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One-way random effects ANOVA (Model II)

Worked example 1

No. adult T. decipiens from grey seals
Western
Isles
ShetlandOrkneysEast
Anglia
East
Scotland
0
21
124
7
13
13
14
1
151
5
3
 
 
 
 
8
29
2
17
0
0
7
5
0
5
 
 
 
 
 
222
390
16
7
543
0
48
204
 
 
 
 
 
 
 
0
0
7
0
0
4
 
 
 
 
 
 
 
 
 
0
0
0
0
0
0
27
3
0
2
0
0
0
0
0
Our first worked example uses data from Young (1972) on the number of nematode parasites (Terranova decipiens) in grey seals from five colonies around the British coast.

In this case it is debatable whether a Model I or Model II ANOVA is most appropriate. This is because we don't know whether the researcher selected those specific colonies for comparison, or whether they were a random (more likely convenience) sample of the colonies available. The author used a fixed effects ANOVA on log transformed data, but (for the sake of this example) we will analyze it using random effects ANOVA. It is unclear exactly how the seals were sampled, but the researcher notes that an attempt was made to obtain a representative sample.

  1. Draw boxplots

    {Fig. 1a}
    Mmod21a.gif

    It is apparent just from looking at the numeric data that distributions are contagious, with a few seals having large numbers of parasites. A log transformation improves the situation for regions A to C - but does not help much for regions C and D.

    Next we look at normal QQ plots both of the raw data and of the log transformed data for the five groups.

    {Fig. 1b}
    Mmod21b.gif

    Groups A - C are (more or less) normalized by a log (Y+1) transformation but groups D and E remain very skewed.

  2. Check homogeneity of variances
    Using
    A Bartlett's test on the raw data demonstrates that the variances are highly heteroscedastic. But after a log transformation, we get a borderline non-significant value (P = 0.06518), so we can (very cautiously) accept that the variances of log transformed data are homogeneous. Note that Bartlett's is probably being too liberal here given the distributions are still skewed.
  3. Carry out analysis of variance on log (Y + 1) transformed data

    SSgroup  =  26.88002  +  15.25042  +  31.80232  +  3.68892  +  5.81712  −  (83.4387)2
    1110861550
       =  80.6493
    SStotal  =  307.5264  −  (83.4387)2  =  168.2861
    50
    SSresidual  =  168.2861  −  80.6493  =  87.6368

    Using
    ANOVA table
    Source of
    variation
    df SS MS F-
    ratio
    P
    Group  4  80.649 20.162 10.355 < 0.001
    Residual 45 87.637 1.947    
    Total 49 168.287      
  4. Estimate variance components & ICC

    Since sample sizes vary markedly between groups, we need to calculate no as an overall measure of sample size:

    no = [1/4][50 − (546/50)] = 9.77

    The added variance component is then calculated as below:
    sA2    =    20.162 − 1.947    =    1.8644
    9.77

    The intraclass correlation coefficient (= percentage variation between groups) is then calculated as:
    ICC    =   100 [ 1.8644 ]    =    48.92%
    1.947 + 1.8644

 

Worked example 2

Cow
No.
PCVTi
1.2.
1333568
2262349
3292958
4323567
5312859
6313162
7313465
8312960
9353368
10212445
Sum   601

We will briefly revisit a worked example from Unit 3 where we took two repeated measurements of packed cell volume from each of 10 cows.

By taking the square root of the mean of the variances we estimated:
Within-subject standard deviation = √(28.5/10) = 1.688

We can obtain the same information by doing a one way analysis of variance:

SScow  =  682  +...  452  −  (601)2
2220
   =  278.45
SStotal  =  18367  −  (601)2  =  306.95
20
SSresidual  =  306.95  −  278.45  =  28.5

ANOVA table
Source of
variation
df SS MS F-
ratio
P
Cow  9  278.45 30.939 10.856 < 0.001
Residual 10 28.5 2.85    
Total 19 306.95      

The mean square residual (MSW) is equal to 2.85. The within subject standard deviation is given by √MSW = 1.688, the same as we obtained using the other computational formula.

The added variance component is calculated as below:
sA2    =    30.939 − 2.85    =    14.0445
2

The intraclass correlation coefficient (percentage variation between cows) is then calculated as:
ICC    =   100 [ 14.0445 ]    =    83.13%
2.85 + 14.0445

This can also be expressed as a percentage measurement error of 16.87%.