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Two-Factor ANOVA model with n = 1 (no replication)

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  • [ "article:topic", "authorname:pauld" ]

    1. Two-factor ANOVA model with n = 1 (no replication)

    • For some studies, there is only one replicate per treatment, i.e., n = 1.
    • ANOVA model for two-factor studies need to be modified, since
      - the degrees of freedom associated with \(SSE\) will be \((n - 1)ab = 0\);
      - thus the error variance \(\sigma^2\) can not be estimated by \(SSE\) anymore.
    • Idea: make the model simpler by assuming the two factors do not interact with each other. Validity of this assumption needs to be checked.

    1.1 Two-factor model without interaction

    With n = 1.

    • Model equation:

    \[Y_{ij} = \mu_{..} + \alpha_i + \beta_j + \epsilon_{ij}, i = 1, ..., a, j = 1, ..., b.\]

    • Identifiability constraints:

    \[\sum_{i=1}^{a}\alpha_i = 0, \sum_{j=1}^{b}\beta_j = 0.\]

    • Distributional assumptions: \(\epsilon_{ij}\) are i.i.d. \(N(0,\sigma^2)\)
    Sum of squares

    Interaction sum of squares now plays the role of error sum of squares.

    \[SSAB = n\sum_{i=1}^{a} \sum_{j=1}^{b}(\overline{Y}_{ij.} - \overline{Y}_{i..} - \overline{Y}_{.j.} + \overline{Y}_{...})^2 = \sum_{i=1}^{a} \sum_{j=1}^{b}(\overline{Y}_{ij} - \overline{Y}_{i.} - \overline{Y}_{.j} + \overline{Y}_{..})^2 \]

    \(MSAB = \frac{SSAB}{(a-1)(b-1)}\) since \(d.f.(SSAB) = (a-1)(b-1)\).

    • In the general two-factor ANOVA model (when n = 1),

    \[E(MSAB) = \sigma^2 + \frac{\sum_{i=1}^{a}\sum_{j=1}^{b} (\alpha\beta)^2_{ij}}{(a-1)(b-1)}\]

    • Under the model without interaction: \(E(MSAB) = \sigma^2\)
    • Thus \(MSAB\) can be used to estimate \(\sigma^2\).
    ANOVA Table

    ANOVA table for two-factor model without interaction and \(n=1\)

    Source of Variation SS df MS
    Factor A \(SSA = b\sum_i(\overline{Y}_{i.} - \overline{Y}_{..})^2\) \(a - 1\) \(MSA\)
    Factor B \(SSB = a\sum_j(\overline{Y}_{.j} - \overline{Y}_{..})^2\) \(b - 1\) \(MSB\)
    Error \(SSAB = \sum_{i=1}^{a}\sum_{j=1}^{b}(\overline{Y}_{ij} - \overline{Y}_{i.} - \overline{Y}_{.j} + \overline{Y}_{..})^2\)                              \((a - 1)(b - 1)\) \(MSAB\)
    Total \(SSTO = \sum_{i=1}^{a}\sum_{j=1}^{b}(\overline{Y}_{ij} - \overline{Y}_{..})^2\)               \(ab - 1\)  

     

    Expected mean squares (under no interaction):

    \[E(MSA) = \sigma^2 + \frac{b\sum_{i=1}^{a}\alpha_i^2}{a - 1}, E(MSB) = \sigma^2 + \frac{a\sum_{j=1}^{b}\beta_j^2}{b - 1}, E(MSAB) = \sigma^2\]

    F tests (for main effects)

    Test factor A main effects: \(H_o: \alpha_1 = ... = \alpha_a = 0\) vs. \(H_a:\) not all \(\alpha_i\)'s are equal to zero.

    • \(F_A^* = \frac{MSA}{MSAB} ~ F_{a - 1, (a - 1)(b - 1)}\) under \(H_o\).
    • Reject \(H_o\) at level of significance \(\alpha\) if observed \(F_A^* > F(1 - \alpha; a - 1, (a - 1)(b - 1))\).

    Test factor B main effects: \(H_o: \beta_1 = ... = \beta_b = 0\) vs. \(H_a:\) not all \(\beta_j\)'s are equal to zero.

    • \(F_B^* = \frac{MSB}{MSAB} ~ F_{b - 1, (a - 1)(b - 1)}\) under \(H_o\).
    • Reject \(H_o\) at level of significance \(\alpha\) if observed \(F_B^* > F(1 - \alpha; b - 1, (a - 1)(b - 1))\).
    Estimation of means

    Estimation of factor level means \(\mu_{i.}\)'s , \(\mu_{.j}\)'s.

    • Proceed as before, viz., use the unbiased estimator \(\overline{Y}_{i.}\) for \(\mu_{i.}\) and \(\overline{Y}_{.j}\) for \(\mu_{.j}\), but replace \(MSE\) by \(MSAB\) and use the degrees of freedom of \(MSAB\), that is \((a - 1)(b - 1)\). Thus, estimated standard errors:

    \[s(\overline{Y}_{i.}) = \sqrt{\frac{MSAB}{b}}, s(\overline{Y}_{.j}) = \sqrt{\frac{MSAB}{a}}.\]

    Estimation of treatment means \(\mu_{ij}\)'s.

