Exercise 1 Section
Researchers are investigating the effect of storage temperature on bacterial growth for two types of seafood. They set up the experiment to evaluate 3 storage temperatures. There were 9 storage units that were available, and so they randomly selected 3 storage units to be used for each storage temperature, and both seafood types were stored in each unit. After 2 weeks, bacterial counts were made. After taking a logarithmic transformation of the counts, they produced the following ANOVA:
Type 3 Analysis of Variance | ||||
---|---|---|---|---|
Source | DF | Sum of Squares | Mean Square | Expected Mean Square |
temp | 2 | 107.656588 | 53.828294 | Var(Residual) + 2 Var(unit(temp)) + Q(temp, temp*seafood) |
seafood | 1 | 3.713721 | 3.713721 | Var(Residual) + Q(seafood, temp*seafood) |
temp*seafood | 2 | 2.647594 | 1.323797 | Var(Residual) + Q(temp*seafood) |
unit(temp) | 6 | 44.050650 | 7.341775 | Var(Residual) + 2 Var(unit(temp)) |
Residual | 6 | 5.590873 | 0.931812 | Var(Residual) |
- For each factor, indicate whether it is a fixed or random effect.
- Identify the treatments and describe (in words) the treatment design.
- Describe the randomization used.
- Compute the F-statistic for the temperature effect in the ANOVA, and determine significance for the effect.
- Temp is fixed, seafood is fixed, and storage unit is random.
- Temperature and seafood. The treatment design is a factorial design. Each seafood type is combined with each temperature level in the experiment.
- Split-plot in a CRD. Temperature levels were assigned (randomly) to storage units. Then the storage unit set at a given temperature is split to accommodate each of the two seafood types.
- Temperature F=53.83/7.342=7.3318 \(F_{critical}=5.14\), Reject \(H_0\).
Exercise 2 Section
Answer the questions based on the following output:
Type 3 Analysis of Variance | ||||
---|---|---|---|---|
Source | DF | Sum of Squares | Mean Square | Expected Mean Square |
group | 3 | 6429.388333 | 2143.129444 | Var(Residual) + 3 Var(blk*group) + Q(group,group*tech_int) |
tech_int | 2 | 881.408750 | 440.704375 | Var(Residual) + Q(tech_int,group*tech_int) |
group*tech_int | 6 | 207.507917 | 34.584653 | Var(Residual) + Q(group*tech_int) |
blk | 3 | 408.985000 | 136.328333 | Var(Residual) + 3 Var(blk*group) + 12 Var(blk) |
blk*group | 9 | 466.543333 | 51.838148 | Var(Residual) + 3 Var(blk*group) |
Residual | 24 | 595.696667 | 24.820694 | Var(Residual) |
- For each factor, indicate whether it is a fixed or random effect.
- Identify the treatments and describe (in words) the treatment design.
- Describe (in words) the randomization used.
- Compute the F-statistic for each effect in the ANOVA, and determine significance (i.e., compare \(F_{calculated}\) to \(F_{critical}\) for each effect).
- Group is fixed, tech_int is fixed, and blk is random.
- Group and tech_int. They are crossed for a factorial treatment design.
- Split-plot in a RCBD, with group as the whole plot treatment and tech_int as the subplot treatment with blk as the blocking factor.
-
- group: \(F=\dfrac{2143.129444}{51.838148} =41.3427\), \(F_{critical} = 3.86\), Reject \(H_0\).
- tech_int: \(F= \dfrac{440.704375}{24.820694} = 17.7555\), \(F_{critical} = 3.40\), Reject \(H_0\).
- group x tech_int: \(F= \dfrac{34.584653}{24.820694} = 1.3934\), \(F_{critical} = 2.51\), Do not reject \(H_0\).
- blk: \(F = \dfrac{136.3283}{51.8381} = 2.6299\), \(F_{critical}= 3.86\), Do Not Reject \(H_0\).
Exercise 3 Section
-
An experimenter wants to compare the yield of three varieties of oats at four different levels of manure. Suppose 6 farmers agree to participate in the experiment and each farmer will designate 3 fields from their farms for the experiment.
- What is the treatment design?
- What is the randomization design?
- Treatment design: 3X4 factorial with oat variety and manure levels as factors having 3 and 4 levels respectively.
- Randomization design: Three oats varieties will be randomly assigned to the 3 fields from each farm using RCBD with farms as blocks. Four manure levels are then randomized within each field using an RCBD. So the randomization design is a split-plot in RCBD.
- In an agricultural setting, an experimenter is applying one of two irrigation methods randomly to 6 plots where all plots are similar in moisture, soil type, slope, fertility, etc. Each plot is then subdivided into 5 portions and 5 levels of nitrogen fertilizer are applied randomly to these portions.
- What is the treatment design?
- What is the randomization design?
- Treatment design: 2X5 factorial with irrigation method and fertilizer levels as factors having 2 and 5 levels, respectively.
- Randomization design: Split-plot in CRD with the whole factor as irrigation method and subplot factor as fertilizer level.
- A survey was conducted to study whether the financial aid package amounts at a certain university differ between male and female athletes. The university offers 3 different types of aid packages: tuition reduction for 4 years, tuition reduction for the first year, or partial coverage of room and board. Within each package type, the package amount for a randomly selected male and female athlete was recorded.
- What is the treatment design?
- What is the randomization design?
- Treatment design: A single factor study with 2 levels - the factor of interest is gender.
- Randomization design: RCBD with package type as the blocking factor.