Design and Analysis of Single Factor Experiments: ANOVA - Part 2

10.1 Completely Randomized Single-Factor Experiments

10.2 The Random-Effects Model

10.2.1 Fixed Versus Random Factors

• In many cases, the factor of interest has a large number of possible levels.

• The goal is to draw conclusions about the entire population of factor levels.

• If we randomly select $a$ levels from the population of factor levels, the factor is considered a random factor.

• It is assumed that factor levels come from a population of infinite size or a size large enough to be considered infinite.

• This differs from the fixed-effects case, where conclusions apply only to the factor levels used in the experiment, not the entire population.

10.2.2 ANOVA and Variance Components

• The linear statistical model is defined as follows: $Y<em>{ij} = \mu + \tau</em>i + \epsilon_{ij}$, where ${i = 1, 2, \cdots , a}$ and ${j = 1, 2, \cdots , n}$.

  • $\tau_i$ represents the treatment effects.

  • $\epsilon_{ij}$ represents the errors.

  • Treatment effects and errors are independent random variables.

• The model's structure is identical to the fixed-effects case, but the parameters have a different interpretation.

• If the variance of the treatment effects $\tau<em>i$ is ${\sigma</em>\tau}^2$, then the variance strength of the response is
  $V(Y<em>{ij}) = {\sigma</em>\tau}^2 + {\sigma}^2$

• The variance components ${\sigma_\tau}^2$ and ${\sigma}^2$ are called variance components.

• The model in Eq. (10.19) is called the components of variance model or the random-effects model.

• To test hypotheses in the random-effects model:

  • The errors $\epsilon_{ij}$ are normally and independently distributed with a mean of zero and variance ${\sigma}^2$.

  • The treatment effects $\tau<em>i$ are normally and independently distributed with a mean of zero and variance ${\sigma</em>\tau}^2$.

  • The {$\tau<em>i$} are independent random variables, which implies that the usual assumption of $\sum</em>{i=1}^{n} \tau_i = 0$ from the fixed-effects model does not apply to the random-effects model.

• Testing the hypothesis that the individual treatment effects are zero is meaningless for the random-effects model.

• It is more appropriate to test hypotheses about ${\sigma_\tau}^2$.

  • $H<em>0: {\sigma</em>\tau}^2 = 0$

  • H1: {\sigma\tau}^2 > 0

• If ${\sigma_\tau}^2 = 0$, all treatments are identical.

• If {\sigma_\tau}^2 > 0, there is variability between treatments.

• The ANOVA decomposition of total variability is still valid:
  $SST = SST_{Treatments} + SSE$

• The expected values of the mean squares for treatments and error are different than in the fixed-effects case.

• In the random-effects model for a single-factor, completely randomized experiment:
  $E(MST<em>{Treatments}) = E(\frac{SST</em>{Treatments}}{a-1}) = {\sigma}^2 + n{\sigma_\tau}^2$
  $E(MSE) = E(\frac{SSE}{a(n-1)}) = {\sigma}^2$

• Both $MSE$ and $MST<em>{Treatments}$ estimate ${\sigma}^2$ when $H</em>0: {\sigma_\tau}^2 = 0$ is true.

• $MSE$ and $MST_{Treatments}$ are independent.

• The ratio $F<em>0 = \frac{MST</em>{Treatments}}{MSE}$ is an $F$ random variable with $a-1$ and $a(n-1)$ degrees of freedom when $H_0$ is true.

• The null hypothesis is rejected at the $\alpha$-level of significance if the computed test statistic value is f0 > f{\alpha, a-1, a(n-1)}.

• The computational procedure and construction of the ANOVA table for the random-effects model are identical to the fixed-effects case.

  • However, the conclusions are different because they apply to the entire population of treatments.

• We usually want to estimate the variance components (${\sigma}^2$ and ${\sigma_\tau}^2$) in the model.

• The procedure to estimate ${\sigma}^2$ and ${\sigma_\tau}^2$ is called the analysis of variance method, because it uses the information in the ANOVA table.

• It does not require the normality assumption on the observations.

