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Nested, or Hierarchical, Designs

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Experimental Design

Abstract

In previous chapters, we have seen one- and two-factor designs and how to analyze them. In a factorial design, such as the one discussed in Chap. 6, the levels of the two factors – for convenience, A and B – are crossed, that is, every level of B will occur with every level of A. In practice, this means that, if we have three levels per factor, we will have nine experimental runs (without considering replicates). In this chapter, we will see a different type of design, called nested designs, where the levels of factor B will occur only at certain levels of A. For instance, we can have three levels of A and nine levels of B, but levels 1–3, 4–6, and 7–9 of B will only occur when the levels of A are 1, 2, and 3, respectively.

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Notes

  1. 1.

    If, in a nested design, there is no replication, the lowest-stage factor in the hierarchy essentially disappears and takes the place of error. Hence, for an explicit set of factors to be studied in a nested design, with an ANOVA that includes them all, replication is necessary.

  2. 2.

    To precisely satisfy the “equal-variance” assumption, theoretically required for the F-test to be appropriate, we would need to have the same number of students in each class, assuming that the σ2 of each student’s score (its propensity to vary from its truth) is identical. For each class to have the same number of students is somewhat unrealistic; however, for each to have approximately the same number of students is not. Given the robustness of the equal-variance assumption, minor differences in class size should not materially affect our results.

  3. 3.

    Purity (%) is calculated as the percentage of sucrose over °Brix (total of soluble solids in solution).

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Appendix

Appendix

Example 7.8 A Larger Example: First United Federal Bank of Boston using SPSS

Excel and SPSS will analyze nested designs in a way similar to that we have used for two-way ANOVA . Note that Excel will display the F-value as we would normally calculate for ANOVA – that is, dividing MSeffect by MSerror – and we will have to calculate manually the “true” F-value. For SPSS, we input the data and start our analysis as we did previously (using Analyze > General Linear Models > Univariate), but we specify fixed and random factors and include the main effect of the fixed factor and an “interaction” term in the model, as shown in Fig. 7.5. Note that SPSS interprets this “interaction ” term as the nested factor. The results are presented in Table 7.8.

Fig. 7.5
figure 5

Steps for analyzing nested design in SPSS

Table 7.8 ANOVA table for bank teller study using SPSS

Alternatively, it is possible to use the New > Syntax to obtain a full ANOVA table for nested design in SPSS . Note that syntax in SPSS can be used for several purposes, and here we illustrate one use – further work with syntax is beyond the scope of this book. After inserting the commands as shown in Fig. 7.6, we click the green triangle or Run > Selection in the syntax menu. The full ANOVA table is presented in Table 7.9.

Fig. 7.6
figure 6

Steps for full ANOVA table for nested design in SPSS

Table 7.9 Full ANOVA table for bank teller study using SPSS

Example 7.9 A Larger Example: First United Federal Bank of Boston using R

There are three ways in which R can recognize a nested design with the same output; they are listed below. In all cases, we will make sure that the independent variables are identified as factors by the software.

> bank <- read.csv(file.path("/Users/documents", "ex7.9.+csv"), header=T)

# Option 1:

> bank1 <- aov(Errors~Branch+Branch:Teller, data=bank)

# Option 2:

> bank2 <- aov(Errors~Branch+Teller%in%Branch, data=bank)

# Option 3:

> bank3 <- aov(Errors~Branch/Teller, data=bank) > summary(bank3)

 

Df

Sum Sq

Mean Sq

F value

Pr(>F)

 

Branch

3

1235

411.6

16.730

2.23e-09

***

Branch:Teller

12

825

68.7

2.794

0.00186

**

Residuals

144

3543

24.6

   

--- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Note that, in these three options, we didn’t specify the random variable and the F-value is not accurate. Considering that “teller” is a random effect, we can calculate the appropriate F- and p-values as follows:

> F <- 16.73/2.794 > F [1] 5.987831 > p-value <- 1- (pf(F, 3, 12, lower.tail=F)) > p-value [1] 0.009799043

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Berger, P., Maurer, R., Celli, G.B. (2018). Nested, or Hierarchical, Designs. In: Experimental Design. Springer, Cham. https://doi.org/10.1007/978-3-319-64583-4_7

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