead to rejection of the null hypothesis of equality of the population proportion to the critical value? Apply the one proportion test to answer these questions.

3.2.2.2.1 Interpret the Results of Step 2 The sample proportion 0.233 is a point estimate of the population proportion of lozenges with total weight greater than 2.85 mg. In our case, we required a one‐sided test; therefore, only the lower bound of the confidence interval is shown. We can be 95% confident that the true proportion is less than 0.36. Notice that the lower bound is less than the critical value 0.40. Because the p‐value (0.031) is less than the significance level α = 0.05, we reject the null hypothesis. There is enough evidence that the proportion is less than 0.40. The difference between the sample proportion 0.233 and the critical value 0.40 is statistically significant.

Test and CI for One Proportion

Method

p: event proportion

Normal approximation method is used for this analysis.

Descriptive Statistics

Test

3.3.1.1 Step 1 – Perform a Descriptive Analysis of the Variable “Open_end” Stratifying by Formulations

For the quantitative variable “Open_end,” use the dotplot (recommended for small samples with less than 50 observations) or the histogram (for moderate or large datasets with sample size greater than 20), stratifying by formulations. Finally, add the boxplot (for moderate or large datasets) and calculate means and measures of variability to complete the descriptive analysis.

To display the dot plots by formulations, go to: Graph > Dotplot

To display the boxplots by formulations, go to: Graph > Boxplot

Under One Y, select the graphical option With Groups and in the next dialog box, select “Open_end” in Graph variables and “Formulation” in Categorical variables for grouping. Then click OK in the dialog box. To change the appearance of a boxplot, use the tips in section 3.2.1.1.

To display the descriptive measures (means, measures of variability, etc.) by formulations, go to:

Select “Open_end” in Variables, and “Formulation” in By variables, then click on Statistics to open a dialog box displaying a range of possible statistics to choose from. In addition to the default options, select Variance, Coefficient of variation, Interquartile range and Range. Click OK in each dialog box.

3.3.1.1.1 Interpret the Results of Step 1 Let's describe the shape, central, and non‐central tendency and variability of the thickness at 30 + 5 mm from open end for the two formulations.

Shape: The distribution is asymmetric to the left for formulation A while formulation B tends to be more symmetric (Stat Tools 1.4, 1.5, 1.11).

Central tendency: The mean is equal to 0.046 mm for group A and 0.048 for group B. Mean and median are close in values for formulation B (Stat Tools 1.6, 1.11).

Non‐central tendency: 25% of condoms of type A have thickness less than 0.044 mm (first quartile Q₁) and 25% greater than 0.048 mm (third quartile Q₃), (Stat Tools 1.7, 1.11). For formulation B, the first and third quartiles are respectively equal to 0.046 mm and 0.050 mm.

Variability: Observing the length of the boxplot (range) and the width of the box (Interquartile range, IQR), formulation B shows slightly higher variability than formulation A. Condom thickness varies from 0.041 mm (minimum) to 0.049 mm (maximum) for group A and from 0.041 mm (minimum) to 0.055 mm (maximum) for group B (Stat Tools 1.8, 1.11).

For formulation A, the average distance of the values from the mean (standard deviation) is 0.002 mm: condom thickness varies on average from (0.046–0.002) = 0.044 mm to (0.046 + 0.002) = 0.048 mm (Stat Tools 1.9). For formulation B, the standard deviation is 0.003 mm: condom thickness varies on average from 0.045 to 0.051 mm (mean ± standard deviation). Remember that the standard deviation is the square root of the variance.

Statistics

From a descriptive point of view, formulation B seems to have slightly higher variability than A. Look at the coefficient of variation (Stat Tool 1.10), which is greater for condoms of type B. The two standard deviations (or variances) seem to be different. Is this difference statistically significant or NOT? Do the sample results lead to rejection of the null hypothesis of equality of the population variances for the two formulations? Apply the two variances test to answer these questions.

3.3.1.2 Step 2 – Assess the Null and the Alternative Hypotheses and Apply the Two Variances Test

The appropriate procedure to determine whether the variability of thickness measured at 30 + 5 mm from open end differs between the two formulations is the two variances test (Stat Tool 3.4). What are the null and alternative hypotheses?

To apply the two variances test, go to:

In the drop‐down menu at the top, choose Both samples are in one column. In Samples, select the variable “Open_end” and in Sample IDs, select the variable “Formulation,” then click Options. Note that the 2 variances procedure performs hypothesis tests and computes confidence intervals for the ratios between two populations' variances or standard deviations. This is why the alternative hypothesis is related to the ratio of the two populations' variances or standard deviations, where a ratio of 1 suggests equality between populations. Under Ratio, leave the selection (sample 1 standard deviation) / (sample 2 standard deviation). Under Alternative hypothesis, choose Ratio ≠ hypothesized ratio, and check the option Use test and confidence intervals based on normal distribution. If your samples have less than 20 observations each, or the distribution for one or both populations is known to be far from the normal distribution, uncheck this option.

Click OK in each dialog box.

3.3.1.2.1 Interpret the Results of Step 2 The sample ratio of standard deviations 0.676 is a point estimate of the population ratio. The confidence interval provides a range of likely values for the population ratio. We can be 95% confident that on average the true ratio is between 0.446 and 1.033. Notice that the CI includes the value 1. Because the p‐value (0.069) of the test is greater than the significance level α = 0.05, we fail to reject the null hypothesis. There is insufficient evidence that the variability is different between the two formulations. The difference between the two standard deviations (or variances) is not statistically significant.

