Answers for factorial experiments questions
Second lecture example on data SET 2
CatalystTemperature (oC) / 0% / 1% / Mean
80 / 24.7 / 28.7 / 26.7
90 / 30.3 / 34.0 / 32.2
100 / 36.3 / 44.3 / 40.3
Mean / 30.4 / 35.7 / 33.1
Analysis of Variance for C4
Source DF SS MS F P
temp 2 567.44 283.72 44.03 0.000
cat 1 122.72 122.72 19.04 0.001
temp*cat 2 17.44 8.72 1.35 0.295
Error 12 77.33 6.44
Total 17 784.94
For interaction
H0: There is no interaction between catalyst and temperature.
H1: There is interaction between catalyst and temperature.
Test statistic:
Decision: Cannot reject H0 at the 5% significance level.
Conclusion: There is no evidence to suggest that there is interaction between catalyst and temperature.
Test for main effect of temperature
H0: Mean yield is the same for all temperatures.
H1: At least one mean yield for temperature differs from another.
Test statistic:
Decision: Reject H0 at the 0.1% significance level.
Conclusion: There is very strong evidence to suggest that at least one mean yield for temperature differs from another.
Test for main effect of catalyst
H0: Mean yield is the same for both amounts of catalyst.
H1: The two mean yields for catalyst differ.
Test statistic:
Decision: Reject H0 at the 1% significance level.
Conclusion: There is strong evidence to suggest that the two mean yields for catalyst differ.
Given that the two main effects are significant, next we should investigate which means differ from one another.
For the main effect of catalyst, we do not need to calculate LSD because there are only two means. Hence since there is evidence that at least one mean differs from another, the difference has to be between these two means. Looking at the table of means above, we can summarize the effect of factor catalyst in the following way. The mean yield with catalyst at 1% (35.7 g) is higher than the mean yield with catalyst at 0% (30.4 g).
For the main effect of temperature, we need to calculate LSD because there are more than two means.
(note we use the same level of precision as the one for the compared means).
The three means for temperature to be compared are in the marginal column. The reason we used n = 6 in the equation for LSD is that each of these means is based on the two means in the same row but each of these two means in turn are based on 3 replicates. So, altogether the replicates for each mean are 2*3 = 6.]
We find that each of the three means for the main effect of temperature is significantly different from each other. We can summarise this as follows. The mean yield increases with each level increase in temperature (26.7 g, 32.2 g and 40.3 g for temperatures of 80, 90 and 100C, respectively.
Tutorial Question 1. (PAGE 113)
The amount of the active ingredient of a drug which can be produced from 500g of the original compound seems to be dependent on the temperature used and on the source of a plant derivative used as in the drug.
To try to determine the optimum mixture of conditions three temperatures (15C, 18C, 21C) and three sources of the plant derivative A B and C were tested with 3 replicates of each combination being used.
The yields were as follows
Plant sourceTempYield(g)
A1523 20 21
A1822 19 20
A2119 18 21
B1522 20 19
B1824 25 22
B2120 19 22
C1518 18 16
C1821 23 20
C2120 22 24
MTB > ANOVA 'yield' = source temp source * temp.
Factor Type Levels Values
source fixed 3 0 1 2
temp fixed 3 15 18 21
Analysis of Variance for yield
Source DF SS MS F P
source 2 8.222 4.111 1.71 0.209
temp 2 20.222 10.111 4.20 0.032
source*temp 4 46.222 11.556 4.80 0.008
Error 18 43.333 2.407
Total 26 118.000
For interaction
H0: there is no interaction between source and temperature
H1: there is interaction between source and temperature
Test statistic F = 4.80
Critical values
Therefore we reject H0 at 1% significance. There is strong evidence of interaction.
Hence we need to concentrate on the interaction means, not the main effect (marginal) means.
Test for main effect of source
H0: mean yield same for all sources
H1: at least one mean differs from the others
Test statistic F = 1.71
Critical values
Hence we cannot reject H0 at 5% significance. There is no evidence of a main effect of source.
Test for main effect of temperature
H0: mean yield same for all temperatures
H1: at least one mean differs from the others
Test statistic F = 4.20
Critical values
Hence we reject H0 at 5% significance. There is some evidence of a main effect of source but since the interaction is significant we do NOT calculate a LSD for the main effect means.
