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Lecture 08: Block Design (2)
Experimental Design in Education
2025-03-07
Example: Tutoring session
- Let’s look at a simple example:
- Research question: What is the effect of time of day of tutoring session on midterm grades?
- The Primary Investigator (PI) of the study wants to control for tutor, believing some tutors may be better in the subject matter than others. Thus, they design a completely randomized block design.
- Each tutor works with students in the morning and afternoon.
- In this example, student select their favorite tutor, but are then randomly assigned to either a morning (AM) or afternoon (PM) session.
- 200 total students: 91 in the morning, 109 in the afternoon
Note
DV: Midterm Score (in cells)
Example: Variables and Null Hypothesis
- IV: Time (a = 2 )
- 2 levels: AM and PM
- Nuisance factor: Tutor (b = 4 )
- 4 levels: Booby, Julia, Monique, and Ned
- DV: Midterm Score
- Null hypothesis pertaining to the IV of interest:
- \(H_0:\mu_{𝐴𝑀} = \mu_{𝑃𝑀} → a= 2\)
- We will also have a null hypothesis pertaining to the blocking factor:
- \(H_0:\mu_{Bobby} = \mu_{Julia} = \mu_{Monique} =\mu_{Ned} → b=4\)
- Two Nulls = two values of \(F_{obs}\), two values of \(F_{crit}\), two decisions
Note
DV: Midterm Score (in cells)
Example: Sum of Squares
IV: Time (𝑎= 2 )
Nuisance factor: Tutor (𝑏= 4 )
DV: Midterm Score
Now:
- \(𝑆𝑆_{Total} =\color{red}{𝑆𝑆_{𝑀𝑜𝑑𝑒𝑙}}+\color{purple}{𝑆𝑆_{𝐵𝑙𝑜𝑐𝑘}}+\color{blue}{𝑆𝑆_{𝐸𝑟𝑟𝑜𝑟}}\)
Thus, we can partition the effects into three parts:
Sum of squares due to treatments (IV = Time),
Sum of squares due to the blocking factor,
and Sum of squares due to error.
We do not model an interaction with blocked designs. (we will talk about it later.)
Example: Mean of Squares and F-statistics
- IV: Time (𝑎= 2 )
- Nuisance factor: Tutor (𝑏= 4 )
- DV: Midterm Score
- Model: \(𝑆𝑆_{Total} =\color{red}{𝑆𝑆_{𝑀𝑜𝑑𝑒𝑙}}+\color{purple}{𝑆𝑆_{𝐵𝑙𝑜𝑐𝑘}}+\color{blue}{𝑆𝑆_{𝐸𝑟𝑟𝑜𝑟}}\)
- ANOVA table:
Example: Marginal Means
- A marginal mean is the mean for one level of the variable, ignoring the other variable
For example:
- The AM marginal mean is the average of all students’ midterm scores in the morning, ignoring who they have as a tutor \[ \bar{y}_{AM.} = 22.95 \]
- The Bobby marginal mean is the average of all students’ midterm scores who had Bobby as a tutor, ignoring time of day \[ \bar{y}_{.Bobby} = 11.91 \]
Example: Details of ANOVA Table
- Then, we can fill out the ANOVA table:
Note
Under \(\alpha=.05\), for “Model” factor – Time, we have \(df_{Model}\) = 1, \(df_{error} = 195\): \(F_{crit}=3.89\) so sig.
Similarly, for “Blocking” - Tutor, we have \(df_{block}\) = 3, \(df_{error} = 195\): \(F_{crit}=2.65\) so sig.
Example: Plant Fertilizer1
In a greenhouse experiment, there was a single factor (fertilizer) with 4 levels (i.e. 4 treatments), six replications, and a total of 24 experimental units (potted plants). Suppose the image below is the greenhouse bench (viewed from above) that was used for the experiment.
To use CBD, we need to randomly assign each of the treatment levels to 6 potted plants. To do this, we first assign numbers to the physical position of the pots on the bench. Each column of plants will be use one Fertilizer.
To further expand it to RCBD, we need to put them into different farms (say we have six farms). Each farms will have 4 plans with each will use one type of fertilizer.
Read in R Data
- After saving the exp1_data.csv into the same directory of your R file. You should be able to import dataset from Files Panel in Rstudio:
Example: Performance of Students at varied environment
- Background: Comparing the performances of students (male and female) blocks in different environments (at home and at college). To represent this experiment in the figure will be as follows:
- Where AC: At College, AH: At Home
stud <- factor(rep(c("male", "female"), each = 2))
perf <- factor(rep(c("ah", "ac" ), times = 2))
perf[1] ah ac ah ac
Levels: ac ahy <- c(5.5, 5,
4, 6.2)
# y is the hours students
# studied in specific places
results <- data.frame(y, stud, perf)
fit <- aov(y ~ perf+stud, data = results)
summary(fit) Df Sum Sq Mean Sq F value Pr(>F)
perf 1 0.7225 0.7225 0.396 0.642
stud 1 0.0225 0.0225 0.012 0.930
Residuals 1 1.8225 1.8225 - Explanation: The value of Mean Sq is
0.7225<<1.8225,i.e, here blocking wasn’t necessary. And as Pr value is 0.642 > 0.05 (5% significance) and the hypothesis is accepted - there is no sufficient evidence suggesting females and males have significant differences in performance.