Test and CI for Two Variances: Open_end versus Formulation

Method

σ₁: standard deviation of Open_end when Formulation = A

σ₂: standard deviation of Open_end when Formulation = B

Ratio: σ₁/σ₂

F method was used. This method is accurate for normal data only.

Descriptive Statistics

Ratio of Standard Deviations

Test

3.3.2.1 Step 1 – Perform a Descriptive Analysis of the Variable “Open_end” stratifying by Formulations

In step 1 of Section 3.3.1 we have already performed the descriptive analysis of variable “Open_end,” stratifying by formulations and displaying the dotplots, the boxplots, and calculating the descriptive measures.

3.3.2.1.1 Interpret the Results of Step 1 We take back the considerations made previously in Section 3.3.1.1.1 about the distribution of the thickness at 30 + 5 mm from open end. In particular, we noticed that the mean was equal to 0.046 mm for group A and 0.048 for group B.

From a descriptive point of view, formulation B seems to have a slightly higher mean than A. The two means seem to differ. Is this difference statistically significant or NOT? Do the sample results lead to rejection of the null hypothesis of equality of the population means for the two formulations? Apply the two‐sample t‐test to answer these questions.

3.3.2.2 Step 2 – Assess the Null and the Alternative Hypotheses and Apply the Two‐Sample t‐Test

The appropriate procedure to determine whether the means of thickness measured at 30 + 5 mm from open end differ between the two formulations is the two‐sample test, as the two random samples are independent (Stat Tools 3.3, 3.5). What are the null and alternative hypotheses?

Next, we'll look at the test results and decide whether to reject or fail to reject the null hypothesis.

To apply the two‐sample t‐test, go to:

In the drop‐down menu at the top, choose Both samples are in one column. In Samples, select the variable “Open_end” and in Sample IDs, select the variable “Formulation,” then click Options. Under Alternative hypothesis, choose Difference ≠ hypothesized difference, to apply a two‐sided test. Check the option Assume equal variances because in Section 3.3.1 after applying the two variances test, we concluded in favor of the equality of variances between the two formulations. If in the application of the two variances test, you did not reject the null hypothesis, uncheck this option. Click OK in each dialog box.

3.3.2.2.1 Interpret the Results of Step 2

The difference between the two sample means −0.002069, is a point estimate of the population mean difference. The confidence interval provides a range of likely values for the population mean difference. We can be 95% confident that on average the true difference is between −0.003769 and − 0.002924. Notice that the CI does not include the value 0. Because the test's p‐value (0.018) is less than the significance level α = 0.05, we reject the null hypothesis. There is enough evidence that the mean of the thickness at 30 + 5 mm from open end is different between the two formulations. The difference between the two means is statistically significant.

Two‐Sample T‐Test and CI: Open_end; Formulation

Method

μ₁: mean of Open_end when Formulation = A

μ₂: mean of Open_end when Formulation = B

Difference: μ₁ − μ₂

Equal variances are assumed for this analysis.

Descriptive Statistics: Open_end

Estimation for Difference

Test

3.3.3.1 Step 1 – Calculate the Sample Proportions of Condoms with Thickness Less than or Equal to 0.045 mm

Let us create a new categorical variable from variable “Open_end” that assumes the two categories: “thickness less than or equal to 0.045,” “thickness greater than 0.045.”

To create a new categorical variable, go to: Data > Recode > To Text

Note that the new variable “Recoded Open_end” will appear at the end of the current worksheet. This variable is a categorical variable assuming two categories: “Less than or equal to 0.045,” “Greater than 0.045.”

To calculate the sample proportion of condoms with thickness less than or equal to 0.045 mm by formulations, go to:

In the drop‐down menu at the top, choose Raw data (Categorical variables). In Rows, select the variable “Recoded Open_end” and in Columns, choose “Formulation.” Then, under Display, check the options Count and Column Percents. Click OK in the dialog box.

To graphically display the sample proportions of condoms with thickness less than or equal to 0.045 mm, go to:

In Categorical variables, select the variable “Recoded Open_end.” Click Labels and in the dialog box select Slice Labels. Check the option Percent and click OK.

In the main dialog box select Multiple Graphs. In the next window, select By variables. In the new dialog box under By variables with groups on same graph, select “Formulation.” Click OK in each dialog box.

3.3.3.1.1 Interpret the Results of Step 1 The table stored in the Session window and the pie charts shows the frequency distribution of the variable “Recoded Open_end” by formulation. For group A, in 8 out of 23 condoms (sample percentage = 34.8%; sample proportion = 0.348), the thickness is less than or equal to 0.045 mm; for formulation B, this sample percentage is 20.0%.

From a descriptive point of view, the two percentages seem to be a little different. Is the difference of 14.8% statistically significant or NOT? Do the sample results lead to rejection of the null hypothesis of equality of the population proportions? Apply the two proportions test to answer these questions.