    • \(\mu_{ij} = E(Y_{ij}) = \mu_{..} + \alpha_i + \beta_j = \mu_{i.} + \mu_{.j} - \mu_{..}\)
      Thus, an unbiased estimator: \(\widehat{\mu}_{ij} = \overline{Y}_{i.} + \overline{Y}_{.j} - \overline{Y}_{..}\)
      Estimated standard error:

    \[s(\widehat{\mu}_{ij}) = \sqrt{MSAB(\frac{1}{b} + \frac{1}{a} - \frac{1}{ab})} = \sqrt{MSAB(\frac{a + b - 1}{ab})}\]

    1.2 Example: Insurance

    An analyst studied the premium for auto insurance charged by an insurance company in six cities. The six cities were selected to represent different sizes (Factor A: small, medium, large) and differentregions of the state (Factor B: east, west). There is only one city for each combination of size and region. The amounts of premiums charged for a specific type of coverage in a given risk category for each of the six cities are given in the following table.

    Table 1: Numbers in parentheses are \(\widehat{\mu}_{ij} = \overline{Y}_{i.} + \overline{Y}_{.j} - \overline{Y}_{..}\)

        Factor B  
      East West  
    Factor A Small     140(135)
    Medium  210(210)
    Large      220(225)
    100(105)
    180(180)
    200(195)
    \(\overline{Y}_{1.} = 120\)
    \(\overline{Y}_{2.} = 195\)
    \(\overline{Y}_{3.} = 210\)
      \(\overline{Y}_{.1} = 190\) \(\overline{Y}_{.2} = 160\) \(\overline{Y}_{..} = 175\)

    Interaction plot based on the treatment sample means \(Y_{ij}\)'s: no strong interactions.

    Sum of squares:
    • Here \(a = 3\), \(b = 2\), \(n = 1\).
    • \(SSA = 2[(120 - 175)^2 + (195 - 175)^2 + (210 - 175)^2] = 9300.\).
    • \(SSB = 3[(190 - 175)^2 + (160 - 175)^2] = 1350\).
    • \(SSAB = (140 - 120 - 190 + 175)^2 + ... + (200 - 210 - 160 + 175)^2 = 100\).
    • \(SSTO = SSA + SSB + SSAB = 10750\).
      Hypothesis testing:
    • Test \(H_o: \mu_{1.} = \mu_{2.} = \mu_{3.}\) (equivalently, \(H_o: \alpha_1 = \alpha_2 = \alpha_3 = 0\)) at level 0.05.
      Table 2: ANOVA Table for Insurance example
      Source of Variation SS df MS
      Factor A \(SSA = 9300\) \(a - 1 = 2\) \(MSA = 4650\)
      Factor B \(SSB = 1350\) \(b - 1 = 1\) \(MSB = 1350\)
      Error \(SSAB = 100\) \((a - 1)(b - 1) = 2\) \(MSAB = 50\)
      Total \(SSTO = 10750\) \(ab - 1 = 5\)  

    \(F_A^* = \frac{MSA}{MSAB} = \frac{4650}{50} = 93\) and \(F(0.95; 2, 2) = 19\). Thus reject \(H_o\) at level 0.05.

    • Estimation of \(\mu_{ij}\): e.g.,
      \(\widehat{\mu}_{11} = \overline{Y}_{1.} + \overline{Y}_{.1} - \overline{Y}_{..} = 120 + 190 - 175 = 135\).
    • Estimation of \(\mu_{i.}\) and \(\mu_{.j}\): e.g.,
      \(\widehat{\mu}_{1.} = \overline{Y}_{1.} = 120\).
      \(s(\overline{Y}_{1.}) = \sqrt{\frac{MSAB}{b}} = \sqrt{\frac{50}{2}} = 5\).
      The 95% C.I. for \(\mu_{1.}\) is:
      \[\overline{Y}_{1.} \pm t(0.975; 2) * s(\overline{Y}_{1.}) = 120 \pm 4.3*5 = (98.5, 141.5).\]

    1.3 Checking for the presence of interaction: Tukey's test for additivity

    For a two-factor study with \(n = 1\), decide whether or not the two factors are interacting.