• The procedure consists of equating the expected mean squares to their observed values in the ANOVA table and solving for the variance components.

• When equating observed and expected mean squares in the one-way classification random-effects model, we obtain:
  $MST<em>{Treatments} = {\sigma}^2 + n{\sigma</em>\tau}^2$ & $MSE = {\sigma}^2$

• Therefore, the estimators of the ANOVA variance components are:
  ${\hat{\sigma}}^2 = MSE$
  ${\hat{\sigma<em>\tau}}^2 = \frac{MST</em>{Treatments} - MSE}{n}$

• Sometimes, the analysis of variance method produces a negative estimate of a variance component when using the above equations.

  • In that case, we set it to zero, re-analyze the model experiment, or assume the linear model does not apply and modify it.

Example 10.4

• In textile manufacturing, the undertaking "Design and Analysis of Experiments" is used to study a single-factor experiment involving the random-effects model.

• A textile manufacturing company weaves a fabric on a large number of looms.

• The company is interested in loom-to-loom variability in tensile strength.

• To investigate this variability, the engineer selects four looms ($a = 4$ treatments) at random and makes four strength determinations ($n = 4$ samples) on fabric samples chosen at random from each loom.

• The data are shown in Table 10.7, and the ANOVA is summarized in Table 10.8.

• Data used in the illustrative example of tensile strength is given in Table 10.7.
  $y<em>{..} = 1527; N = 16, y</em>{i.} = 390, 366, 383 \&amp; 388$
  $a = 1, 2, 3, 4$: four treatments, $n = 4$ (samples), $N = a \times n = 4 \times 4 = 16$

• This example depicts an experiment that is a completely randomized design.

• Use the analysis of variance (ANOVA) to test the hypothesis that loom-to-loom mean tensile strength. Use $\alpha = 0.01$.
  $H<em>0: \tau</em>1 = \tau<em>2 = \tau</em>3 = \tau<em>4 = 0$ & $H</em>1: \tau_i \neq 0$ for at least one $i$

• The sums of squares for the analysis of variance are computed from Eqs. (10.8), (10.9) & (10.10) as follows:
  $SST = \sum<em>{i=1}^{a} \sum</em>{j=1}^{n} y<em>{ij}^2 - \frac{y</em>{..}^2}{N} = \sum<em>{i=1}^{4} \sum</em>{j=1}^{4} y<em>{ij}^2 - \frac{y</em>{..}^2}{N} = \sum<em>{i=1}^{4} y</em>{i1}^2 + y<em>{i2}^2 + y</em>{i3}^2 + y<em>{i4}^2 - \frac{y</em>{..}^2}{N}$

• Sum up the squares for row by row or treatment.
  $SST = 98^2 + 97^2 + 99^2 + 96^2 + 91^2 + 90^2 + 93^2 + 92^2 + 96^2 + 95^2 + 97^2 + 95^2 + 95^2 + 96^2 + 99^2 + 98^2 - \frac{1527^2}{16}$
  $= 38030 + 33494 + 36675 + 37646 - \frac{1527^2}{16} = 145845 - 145733.063$
  $SST = 111.9375$
  $y<em>{i.} = \text{Totals}= 390, 366, 383, 388 \text{ and } a = 4$ $SST</em>{Treatments} = \sum<em>{i=1}^{a} \frac{y</em>{i.}^2}{n} - \frac{y_{..}^2}{N}$

• Thus,
  $SST<em>{Treatments} = \sum</em>{i=1}^{4} \frac{y<em>{i.}^2}{n} - \frac{y</em>{..}^2}{N} = \frac{y<em>{1.}^2 + y</em>{2.}^2 + y<em>{3.}^2 + y</em>{4.}^2}{n} - \frac{y<em>{..}^2}{N} = \frac{390^2 + 366^2 + 383^2 + 388^2}{4} - \frac{1527^2}{24}$ $= \frac{583289}{4} - \frac{1527^2}{16} = 89.1875$ $SST</em>{Treatments} = 89.1875$