Tabulated Statistics: Recoded Open_end; Formulation

Rows: Recoded Open_endColumns: Formulation

3.3.3.2 Step 2 – Assess the Null and the Alternative Hypotheses and Apply the Two Proportions Test

The appropriate procedure to determine whether the proportions of condoms with thickness less than or equal to 0.045 mm is different between the two formulations, is the two proportions test (Stat Tool 3.6). What are the null and alternative hypotheses?

Look at the alternative hypothesis: The group A proportion can be greater or less than the group B proportion. This is a bidirectional alternative hypothesis. In this case, the test is called a two‐tailed (or two‐sided) test.

Next we'll look at the test results and decide whether to reject or fail to reject the null hypothesis.

To apply the two proportions test, go to:

In the drop‐down menu at the top, choose Summarized data. Under Sample 1, in Number of events, enter 8 (the number of condoms not exceeding 0.045 mm as thickness for formulation A) and in Number of trials, enter 23 (the sample size). Under Sample 2, in Number of events, enter 5 (the number of condoms not exceeding 0.045 mm as thickness for formulation B) and in Number of trials, enter 25 (the sample size). Click Options. Under Alternative hypothesis, choose Difference ≠ hypothesized difference, to apply a two‐sided test, and under Test Method leave the default option Estimate the proportions separately. Click OK in each dialog box.

3.3.3.2.1 Interpret the Results of Step 2 The difference between the two‐sample proportions 0.148 is a point estimate of the population difference. We can be 95% confident that on average the true difference is between −0.102119 and 0.397771. Notice that the CI includes the value 0.

Minitab uses the normal approximation method and Fisher's exact method to calculate the p‐values for the two proportions test. If the number of events and the number of non‐events is at least 5 in both samples, use the smaller of the two p‐values. If either the number of events or the number of non‐events is less than 5 in either sample, the normal approximation method may be inaccurate. Because the test's p‐value (0.246) is greater than the significance level α = 0.05, we fail to reject the null hypothesis. There is not enough evidence that the proportions of condoms with thickness less than or equal to 0.045 mm is different between the two formulations. The difference between the two proportions is not statistically significant.

Test and CI for Two Proportions Method

p₁: proportion where Sample 1 = Event

p₂: proportion where Sample 2 = Event

Difference: p₁ – p₂

Descriptive Statistics

Estimation for Difference

CI closed on normal approximation

Test

3.4.1.1 Step 1 – Descriptive Analysis of “Appropriateness” Stratified by “Fragrance”

Let's start the descriptive analysis of the variable “Appropriateness” stratified by the categorical variable “Fragrance,” by displaying the frequency distribution of the “Appropriateness” by “Fragrance” through graphs that help to emphasize trend, pattern, and other important aspects of data distributions. “Appropriateness” being a quantitative discrete variable, use the dotplot or the histogram to display the distribution of the data. Add the boxplot and calculate means and measures of variability to complete the analysis.

To display the histogram, go to: Graph > Histogram

Use the tips in Chapter 2, section 2.2.2.1 to change the appearance of the histogram:

To display the Dotplot, go to: Graph > Dotplot

Choose the graphical option With Groups to display one dotplot of Appropriateness for each fragrance. In the next dialog box, select “Appropriateness” in Graph variables and “Fragrance” in Categorical variables for grouping.

Use the tips previously referred to histograms to change the appearance of a dotplot.

To display the boxplot, go to: Graph > Boxplot

Choose the graphical option With Groups to display one boxplot of Appropriateness for each fragrance. In the next dialog box, select “Appropriateness” in Graph variables and “Fragrance” in Categorical variables for grouping.

To change the appearance of a boxplot, use the tips previously referred to histograms, as well as the following ones:

• To display the boxplot horizontally: double‐click on any scale value on the horizontal axis. A dialog box will open with several options to define. Select from these options: Transpose value and category scales.
• To add the mean value to the boxplot: right‐click in the area inside the border of the boxplot, then select Add > Data display > Mean symbol.

To display the descriptive measures (means, measures of variability, etc.), go to:

In the dialog box, under Variables select “Appropriateness” and under By variables select “Fragrance,” then click on Statistics to open a dialog box displaying a range of possible statistics to choose from. In addition to the default options, select Coefficient of variation, Interquartile range and Range.

3.4.1.1.1 Interpret the Results of Step 1 From a descriptive point of view, let us compare the shape, central and non‐central tendency and variability of the distribution of appropriateness between the two fragrances.

Shape. The distributions of appropriateness are skewed to the left for both fragrances: middle and high scores are more frequent than low scores. For fragrance A, the distribution is unimodal with mode equal to 8. For fragrance B, the distribution is reasonably spread out with no concentration on a unique value; scores 6, 9, and 10 show the higher frequencies (Stat Tools 1.4–1.5, 1.11).

Central tendency. For both fragrances, the central tendency of appropriateness (the median score) is equal to 7: about 50% of the scores is less than or equal to 7 and 50% greater than or equal to 7. Mean and median overlap for fragrance A, while for fragrance B, with a distribution more skewed to the left, mean and median are not close in value. The median score is a better indicator than the mean (Stat Tools 1.6, 1.11).

Non‐central tendency. For fragrance A, 25% of the scores are less than 6 (first quartile Q₁) and 25% are greater than 8 (third quartile Q₃). For fragrance B, 25% of the scores are less than 4.75 (first quartile Q₁) and 25% are greater than 9 (third quartile Q₃) (Stat Tools 1.7, 1.11).

