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8.4: Try It!

  • Page ID
    33904
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    Exercise \(\PageIndex{1}\)

    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)

    a) For each factor, indicate whether it is a fixed or random effect.

    b) Identify the treatments and describe (in words) the treatment design.

    c) Describe the randomization used.

    d) Compute the \(F\)-statistic for the temperature effect in the ANOVA, and determine significance for the effect.

    Show Solution

    a) temp=fixed, seafood=fixed, storage unit=random

    b) Temperature and Seafood, factorial design. Each seafood type is combined with each temperature level in the experiment.

    c) 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.

    d) \(F_{Temperature} =53.83/7.342=7.3318\). \(F_{critical}=5.14\), so reject \(H_{0}\).

    Exercise \(\PageIndex{2}\)

    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)

    a) For each factor, indicate whether it is a fixed or random effect

    b) Identify the treatments and describe (in words) the treatment design.

    c) Describe (in words) the randomization used.

    d) Compute the \(F\)-statistic for each effect in the ANOVA, and determine significance (i.e., compare \(F_{calculated}\) to \(F_{critical}\) for each effect).

    Show Solution

    a) group = fixed, tech_int = fixed, blk = random

    b) group and tech_int, crossed for a factorial treatment design

    c) 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.

    d) 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 \(\times\) 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 \(\PageIndex{3}\)

    1. 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.

    1. What is the treatment design?
    2. What is the randomization design?
    Show Solution

    a) Treatment design: \(3 \times 4\) factorial with oat variety and manure levels as factors having 3 and 4 levels respectively

    b) 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.

    2. 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.

    1. What is the treatment design?
    2. What is the randomization design?
    Show Solution

    a) Treatment design: \(2 \times 5\) factorial with irrigation method and fertilizer levels as factors having 2 and 5 levels respectively

    b) Randomization design: Split-plot in CRD with the whole factor as irrigation method and subplot factor as fertilizer level

    3. A survey was conducted among 100 high schoolers who were potential athletes to learn about their preferences on financial benefits. The sample consisted of an equal number of male and female students and 3 incentive types were offered: a 20% tuition reduction for all 4 years; a 50% tuition reduction in the first year, but renewable based on freshman GPA; and full room and board for all 4 years.

    1. What is the treatment design?
    2. What is the randomization design?
    Show Solution

    a) Treatment design: A single factor study with 3 levels; the factor of interest is incentive type

    b) Randomization design: RCBD with gender as the blocking factor


    This page titled 8.4: Try It! is shared under a CC BY-NC 4.0 license and was authored, remixed, and/or curated by Penn State's Department of Statistics.

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