• From Eq. (10.10), we have $SSE = SST - SST_{Treatments}$
  $SSE = <a target="_blank" rel="noopener noreferrer nofollow" class="link" href="tel:111.9375" data-prevent-progress="true">111.9375</a> - 89.1875 = 22.7500$

• From Eq. (10.7), we calculate the test statistic as $F<em>0 = \frac{SST</em>{Treatments} / a-1}{SSE / a(n-1)} = \frac{MST<em>{Treatments}}{MSE}$ $F</em>0 = \frac{89.1875}{4-1} = 29.7291667 \approx 29.73$
  $MSE = \frac{SSE}{a (n-1)} = \frac{22.7500}{4 (4-1)} = 1.89583333 = {\sigma}^2$
  $MSE = {\sigma}^2 = 1.89583333$

• From Eq. (10.7) we calculate the test statistic as
  $f<em>0 = \frac{MST</em>{Treatments}}{MSE} = \frac{29.7291667}{1.89583333} = 15.6813187$

• We note that $f_0$ is an $F$-distribn with $a-1 = 4-1 = 3$ and $a(n-1) = 4(4-1) = 12$ degrees of freedom.

• Find $f<em>{0.01,3,12}$ from Table VI ~ $\alpha = 0.01: f</em>{\alpha, a-1, a(n-1)} = f_{0.01,3,12} = 5.95$

• Reject $H<em>0$ if f0 = 11.99 > f{\alpha, a-1, a(n-1)} = f{0.01,3,12} = 5.95

• $P$-value for this test statistic (via software) is P = P(F_{3,12} > 15.6813187) = 1.8 \times 10^{-4}

• Because $P \simeq 2.33361 \times 10^{-4}$ is considerably smaller than $\alpha = 0.01$, we have strong evidence to conclude that $H_0$ is not true.

• With mean square error, $MSE = {\sigma}^2 = 1.8958333$

• $\hat{{\sigma}}^2 = MSE$, via Eq. (10.25) we have ${\hat{\sigma<em>\tau}}^2 = \frac{MST</em>{Treatments} - MSE}{n} = \frac{29.7291667 - 1.89583333}{4} = 6.95833334$
  ${\hat{\sigma_\tau}}^2 \cong 6.96$

• From the analysis of variance, we conclude that the looms in the plant differ significantly in their ability to produce fabric of uniform strength.

• The variance components are estimated via Eqs. (10.24) & (10.25) to be ${\hat{\sigma}}^2 = MSE = 1.8958 \dots \cong 1.90$ & ${\hat{\sigma_\tau}}^2 \cong 6.96$

• Therefore, the variance of strength in the manufacturing process is estimated by Eq. (10.19) as $V(Y<em>{ij}) = {\sigma</em>\tau}^2 + {\sigma}^2 \approx 8.86$

• Most of the variability in strength in the output product is attributable to differences between looms.

• ANOVA for the Tensile Strength of looms via software:

  • $SST = <a target="_blank" rel="noopener noreferrer nofollow" class="link" href="tel:111.9375" data-prevent-progress="true">111.9375</a>, SST_{Treatments} = 89.1875$

  • $MST_{Treatments} = 29.7291667, MSE = 1.89583333$

  • (numerator) $dF = a - 1 = 3$, (denominator) $dF = a(n - 1) = 12$, (total) $dF = 15$

Work through exercises 13.34 to 13.41 in section 13.3

*Follow the definitions especially defining expressions (the equations).

*Complicated problems are what engineers solve.

*Most problems will follow examples on how to use the equations.

*Practice, and practice!

10.3 Randomized Complete Block Design (RCBD) Model

10.3.1 Design and Statistical Analysis

• In experimental design problems, it is necessary to design the experiment so that the variability arising from a nuisance factor can be controlled.

• The paired $t$-test is a procedure for comparing two treatment means when all experimental runs cannot be made under homogeneous conditions.

• The paired t-test is viewed as a method for reducing the background noise in the experiment by blocking out a nuisance factor effect.

• The randomized block design is an extension of the paired $t$-test to situations where the factor of interest has more than two levels; i.e., more than two treatments must be compared.