Variability. Observing the length of the boxplots (ranges) and the width of the boxes (Interquartile ranges, IQRs), fragrance B shows more variability. Its scores vary from 1 (minimum) to 10 (maximum). For fragrance A, no respondent gave a score less than 3 and the maximum observed difference between evaluations was 7 (Range). 50% of middle evaluations vary from 6 to 8 for fragrance A (IQR = maximum observed difference between evaluations equal to 2) and from 4.75 to 9 for fragrance B (IQR = 4.25) (Stat Tools 1.8, 1.11).

The average distance of the scores from the mean (standard deviation) is 1.93 for fragrance A (scores vary on average from 7 – 1.93 = 5.07 to 7 + 1.93 = 8.93) and 2.56 for fragrance B (scores vary on average from 6.77 – 2.56 = 4.21 to 6.77 + 2.56 = 9.33) (Stat Tools 1.9).

The coefficient of variation is less for fragrance A (CV = 27.57%) confirming the lower variability of the scores for this fragrance (Stat Tool 1.10).

Statistics

3.4.1.2 Step 2 – Descriptive Analysis of “Difference:A_B”

Let's calculate the differences between the score for fragrance A and the score for fragrance B for each panelist, creating a new variable “Difference:A_B,” and perform a synthetic descriptive analysis to get a sense of the mean and variability in the differences between scores. “Difference:A_B” being a quantitative continuous variable, use the boxplot and calculate means and measures of variability to complete the analysis.

To split the column “Appropriateness” into separate columns according to the values of “Fragrance,”

To calculate the differences between the new variables, go to: Calc > Calculator

In the dialog box, write the name of variable “Difference:A_B” in Store result in variable, and type the expression under Expression, selecting variables “Appropriateness_A” and “Appropriateness_B” from the list on the left and using the numeric keypad. The worksheet will include a new variable “Difference:A_B.”

To display the boxplot, go to: Graph > Boxplot

Choose the graphical option Simple under One Y. In the next dialog box, select “Difference:A_B” in Graph variables.

To display the descriptive measures (means, measures of variability, etc.), go to:

In the dialog box, under Variables select “Difference:A_B” and leave By variables blank, then click on Statistics to open a dialog box displaying a range of possible statistics to choose from. In addition to the default options, select Interquartile range and Range and uncheck Coefficient of variation.

3.4.1.2.1 Interpret the Results of Step 2 We may see that on average the difference in the score of appropriateness of fragrances is 0.23 (median = 0) (Stat Tools 1.6, 1.11). 50% of middle differences vary from −1.25 to 2.25 (IQR = 3.5) (Stat Tools 1.8, 1.11). The average distance of the differences from their mean (standard deviation) is 0.44 (differences vary on average from 0.23–0.44 = −0.21 to 0.23 + 0.44 = 0.67) (Stat Tool 1.9).

Statistics

From a descriptive point of view, the two fragrances show similar appropriateness, but we need to analyze the paired t‐test results to determine whether the mean difference between the two fragrances is statistically different from 0.

3.4.1.3 Step 3 – Paired t‐Test on Mean Difference

Let's apply the paired t‐test (Stat Tool 3.7). The paired t‐test results include both a confidence interval and a p‐value. Use the confidence interval to obtain a range of reasonable values for the population mean difference. Use the p‐value to determine whether the mean difference is statistically equal or not equal to 0. If the p‐value is less than α, reject the null hypothesis that the mean difference equals 0.

To apply the paired t‐test, go to: Stat > Basics Statistics > Paired t

3.4.1.3.1 Interpret the Results of Step 3

The sample mean difference 0.233 is an estimate of the population mean difference. The confidence interval provides a range of likely values for the mean difference. We can be 95% confident that on average the true mean difference is between −0.674 and 1.141. Notice that the CI includes the value 0.

Because the p‐value (0.603) is greater than 0.05, we fail to reject the null hypothesis. There is not enough evidence that the mean difference is different from 0. The mean difference of 0.233 between the appropriateness of fragrances A and B is not statistically significant. The two fragrances show similar appropriateness score means.

Paired T‐Test and CI: Appropriateness_A; Appropriateness_B

Descriptive Statistics

Estimation for Paired Difference

μ_difference: mean of (Appropriateness_A – Appropriateness_B)

Test

3.5.1.1 Step 1 – Create a Full Factorial Design

Let's begin to create a full factorial design, considering three levels for each factor.

To create a full factorial design, go to: Stat > DOE > Factorial > Create Factorial Design

Multilevel Factorial Design

Design Summary

Number of levels: 3; 3; 3; 3

3.5.1.2 Step 2 – Reduce the Full Factorial Design to an Optimal Design

Investigators need to arrange the experiment in such a way that only 40 different combinations of factor levels will be tested. We proceed by reducing the previous full factorial design to obtain an optimal design with 40 runs.

