An industrial engineer is conducting an experiment on eye focus time. He is inte
ID: 3292592 • Letter: A
Question
An industrial engineer is conducting an experiment on eye focus time. He is interested in the effect of the distance of the object from the eye on the focus time. Four different distances, 4, 6, 8, 10 (ft) are of interest. Six subjects in his company are randomly selected for the experiment. Each subject is tested four times at the different distances in a random order, and the values of focus time are recorded. It is suspected that there may be differences among individuals.
1. Explain why it makes sense to treat subject as a random blocking factor?
2. What is the name of this design of experiment? Write down the model for analyzing this experiment.
3. Complete the ANOVA table below.
Source
DF
Sum of Squares
Mean Square
F Value
Distance
178.17
59.39
Subject
38.45
Err
7.33
Corrected Total
23
480.31
4. Does distance have a significant effect on the focus time? Use a= 5% and critical value 3.29.
5. Estimate the variance component for subject.
Source
DF
Sum of Squares
Mean Square
F Value
Distance
178.17
59.39
Subject
38.45
Err
7.33
Corrected Total
23
480.31
Explanation / Answer
1. Explain why it makes sense to treat subject as a random blocking factor?
Because it is suspected that there may be differences among individuals.
2. What is the name of this design of experiment? Write down the model for analyzing this experiment.
This is an example of randomized block design. The “treatments” are the distances, and the “blocks” are the subjects.
Null Hypothesis H0: Mean focus time for all 4 distances are equal.
Alternative Hypothesis H1: Mean focus time of any one of distances are not equal.
3. Complete the ANOVA table below.
First calculate DF for Distance,
We know that, Mean Square = Sum of squares / DF
So,
DF for Distance = Sum of squares / Mean Square = 178.17 / 59.39 = 3
Also, DF for Distance = Number of Distances - 1 = 4 - 1 = 3
DF for Subjects = Number of Subjects - 1 = 6 - 1 = 5
We know that, Mean Square = Sum of squares / DF
So, Sum of squares of Subject = Mean Square * DF = 38.45 * 5 = 192.25
DF for error = (Number of Subjects - 1) * (Number of Distances - 1) = 5 * 3 = 15
So, Sum of squares of Error = Mean Square * DF = 7.33 * 15 = 109.95
F value for Distance = Mean Square of distance / Mean square for error = 59.39 / 7.33 = 8.1023
F value for Subject = Mean Square of subject / Mean square for error = 38.45 / 7.33 = 5.2456
4. Does distance have a significant effect on the focus time? Use a= 5% and critical value 3.29.
As, the observed value of F (8.1023) is greater than the critical value (3.29) , we reject the null hypothesis and conclude that distance have a significant effect on the focus time.
5. Estimate the variance component for subject.
Sum of squares for subject = 192.25
So, the variance component of subject is 192.25
Source DF Sum of Squares Mean Square F Value Distance 3 178.17 59.39 8.1023 Subject 5 192.25 38.45 5.2456 Err 15 109.95 7.33 Corrected Total 23 480.31 20.88