• For example, suppose that three methods could be used to evaluate the strength readings on steel plate girders (Z-like structural elements).

• These are considered as three treatments, say $t<em>1, t</em>2$, and $t_3$.

• If we use four girders as the experimental units, a randomized complete block design (RCBD) would appear as shown in Fig. 10.9.

• The design is called a RCBD because:

  • Each block is large enough to hold all the treatments.

  • The actual assignment of each of the three treatments within each block is done randomly.

• Experiment data is recorded in a table, such as is shown in Table 10.9.

• The observations in this table, say, $y_{ij}$, represent the response obtained when method $i$ is used on girder $j$.

• The general procedure for a RCBD consists of selecting $b$ blocks and running a complete replicate of the experiment in each block.

• The data that result from running a RCBD for investigating a single factor with $a$ levels and $b$ blocks are shown in Table 10.10.

• There are $a$ observations (one per factor level) in each block.

• The order in which these observations are run is randomly assigned within the block.

• The statistical analysis for the RCBD:

  • Suppose that a single factor with $a$ levels is of interest and that the experiment is run in $b$ blocks.

  • The observations are here represented by the linear statistical model: $Y<em>{ij} = \mu + \tau</em>i + \beta<em>j + \epsilon</em>{ij}$, where ${i = 1, 2, \cdots , a}$ and ${j = 1, 2, \cdots , b}$

    • $\mu$ is an overall mean.

    • $\tau_i$ is the effect of the $i$th treatment.

    • $\beta_j$ is the effect of the $j$th block.

    • $\epsilon_{ij}$ is the random error term, which is assumed to be normally and independently distributed with mean zero and variance ${\sigma}^2$.

  • Furthermore, the treatment and block effects are defined as deviations from the overall mean, so $\sum<em>{i=1}^{a} \tau</em>i = 0$ and $\sum<em>{j=1}^{b} \beta</em>j = 0$.

  • This was the same type of definition used for completely randomized experiments.

  • It is assumed that treatments and blocks do not interact.

  • The effect of treatment $i$ is the same regardless of which block (or blocks) in which it is tested.

  • It is of interest to test the equality of the treatment effects.
    $H<em>0: \tau</em>1 = \tau<em>2 = \cdots = \tau</em>a$
    $H<em>1: \tau</em>i \neq 0 \text{ for at least one }$

• The analysis of variance can be extended to the RCBD.

• The procedure uses a sum of squares identity that partitions the total sum of squares into three components.

• ANOVA Sums of Squares Identity: Randomized Complete Block Experiment

• The sum of squares identity for the randomized complete block design is:
  $\sum<em>{i=1}^{a} \sum</em>{j=1}^{b} (y<em>{ij} - \bar{y}</em>{..})^2 = b\sum<em>{i=1}^{a} (\bar{y}</em>{i.} - \bar{y}<em>{..})^2 + a\sum</em>{j=1}^{b} (\bar{y}<em>{.j} - \bar{y}</em>{..})^2 + \sum<em>{i=1}^{a} \sum</em>{j=1}^{b} (y<em>{ij} - \bar{y}</em>{.j} - \bar{y}<em>{i.} + \bar{y}</em>{..})^2$
  $SST = SST<em>{Treatments} + SSB</em>{Blocks} + SSE$

• The degrees of freedom corresponding to these sums of squares are:
  $ab - 1 = a - 1 + b - 1 + (a - 1)(b - 1)$

• For the randomized block design, the relevant mean squares are:
  $MST<em>{Treatments} = \frac{SST</em>{Treatments}}{a - 1}$
  $MSB<em>{Blocks} = \frac{SSB</em>{Blocks}}{b - 1}$
  $MSE = \frac{SSE}{(a - 1)(b - 1)}$

• The expected values of these mean squares can be shown to be as follows:
  $E(MST<em>{Treatments}) = {\sigma}^2 + \frac{b \sum</em>{i=1}^{a} {\tau<em>i}^2}{a - 1}$ $E(MSB</em>{Blocks}) = {\sigma}^2 + \frac{a \sum<em>{i=1}^{b} {\beta</em>j}^2}{a - 1}$
  $E(MSE) = {\sigma}^2$

• If the null hypothesis $H<em>0$ is true, so that all treatment effects $\tau</em>i = 0$, $MST_{Treatments}$ is an unbiased estimator of ${\sigma}^2$.