To create an optimal factorial design, go to:

Optimal Design: A; B; C; D

Factorial design selected according to D‐optimality

Number of candidate design points: 81

Number of design points in optimal design: 40

Model terms: A; B; C; D; AB; AC; AD; BC; BD; CD

Initial design generated by sequential method

Initial design improved by exchange method

Number of design points exchanged is 1

Optimal Design

Row number of selected design points: 27; 63; 75; 52; 70; 76; 45; 51; 69; 80; 3; 8; 20; 56; 40; 15; 16; 22; 33; 34; 39; 44; 46; 50; 58; 64; 68; 6; 23; 1; 11; 29; 54; 72; 79; 7; 17; 32; 21; 78

To request two replicates of each run, go to:

Under Modification, choose Replicate design, then click on Specify. In the next dialog box, under Number of replicates to add, specify 1. Click OK in each dialog box and Minitab shows the replicated optimal design in the previous worksheet. The second replicates for each run will have the value 2 under the column Block.

Before proceeding with the randomization, renumber the design by going to:

Under Modification, choose Renumber design, then click on Put the modified design in a new worksheet. Click OK in each dialog box and Minitab shows the replicated optimal design in a new worksheet.

Let's complete the creation of the optimal design with the randomization of the runs. Go to:

Under Modification, choose Randomize runs, then click on Specify. In the next dialog box, select Randomize entire design. Click OK in each dialog box and Minitab shows the randomized optimal design in the previous worksheet (only the first five runs are shown here as an example).

3.5.1.3 Step 3 – Assign the Designed Factor Level Combinations to the Experimental Units and Collect Data for the Response Variable

Collect the data for the response variables (reflectance measures) in the order set out in the column RunOrder in the optimal design. For example, the first formulation to test will be that with component A at level 40, component B at level 0.30, C at 0 and D at 10. Each formulation will be tested on two fabric specimens given we are considering two replicates. Note that the column “Blocks” has two values: 1 for the first replicate and 2 for the second.

Using a spectrophotometer, reflectance data (cleaning performance scores) will be measured in relation to several stains. To take into account the inherent variability in the measurement system, the cleaning performance for each stain will be measured at two different points on each fabric specimen. Remember that the individual measurements for each point on the piece of fabric are not replicates but repeated measurements because the two positions will be processed together in the experiment, receiving the same formulation simultaneously. Their average will be the response variable to analyze (Stat Tool 2.5).

Once all the designed factor level combinations have been tested, enter the recorded response values (cleaning performance scores) onto the worksheet containing the design. Now you are ready to proceed with the statistical analysis of the collected data.

3.5.2.1 Step 1 – Perform a Descriptive Analysis of the Response Variables

Firstly calculate the average of the two repeated measurements for each stain, creating two new variables that will become the response variables to analyze.

To calculate the average of the two repeated measurements for each stain, go to: Calc > Calculator

The responses are quantitative variables, so let's use a boxplot to represent how the cleaning performance scores occurred in our sample, and calculate means and measures of variability to complete the descriptive analysis of each response.

To display the boxplot, go to: Graph > Boxplot

To change the appearance of a boxplot, use the tips in Chapter 2, Section 2.2.2.1 in relation to histograms, as well as the following ones:

• To display the boxplot horizontally: double‐click on any scale value on the horizontal axis. A dialog box will open with several options to define. Select from these options: Transpose value and category scales.
• To add the mean value to the boxplot: right‐click on the area inside the border of the boxplot, then select Add > Data display > Mean symbol.

To display the descriptive measures (means, measures of variability, etc.), go to:

In the dialog box, under Variables select “Mustard” and “Curry,” then click on Statistics to open a dialog box displaying a range of possible statistics to choose from. In addition to the default options, select Interquartile range, Range, and Coefficient of variation.

3.5.2.1.1 Interpret the Results of Step 1 Reflectance data is fairly symmetric for the curry stain and slightly skewed to the left with the presence of an outlier for the mustard stain (Stat Tools 1.4–1.5).

Regarding the central tendency, the mean and median overlap for cleaning performances on curry and mustard stains, with a higher mean value for curry (Stat Tools 1.6, 1.11).

With respect to variability, observing the length of the boxplots (ranges) and the width of the boxes (interquartile ranges, IQRs), the performance shows moderate spread for both stains. The average distance of the scores from the mean (standard deviation) is about 1.7/1.8, respectively, for Mustard and Curry (Stat Tools 1.9). Additionally, the coefficient of variation is about 2.1%/2.2% respectively for Mustard and Curry, confirming the similar spread of the cleaning scores for the two stains (Stat Tool 1.10).

Statistics

3.5.2.2 Step 2 – Fit a Response Surface Model

When you have at least one quantitative factor in your general factorial design, you can analyze it by fitting a response surface model. We have added two new variables, “Mustard” and “Curry,” to the worksheet including the factorial design. Therefore, before analyzing it, we must specify our factors.

To define the response surface design, go to:

Under Continuous Factors, select the components A, B, C, D. If some of your factors are categorical, these must be specified under Categorical Factors. Click on Low/High. For each factor in the list, verify that the high and low settings are correct and if these levels are in the original units of the data, check the option Uncoded under Worksheet Data Are. Click OK in each dialog box. Now you are ready to analyze the response surface design.

Note that in our example we don't have blocking factors (Stat Tool 2.4). Should there be any, in the main dialog box click on Designs. At the bottom of the screen, under Blocks, check the option Specify by column, and in the available space select the column representing the blocking variable.

Click OK in each dialog box. You would now be ready to analyze the response surface design.