    • In the no-interaction model, we assume that all \((\alpha\beta)_{ij} = 0\).
    • Idea: use a less severe restriction on the interaction effects, by assuming
      \[(\alpha\beta)_{ij} = D\alpha_i\beta_j,   i = 1, ... , a,   j = 1, ... , b,\]
      where \(D\) is an unknown parameter.
    • The model becomes:
      \[Y_{ij} = \mu_{..} + \alpha_i + \beta_j + D\alpha_i\beta_j + \epsilon_{ij},   i = 1, ... , a,    j = 1, ..., b,\]
      under the constraints that
      \[\sum_{i = 1}^{a}\alpha_i = \sum_{j = 1}^{b}\beta_j = 0.\]
    Estimation of \(D\)
    • Multiply \(\alpha_i\beta_j\) on both sides of the equation:
      \[\alpha_i\beta_jY_{ij} = \mu_{..}\alpha_i\beta_j + \alpha_i^2\beta_j + \alpha_i\beta_j^2 + D\alpha_i^2\beta_j^2 + \epsilon_{ij}\alpha_i\beta_j\]
    • Sum over all pairs (i, j):
      \[\sum_{i=1}^{a}\sum_{j=1}^{b}\alpha_i\beta_jY_{ij}=D\sum_{i=1}^{a}\sum_{j=1}^{b}\alpha_i^2\beta_j^2 + \sum_{i=1}^{a}\sum_{j=1}^{b}\epsilon_{ij}\alpha_i\beta_j\]
    • Then
      \[\widetilde{D} := \frac{\sum_{i=1}^{a}\sum_{j=1}^{b}\alpha_i\beta_jY_{ij}}{(\sum_{i=1}^{a}\alpha_i^2)(\sum_{j=1}^{b}\beta_j^2)} \approx D\]
    • We have the following estimates:
      \[\widehat{\alpha}_i = \overline{Y}_{i.} - \overline{Y}_{..}, \widehat{\beta}_j =  \overline{Y}_{.j} - \overline{Y}_{..}\]
    • Thus, an estimator of \(D\) (which is also the least squares and the maximum likelihood estimator) is given by
      \[\widehat{D} = \frac{\sum_{i=1}^{a}\sum_{j=1}^b ( \overline{Y}_{i.} - \overline{Y}_{..})( \overline{Y}_{.j} - \overline{Y}_{..})Y_{ij}}{(\sum_{i=1}^{a}( \overline{Y}_{i.} - \overline{Y}_{..})^2)(\sum_{j=1}^{b}( \overline{Y}_{.j} - \overline{Y}_{..})^2)}.\]

    ANOVA decomposition

    \[SSTO = SSA + SSB + SSAB* + SSRem*.\]

    • Interaction sum of squares
      \[SSAB* = \sum_{i=1}^{a}\sum_{j=1}^{b}\widehat{D}^2\widehat{\alpha}_i^2\widehat{\beta}_j^2 = \frac{(\sum_{i=1}^{a}\sum_{j=1}^b ( \overline{Y}_{i.} - \overline{Y}_{..})( \overline{Y}_{.j} - \overline{Y}_{..})Y_{ij})^2}{(\sum_{i=1}^{a}( \overline{Y}_{i.} - \overline{Y}_{..})^2)(\sum_{j=1}^{b}( \overline{Y}_{.j} - \overline{Y}_{..})^2)}\]
    • Remainder sum of squares
      \[SSREM^* = SSTO - SSA - SSB - SSAB^*\]
    • Decomposition of degrees of freedom
      \[df(SSTO) = df(SSA) + df(SSB) + df(SSAB^*) + df(SSRem^*)\]
      \[ab - 1 = (a - 1) + (b - 1) + 1 + (ab - a - b)\]
    • Tukey's one degree of freedom test for additivity: \(H_o: D = 0\) (i.e., no interaction) vs. \(H_a: D \neq 0\).
    • \(F\) ratio \(F_{Tukey}^{*} = \frac{SSAB^*/1}{SSRem^*/(ab - a - b)}\sim F_{1, ab - a - b}\) under \(H_o\).
    • Decision rule: reject \(H_o: D = 0\) at level of significance \(\alpha\) if \(F_{Tukey}^{*} > F(1 - \alpha; 1, ab - a - b)\).

    Example: Insurance

    • \(\sum_{ij}(\overline{Y}_{i.} - \overline{Y}_{..})( \overline{Y}_{.j} -  \overline{Y}_{..})Y_{ij} = -13500.\)
    • \(\sum_{i=1}^{a}( \overline{Y}_{i.} -  \overline{Y}_{..})^2 = 4650\), and \(\sum_{j=1}^{b}( \overline{Y}_{.j} -  \overline{Y}_{..})^2 = 450.\)
    • \(SSAB^* = \frac{(-13500)^2}{4650 * 450} = 87.1.\)
    • \(SSRem^* = 10750 - 9300 - 1350 - 87.1 = 12.9.\)
    • \(ab - a - b = 3*2 - 3 - 2 = 1.\)
    • \(F\)-ratio for Tukey's test:
      \[F_{Tukey}^{*} = \frac{SSAB^*/1}{SSRem^*/1} = \frac{87.1}{12.9} = 6.8.\]
    • When \(\alpha = 0.05, F(0.95; 1, 1) = 161.4 > 6.8.\)
    • Thus, we can not reject \(H_o: D = 0\) at the 0.05 level, and we conclude that there is no significant interaction between the two factors.
    • Indeed, the p-value is \(p = P(F_{1,1} > 6.8) = 0.23\) which is not at all significant.

    Contributors

    • Scott Brunstein (UCD)
    • Debashis Paul (UCD)