• If $H<em>0$ is false, $MST</em>{Treatments}$ overestimates ${\sigma}^2$.

• The mean square for error is always an unbiased estimate of ${\sigma}^2$.

• To test the null hypothesis that the treatment effects are all zero, we use the ratio
  $F<em>0 = \frac{MST</em>{Treatments}}{MSE}$

• $F_0$ has an $F$-distribn with $a - 1$ and $(a - 1)(b - 1)$ degrees of freedom if the null hypothesis is true.

• Reject the null hypothesis at the $\alpha$-level of significance if the computed value of the test statistic is f0 > f{\alpha, a-1, (a-1)(b-1)}.

• Compute $SST, SST<em>{Treatments}$ and $SSB</em>{Blocks}$ to obtain the error sum of squares $SSE$ by subtraction.

• Computing Formulas for ANOVA: Randomized Block Experiment (RCBD)
  $SST = \sum<em>{i=1}^{a} \sum</em>{j=1}^{b} y<em>{ij}^2 - \frac{y</em>{..}^2}{ab}$
  $SST<em>{Treatments} = \frac{1}{b} \sum</em>{i=1}^{a} y<em>{i.}^2 - \frac{y</em>{..}^2}{ab}$
  $SSB<em>{Blocks} = \frac{1}{a} \sum</em>{j=1}^{b} y<em>{j.}^2 - \frac{y</em>{..}^2}{ab}$
  $SSE = SST - SST<em>{Treatments} - SSB</em>{Blocks}$

• The computations are usually arranged in an ANOVA table.

• Computer software is used to perform the analysis of variance for a RCBD because it is easier at data handling.

Example 10.5

• An experiment was performed to determine the effect of four different chemicals on the strength of a fabric.

• The chemicals are used as part of the permanent-press finishing process.

• Five fabric samples were selected, and a RCBD was run by testing each chemical type once in random order on each fabric sample.

• It is required to test for differences in means using an ANOVA with $\alpha = 0.01$ significance level.

• Data analysis and interpretation:

  • $y<em>{..} = 39.2 ; \bar{y}</em>{..} = 1.96$

  • $y<em>{i.} = 5.7, 8.8, 6.9 \&amp; 17.8; \bar{y}</em>{i.} = 1.14, 1.76, 1.38 \&amp; 3.56$

  • $y_{.j} = 9.2, 10.1, 3.5, 8.8 \&amp; 7.6$

  • $\bar{y}_{.j} = 2.3, 2.525, 0.875, 2.2 \&amp; 1.9$

  • $N = 20, a = 1, 2, 3, 4$: four treatments; $b = 5$ (blocks, samples), $N = a \times b = 4 \times 5 = 20$

  • $dF: a - 1 = 3$ (numerator) & $(a - 1)(b - 1) = 12$ (denominator)

• This example depicts an experiment that is a completely randomized design.

• Use the analysis of variance (ANOVA) to test the hypothesis on the means using $\alpha = 0.01$.
  $H<em>0: \tau</em>1 = \tau<em>2 = \tau</em>3 = \tau<em>4 = 0$ $H</em>1: \tau_i \neq 0 \text{ for at least one } i$

• The sums of squares for the analysis of variance are computed from:
  $SST = \sum<em>{i=1}^{4} \sum</em>{j=1}^{5} y<em>{ij}^2 - \frac{y</em>{..}^2}{ab}$