To analyze the response surface design, go to:

Stat > DOE > Response Surface > Analyze Response Surface Design

Select the first response variable “Mustard” under Responses and click on Terms. At the top of the next dialog box, in Include the following terms, specify Full quadratic to study all main effects (both linear and quadratic effects) and the two‐way interactions between factors. As stated previously, we don't have blocking factors in our example (Stat Tool 2.4). If present, the grayed‐out Include blocks in the model option would be selectable to consider blocks in the analysis. Click OK and now choose Graphs in the main dialog box. Under Effects plots choose Pareto and Normal. These graphs help identify the terms (linear and quadratic factor effects and/or interactions) that influence the response and compare the relative magnitude of the effects, along with their statistical significance. Click OK in the main dialog box and Minitab shows the results of the analysis, both in the Session window and through the relevant graphs.

3.5.2.2.1 Interpret the Results of Step 2

Let us first consider the Pareto chart that shows which terms contribute the most to explaining the clinical performance on the mustard stain. Any bar extending beyond the reference line (considering a significance level equal to 5%) is related to a significant effect.

In our example, the main effects of three components B, C and D and their interactions are statistically significant at the 0.05 level. Also, the quadratic effect of factor B is marked as statistically significant.

From the Pareto chart, you can detect which effects are statistically significant, but you have no information on how these effects affect the response.

Use the normal plot of the standardized effects to evaluate the direction of the effects.

In the normal plot, the line represents the situation in which all the effects (main effects and interactions) are 0. Effects that are further from 0 and from the line are statistically significant. Minitab shows statistically significant and nonsignificant effects by giving different colors and shapes to the points. In addition, the plot indicates the direction of the effect. Positive effects (displayed on the right of the graph) increase the response when the factor moves from its low value to its high value.

Negative effects (displayed on the left of the line) decrease the response when moving from the factor's low value to its high value.

In our example, the main effects for B, C, D are statistically significant at the 0.05 level.

The cleaning scores seem to increase with higher levels of each component. Furthermore, the interactions between factor pairs are marked as statistically significant, but the interaction between components C and D lies on the left of the graph along with the quadratic effect of component B. We will clarify the meaning of this later.

In the ANOVA table we examine the p‐values to determine whether any factors or interactions, or the blocks if present, are statistically significant. Remember that the p‐value is a probability that measures the evidence against the null hypothesis. Lower probabilities provide stronger evidence against the null hypothesis.

For the main effects, the null hypothesis is that there is no significant difference in the mean performance score across levels of each factor. As we estimate a full quadratic model, we have two tests for each factor that analyze its linear and quadratic effect on the response.

For the two‐factor interactions, H₀ states that the relationship between a factor and the response does not depend on the other factor in the term.

Usually we consider a significance level alpha equal to 0.05, but in an exploratory phase of the analysis, we may also consider a significance level equal to 0.10.

When the p‐value is greater than or equal to alpha, we fail to reject the null hypothesis. When it is less than alpha, we reject the null hypothesis and claim statistical significance.

So, setting the significance level alpha to 0.05, which terms in the model are significant in our example? The answer is: the linear effects of B, C, D; the two‐way interactions among these three components and the quadratic effect of B. Now, you may want to reduce the model to include only significant terms.

Response Surface Regression: Mustard versus A; B; C; D

Analysis of Variance

When you use statistical significance to decide which terms to keep in a model, it is usually advisable not to remove entire groups of nonsignificant terms at the same time. The statistical significance of individual terms can change because of other terms in the model. To reduce your model, you can use an automatic selection procedure, namely the stepwise strategy, to identify a useful subset of terms, choosing one of the three commonly used alternatives (standard stepwise, forward selection, and backward elimination).

3.5.2.3 Step 3 – If Need Be, Reduce the Model to Include the Significant Terms

To reduce the model, go to:

Select the numeric response variable “Mustard” under Responses and click on Terms. In Include the following terms, specify Full Quadratic, to study all main effects (both linear and quadratic effects) and the two‐way interactions between factors. Note that in our example we don't have blocking factors (Stat Tool 2.4). If present, the grayed‐out Include blocks in the model option would be selectable to consider blocks in the analysis. Click OK and in the main dialog box choose Stepwise. In Method select Backward elimination and in Alpha to remove specify 0.05, then click OK. In the main dialog box choose Graphs. Under Residual plots choose Four in one. Minitab will display several residual plots to examine whether your model meets the assumptions of the analysis (Stat Tools 2.8, 2.9). Click OK in the main dialog box.

Complete the analysis by adding the factorial plots that show the relationships between the response and the significant terms in the model, displaying how the response mean changes as the factor levels or combinations of factor levels change.

To display factorial plots, go to:

3.5.2.3.1 Interpret the Results of Step 3 Setting the significance level alpha to 0.05, the ANOVA table shows the significant terms in the model. Take into account that the stepwise procedure may add nonsignificant terms in order to create a hierarchical model. In a hierarchical model, all lower‐order terms that comprise a higher‐order term also appear in the model.

Response Surface Regression: Mustard versus A; B; C; D

Backward Elimination of Terms

α to remove = 0.05

Analysis of Variance

Note that in the ANOVA table, Minitab displays the lack‐of‐fit test when data contain replicates. This test helps investigators to determine whether the model accurately fits the data. Its null hypothesis states that the model accurately fits the data. As usual, compare the p‐value to your significant level α. If the p‐value is less than α, reject the null hypothesis and conclude that the lack of fit is statistically significant. If the p‐value is larger than α, you fail to reject the null hypothesis and you can conclude that the lack of fit is not statistically significant.

