R help - Sep 2023 - prop.trend.test question

Colleagues,

The code for prop.trend.test is given by:

function (x, n, score = seq_along(x)) 
{
    method <- "Chi-squared Test for Trend in Proportions"
    dname <- paste(deparse1(substitute(x)), "out of",
deparse1(substitute(n)), 
        ",\n using scores:", paste(score, collapse = " "))
    x <- as.vector(x)
    n <- as.vector(n)
    p <- sum(x)/sum(n)
    w <- n/p/(1 - p)
    a <- anova(lm(freq ~ score, data = list(freq = x/n, score
as.vector(score)),
        weights = w))
    chisq <- c(`X-squared` = a["score", "Sum Sq"])
    structure(list(statistic = chisq, parameter = c(df = 1), 
        p.value = pchisq(as.numeric(chisq), 1, lower.tail = FALSE), 
        method = method, data.name = dname), class = "htest")
}

It seems to me that the direction of the trend is found using the weighted
regression lm(freq ~ score, data = list(freq = x/n, score as.vector(score)), 
        weights = w))

Am I on the right track here?

Thomas Subia

In a word: Yes.

We discussed this about 2w ago. Basically, the lm() fits a local Linear
Probability Model and the coef to "score" gives you the direction of
the effect.

In the same thread it was discussed (well, readable between the lines, maybe)
that if you change the lm() to a Gaussian glm() and use summary(...,
dispersion=1), you can extract the z-test version of the trend test. (I think
that the reason for using the chisquare version was to match the example in
Altman: Practical Statistics for Medical Research.)

Arguably the code should be updated to use the z and at the same time include
alternative=c("two.sided", "less", "greater") like
other 1df tests have. Just a matter of these darn little round tuits that you
seem never to be able to get... Also slightly tricky whether/how to make it
backwards compatible.

- pd
> On 25 Sep 2023, at 04:10 , tgs77m--- via R-help <r-help at
r-project.org> wrote:
> 
> Colleagues,
> 
> The code for prop.trend.test is given by:
> 
> function (x, n, score = seq_along(x)) 
> {
>    method <- "Chi-squared Test for Trend in Proportions"
>    dname <- paste(deparse1(substitute(x)), "out of",
> deparse1(substitute(n)), 
>        ",\n using scores:", paste(score, collapse = "
"))
>    x <- as.vector(x)
>    n <- as.vector(n)
>    p <- sum(x)/sum(n)
>    w <- n/p/(1 - p)
>    a <- anova(lm(freq ~ score, data = list(freq = x/n, score >
as.vector(score)),
>        weights = w))
>    chisq <- c(`X-squared` = a["score", "Sum Sq"])
>    structure(list(statistic = chisq, parameter = c(df = 1), 
>        p.value = pchisq(as.numeric(chisq), 1, lower.tail = FALSE), 
>        method = method, data.name = dname), class = "htest")
> }
> 
> It seems to me that the direction of the trend is found using the weighted
> regression lm(freq ~ score, data = list(freq = x/n, score >
as.vector(score)),
>        weights = w))
> 
> Am I on the right track here?
> 
> Thomas Subia
> 
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-- 
Peter Dalgaard, Professor,
Center for Statistics, Copenhagen Business School
Solbjerg Plads 3, 2000 Frederiksberg, Denmark
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Email: pd.mes at cbs.dk  Priv: PDalgd at gmail.com