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I'm working on lung dataset of ISwR package in R.

A book by Peter Dalgaard, "Introductory Statistics with R" has assigned me to answer this question: 7.2 In the lung data, do the three measurement methods give systematically different results? If so, which ones appear to be different?

In this data, volume is a numeric vector and there are two factor variables: method(A, B, C) and subject(1, 2, 3, 4, 5, 6).

So What I did was simply,

anova(lm(volume ~ method, data=lung))

Its result was just an insignificant one.

I found out in the answer section of the book that I also have to put in 'subject', as in

anova(lm(volume ~ method + subject, data=lung))

And it gave me the right answer, showing significance for both method and subject.

But I cannot understand the result; Why is it insignificant with the one-way anova? And why did it become significant with one more variable added?

To faithfully answer the question, don't I have to input only method?

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    $\begingroup$ Did each subject get assessed with all three methods? If so, you need to account for this in your analysis. $\endgroup$
    – Harvey Motulsky
    Commented Dec 14, 2023 at 16:28
  • $\begingroup$ By any reasonable account, the evidence that gives "just" not significant and just significant are pretty much the same. What were the p-values? Do you really want an all-or-none binary result from your analyses? $\endgroup$
    – Michael Lew
    Commented Apr 20 at 1:24

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In your model with method as only predictor variable, you do not take into account that the three data points from each subject are not independent from each other. That is, you treat your predictor variable as a between-subjects variable while it is in fact a within-subjects variable. This reduces the power of your analysis because you do not factor out the variance explained by overall differences between subjects.

I'm not completely sure about the maths behind this, but adding subject as a predictor variable makes your analysis treat method as a within-subjects variable. You can see this when you run the ANOVA with a package that let's you directly specify within-subjects variables, e.g. the ez package:

ezANOVA(lung, volume, wid=subject, within=.(method))

That should give you the numerically identical result for method as in the textbook solution.

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