Model Summary

In our example, the p‐value (0.183) is greater than 0.05: we conclude that there is no strong evidence of lack of fit. Lack‐of‐fit can occur if important terms such as interactions or quadratic terms are not included in the model or if the fitted model shows several, unusually large residuals (Stat Tool 2.9).

In addition to the results of the analysis of variance, Minitab displays some other useful information in the Model Summary table.

The quantity R‐squared (R‐sq, R²) is interpreted as the percentage of the variability among reflectance scores, explained by the terms included in the ANOVA model.

The value of R² varies from 0 to 100%, with larger values being more desirable.

The adjusted R² (R‐sq(adj)) is a variation of the ordinary R² that is adjusted for the number of terms in the model. Use adjusted R² for more complex experiments with several factors, when you want to compare several models with different numbers of terms.

The value of S is a measure of the variability of the errors that we make when we use the ANOVA model to estimate the reflectance scores. Generally, the smaller it is, the better the fit of the model to the data.

Before proceeding with the results reported in the Session window, take a look at the residual plots. A residual represents an error that is the distance between an observed value of the response and its estimated value by the ANOVA model. The graphical analysis of residuals based on residual plots helps you to discover possible violations of the ANOVA underlying assumptions (Stat Tools 2.8, 2.9).

A check of the normality assumption could be made by looking at the histogram of the residuals (bottom left), but with small samples, often the histogram has an irregular shape.

The normal probability plot of the residuals may be more useful. Here we can see a tendency of the plot to bend slightly upward on the right, but the plot is not in any case grossly non‐normal. In general, moderate departures from normality are of little concern in the ANOVA model with fixed effects.

The other two graphs (Residuals versus Fits and Residuals versus Order) seem unstructured, thus supporting the validity of the ANOVA assumptions.

Coded Coefficients

Regression Equation in Uncoded Units

Returning to the Session window, we find a table of coded coefficients followed by a regression equation in uncoded units.

In ANOVA, we have seen that many of the questions the experimenter wishes to answer can be solved by several hypothesis tests.

It is also helpful to present the results in terms of a regression model, i.e. an equation derived from the data that expresses the relationship between the response and the important design factors.

In the regression equation, we have a coefficient for each term in the model. They are expressed in uncoded units, i.e. in the original measurement units. These coefficients are also displayed in the Coded Coefficients table, but in coded units. Earlier, we learned how to examine the p‐values in the ANOVA table to determine whether a factor is statistically related to the response. In the Coded Coefficients table, we find the same results in terms of tests on coefficients and we can evaluate the p‐values to determine whether they are statistically significant. The null hypothesis states the absence of any effect on the response. Setting the significance level α at 0.05, if a p‐value is less than α, reject the null hypothesis and conclude that the effect is statistically significant.

Look at the factorial plots to better understand how the mean reflectance score varies as factor levels change.

The presence of the quadratic term B*B makes a curvature in the factorial plots involving component B. Considering the interaction B*C, the best result (the higher mean cleaning score) is obtained when B and C are increasing in value. For the interaction C*D, when C is equal to its highest level (3), the mean cleaning score does not vary much across levels of component D. When C is absent, the mean cleaning score tends to increase with increasing levels of D. Note that the interaction C*D lies on the left side of the normal plot of the standardized effects.

Now repeat steps 2 and 3 for the second response variable “Curry.” For brevity, we show only the reduced model with residual and factorial plots.

Response Surface Regression: Curry versus A; B; C; D

Backward Elimination of Terms

α to remove = 0.05

Analysis of Variance

Model Summary

3.5.2.4 Step 4 – Optimize the Responses

After fitting the response surface models, additional analyses can be carried out to find the optimum set of formulation components that optimizes the response variables (Stat Tool 3.8).

To optimize the response variables, go to:

In the table, under Goal, select one of the following options for each response:

• Do not optimize: Do not include the response in the optimization process.
• Minimize: Lower values of the response are preferable.
• Target: The response is optimal when values meet a specific target value.
• Maximize: Higher values of the response are preferable.

In our case study, the goal is to maximize the reflectance data: choose Maximize from the drop‐down list and click Setup. In the next window you need to specify a response target and lower or upper bounds, depending on your goal. Minitab automatically sets the Lower value and the Upper value to the smallest and largest values in the data, but you can change these values depending on your response goal (Stat Tool 3.8).

• If your goal is to maximize the response (larger is better), you need to specify the lower bound and a target value. You may want to set the lower value to the smallest acceptable value and the target value at the point of diminishing returns, i.e. a point at which going above its value does not make much difference. If there is no point of diminishing returns, use a very high target value.
• If your goal is to target a specific value, you should set the lower value and upper value as points of diminishing returns.
• If the goal is to minimize (smaller is best), you need to specify the upper bound and a target value. You may want to set the upper value to the largest acceptable value and the target value at the point of diminishing returns, i.e. a point at which going below its value does not make much difference. If there is no point of diminishing returns, use a very small target value.

In our example, we leave the default values with the lower bound set to the minimum for each response and the target and upper values set to the maximum.