• Thus,
  $SST = 1.3^2 + 1.6^2 + 0.5^2 + 1.2^2 + 1.1^2 + 2.2^2 + 2.4^2 + 0.4^2 + 2.0^2 + 1.8^2 + 1.8^2 + 1.7^2 + 0.6^2 + 1.5^2 + 1.3^2 + 3.9^2 + 4.4^2 + 2.0^2 + 4.1^2 + 3.4^2 - \frac{39.2^2}{20}$
  $= 7.15 + 18.00 + 10.43 + 66.94 - \frac{39.2^2}{20}$
  $SST = 102.52 - 76.832 = 25.688$

  • $y<em>{i.} = \text{Totals}= 5.7, 8.8, 6.9, 17.8$ $SST</em>{Treatments} = \sum<em>{i=1}^{a} \frac{y</em>{i.}^2}{b} - \frac{y<em>{..}^2}{ab}$ $SST</em>{Treatments} = \sum<em>{i=1}^{4} \frac{y</em>{i.}^2}{b} - \frac{y<em>{..}^2}{ab} = \frac{y</em>{1.}^2 + y<em>{2.}^2 + y</em>{3.}^2 + y<em>{4.}^2}{5} - \frac{y</em>{..}^2}{ab} = \frac{5.7^2 + 8.8^2 + 6.9^2 + 17.8^2}{5} - \frac{39.2^2}{20}$
    $= \frac{474.38}{5} - \frac{39.2^2}{20} = 18.044$
    $SST_{Treatments} = 18.044$

• With $y_{.j} =9.2, 10.1, 3.5, 8.8 \&amp; 7.6$

  $SSB<em>{Blocks} = \sum</em>{j=1}^{5} \frac{y<em>{.j}^2}{a} - \frac{y</em>{..}^2}{ab} = \frac{y<em>{.1}^2 + y</em>{.2}^2 + y<em>{.3}^2 + y</em>{.4}^2 + y<em>{.5}^2}{5} - \frac{y</em>{..}^2}{ab} = \frac{9.2^2 + 10.1^2 + 3.5^2 + 8.8^2 + 7.6^2}{4} - \frac{39.2^2}{20}$
  $= \frac{334.10}{4} - \frac{39.2^2}{20} = 6.693$
  $SSB_{Blocks} = 6.693$

  $SSE = SST - SSB<em>{Blocks} - SST</em>{Treatments} = 25.688 - 6.693 - 18.044 = 0.951$

• From Eq. (10.7) we calculate the test statistic:
  $F<em>0 = \frac{SST</em>{Treatments} / a-1}{SSE / (a-1)(b-1)} = \frac{MST<em>{Treatments}}{MSE}$ $F</em>0 = \frac{18.044}{ 4 - 1} = \frac{0.951}{(4 - 1) (5 - 1)} = 75.89484753$

• Note that $f_0$ is an $F$-distribn with $a - 1 = 4 - 1 = 3$ and $a - 1 b - 1 = (4 - 1) (5 - 1) = 12$ degrees of freedom

• Find $f<em>{0.01,3,12}$ from Table VI ~ $\alpha = 0.01: f</em>{\alpha, a-1, a(n-1)} = f_{0.01,3,12} = 5.95$

• Reject $H<em>0$ if f0 = 75.89 > f{\alpha, a-1, a(n-1)} = f{0.01,3,12} = 5.95 where $f<em>0$ is the computed value of $F</em>0$ from Eq. (10.7).

• Find a $P$-value for this test statistic:
  P = P(F_{3,12} > 75.89) = 4.5 \times 10^{-8}

• Because $P \simeq 4. 5 \times 10^{-8}$ (calculator) is considerably smaller than $\alpha = 0.01$, we conclude that there is a significant difference in the chemical types so far as their effect on strength is concerned.
  MathML
  $MST<em>{Treatments} = \frac{SST</em>{Treatments}}{a - 1} = \frac{18.044}{4 - 1} = 6.0147$
  $MSE = \frac{SSE}{(a - 1)(b - 1)} = \frac{0.951}{12} = 0.07925$
  $MSB<em>{Blocks} = \frac{SSB</em>{Blocks}}{b - 1} = \frac{6.693}{5 - 1} = 1.67325$
  *Practice problems at the end of section 13.4 in the textbook.

Work through exercises 13.42 to 13.69 in section 13.2.