Having done all required F-test now go on to find the appropriate LSD. In this case for the interaction means
Table of means
Temperature
15oC 18oC 21oC Mean
A 21.333 20.333 19.333 20.333
B 20.333 23.667 20.333 21.444
C 17.333 21.333 22.000 20.222
Mean 19.667 21.778 20.556 20.667
For plant source A, the change in temperature makes no difference. For plant source C 18oC gives a significantly higher yield than either of the other temperatures. For plant source C 18oC and 21oC gives a higher yield than 15oC.
You could compare down the columns so that you start ‘ At 15oC plant source C gives a lower yield then sources A and B’ and so on. But you do not need to do both. In practice you would know which was the more important for the experiment.
2. Lettuce question PAGE 114
Fresh weight
First the required F-tests
This output lacks the F-values but I did them in the lecture. Remember that you always divide by the error mean square.
SourcedfSSMSF-ratio
date311260.63753.581.41
solution3 10.7 3.6 0.08
interaction91015.9112.9 2.44
error642049.146.1
total795236.3
Interaction
H0 There is no interaction between date and solution
H1 There is interaction
Test value F = 2.44
Critical values:
Therefore reject H0 at 5% significance. There is weak evidence of interaction. This says that the pattern of change with harvest date is not the same for all solutions
Main effect of date
H0 mean fresh weight is same for all dates
H1 At least one mean differs
Test value F=81.41
Critical values
Therefore reject H0 at 0.1% significance.This means that there were some differences in the fresh weight averaged over the 4 solutions for different harvest dates. This is as expected. All it tells us, is that the lettuces grew!
Main effect of solution
H0 mean fresh weight is same for all solution
H1 At least one mean differs
Test value F=0.08
Critical values
Therefore do not reject H0 . This say s that on average solution made no difference.
Least Significant Differences
Since there is evidence of interaction we need a LSD for comparing within the table of means. Even though there is a massive effect of date we cannot use the marginal means ('edges of table') as this would average out the different patterns within rows or columns which have been flagged by the significant F value or interaction.
Mean fresh leaf weight
datePersilEcover1Ecover2Water
1 4.045 4.740 1.641 3.910 3.559
217.06012.079 6.44116.84613.106
330.98726.46227.56023.42027.107
426.77036.15540.12632.93833.997
19.71519.85918.94219.25319.442
LSD for interaction means
We now look at where there are differences between means that are larger than the LSD. To be sensible we look either along rows or down columns. Which is better, depends on the purpose of the experiment and it may be necessary to do both. However it is not a good idea to jump all over like a demented rabbit. Think about the science involved.
For Persil the plants grew up to date 3 in a manner comparable with water but there was no significant growth between date 3 and date 4. Ecover1 appears to have a growth pattern similar to Water throughout. Ecover2 was slow at the start with significantly less fresh weight at date 2 than water or Persil but after that grew very fast ending significantly larger than the lettuces on Persil and making much the largest growth between date 3 and date 4.
Dry weight
SourcedfSSMSF-ratio
date 3264.6988.2336.31
solution 3 39.3913.13 5.40
interaction 9 37.174.13 1.70
error64155.502.43
total79496.75
Interaction
H0: There is no interaction between date and solution
H1 : There is interaction
Test value F = 1.70
Critical values
Therefore do not reject H0 at 5% significance. There is no evidence of interaction.
Main effect of date
H0: mean dry weight is same for all dates
H1: At least one mean differs
Test value F = 36.31
Critical values:
Therefore reject H0 at 0.1% significance. Very strong evidence of an effect due to date.
Main effect of solution
H0: mean dry weight is same for all solution
H1: At least one mean differs
Test value F = 5.40
Critical values
Therefore reject H0at 1% significance. Strong evidence of an effect of solution.
Least Significant Differences
Here we have no interaction. So we can assume that the pattern of response to time is the same for all solutions. Here we have two significant main effects (date and solution) so need LSD's for the main effect means i.e. the
means on the edges of the tables.
Table of means
Mean dry weight
datePersilEcover1Ecover2WaterMean
10.48500.75100.11460.5118|0.4656
21.52721.60980.72181.6132|1.3680
32.64285.07303.24664.5130|3.8688
42.50827.12824.96862.2629|4.9597
Mean1.79083.64052.26292.9680|2.6655
Comparisons this time are easier as we are looking just at the right hand column for date and just at the bottom row for solution.
For date there was no significant difference in dry weight between dates 1 and 2. After that there was significant gain at date 3 and again at date 4.
For solution lettuces in Persil had a significantly lower mean dry weight than those in water and Ecover1 and Ecover2 had a lower mean dry weight than Ecover1.
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