Note that in the process of response optimization, Minitab calculates and uses desirability indicators from 0 to 1 to represent how well a combination of factor levels satisfies the goals you have for the responses. Specifying a different weight varying from 0.1 to 10, you can emphasize or deemphasize the importance of attaining the goals. Furthermore, when you have more than one response, Minitab calculates individual desirabilities for each response and a composite desirability for the entire set of responses. With multiple responses, you may assign a different level of importance to each response by specifying how much effect each response has on the composite desirability.

In our example, leave Weight and Importance equal to 1 and click OK. In the main dialog box, select Results and in Number of solutions to display, specify 5. Click OK in each dialog box. In the optimization plot, Minitab will display the optimal solution, that is the factor setting which maximizes the composite desirability, while in the Session window we will find a list of five different factor combinations close to the optimal one, which is the first on the list.

The optimization plot shows how different factor settings affect the responses. The vertical red lines on the graph represent the current factor settings. The numbers displayed at the top of a column show the current factor level settings (in red). The horizontal blue lines and numbers represent the responses for the current factor level. The optimal solution is plot on the graph and serves as the initial point. The settings can then be modified interactively, to determine how different combinations of factor levels affect responses, by moving the red vertical lines corresponding to each factor. To return to the initial settings, right‐click and select Navigation and then Reset to Initial Settings.

3.5.2.4.1 Interpret the Results of Step 4 The optimization plot is a useful tool to explore the sensitivity of response variables to changes in the factor settings and in the vicinity of a local solution – for example, the optimal one.

Response Optimization: Curry; Mustard

Parameters

Solutions

Multiple Response Prediction

The Session window displays:

• Information about the boundaries, weight, and importance for each response variable (Parameters).
• The factor settings, the fitted responses, and the composite desirability for the required 5 solutions. The first in the list is the optimal solution (Solutions).
• The factor settings for the optimal solution and the fitted responses with their confidence intervals (Stat Tool 1.14) and prediction intervals (Multiple Response Prediction).

For reflectance data, the optimal solution is the formulation with component A set to 40, B to 0.30, C to 3, and D to 10. For this setting, the fitted response is equal to 84.59 (95% CI: 83.90–85.27) for the Curry stain and 83.16 (95% CI: 82.32–83.99) for Mustard. The composite desirability (0.83) is reasonably close to 1, denoting a satisfactory overall achievement of the specified goals.

3.5.2.5 Step 5 – Examine the Shape of the Response Surface and Locate the Optimum

Once investigators have found the optimal solutions, it is usually convenient to study the response surface in the vicinity of these points. Graphic techniques such as contour, surface, and overlaid plots help you to examine the shape of the response surface in regions of interest. Through these plots, you can explore the potential relationship between pairs of factors and the responses. You can display a contour, a surface, or an overlaid plot considering two factors at a time, while setting the remaining factors to predefined levels. With contour and surface plots, you consider a single response; in overlaid plots, you consider all the responses together.

To display the contour plots, go to: Stat > DOE > Response Surface > Contour Plot

To display the surface plots, go to: Stat > DOE > Response Surface > Surface Plot

In the dialog box, select the first response “Mustard” in Response, and check the option Generate plots for all pairs of continuous variables. Then click Settings, and set the factors at levels shown in the first optimal solution found previously in the Response Optimizer (A = 40, B = 0.30, C = 3, D = 10). Click OK in each dialog box. Do the same for the response variable “Curry.”

To change the appearance of the surface plots, double‐click on any point inside a surface. In Attributes, under Surface Type, select the option Wireframe instead of the default Surface.

To display the overlaid plots, go to: Stat > DOE > Response Surface > Overlaid Contour Plot

Select all the responses and click Contours. Note that we are building the overlaid contour plot for the pair A and B. In the next window, for each response, you need to define a low and a high contour, depending on your goal for the responses. For example, you may want to set the Low value and the High value at the points of diminishing returns, or you can simply consider the fitted response values obtained with the optimal solution and specify two points close to the fitted values. In step 4, the fitted response values for the optimal solution were 83.157 for Mustard and 84.586 for Curry. Specify the interval 82–84 for Curry and 81–83 for Mustard as Contours and then click OK. In the main dialog box, select Settings. Set the remaining factors (C and D) at levels shown in the first optimal solution found previously in the Response Optimizer (A = 40, B = 0.30, C = 3, D = 10). Click OK in each dialog box. Repeat the process for the other factor pairs to create all overlaid contour plots.

3.5.2.5.1 Interpret the Results of Step 5 Contour and surface plots are useful tools to explore how a response variable changes while increasing or decreasing in factor levels. In contour plots, two factors are plotted on the x‐ and y‐ axes and the reflectance performance is represented by different level curves, while the remaining factors are set to predefined levels (Hold values). Double‐click on any point in the frame and select Crosshairs to interactively view variations in the response. In surface plots the reflectance performance is represented by a smooth surface.

In overlaid contour plots we simultaneously consider both responses in order to explore factor settings that to a certain extent optimize all responses or keep them in desired ranges of values. Remember that Minitab builds an overlaid contour plot for each factor pair. For brevity, here we show only one of these plots, namely that related to factors C and D, respectively plotted on the x‐ and y‐axes.

Double‐click on any point in the frame and select Crosshairs to interactively view variations in the responses. The white region shows the ranges of values for component C and D where the criteria (low and high boundaries previously defined in the Settings dialog box) are satisfied for both response variables.