Às 10:06 de 08/09/2023, peter dalgaard escreveu:
Yes, this was written a bit bone-headed (as I am allowed to say...)
If you look at the code, you will see inside:
a <- anova(lm(freq ~ score, data = list(freq = x/n, score =
as.vector(score)),
weights = w))
and the lm() inside should give you the direction via the sign of the regression
coefficient on "score".
So, at least for now, you could just doctor a copy of the code for your own purposes, as in
fit <- lm(freq ~ score, data = list(freq = x/n, score = as.vector(score)),
weights = w)
a <- anova(fit)
and arrange to return coef(fit)["score"] at the end. Something like structure(... estimate=c(lpm.slope=coef(fit)["score"]) ....)
(I expect that you might also extract the t-statistic from coef(summary(fit))
and find that it is the signed square root of the Chi-square, but I won't have
time to test that just now.)
-pd
On 8 Sep 2023, at 07:22 , Thomas Subia via R-help <r-help@r-project.org> wrote:
Colleagues,
Thanks all for the responses.
I am monitoring the daily total number of defects per sample unit.
I need to know whether this daily defect proportion is trending upward (a bad
thing for a manufacturing process).
My first thought was to use either a u or a u' control chart for this.
As far as I know, u or u' charts are poor to detect drifts.
This is why I chose to use prop.trend.test to detect trends in proportions.
While prop.trend.test can confirm the existence of a trend, as far as I know,
it is left to the user
to determine what direction that trend is.
One way to illustrate trending is of course to plot the data and use
geom_smooth and method lm
For the non-statisticians in my group, I've found that using this method along
with the p-value of prop.trend.test, makes it easier for the users to determine
the existence of trending and its direction.
If there are any other ways to do this, please let me know.
Thomas Subia
On Thursday, September 7, 2023 at 10:31:27 AM PDT, Rui Barradas
<ruipbarra...@sapo.pt> wrote:
Às 14:23 de 07/09/2023, Thomas Subia via R-help escreveu:
Colleagues
Consider
smokers <- c( 83, 90, 129, 70 )
patients <- c( 86, 93, 136, 82 )
prop.trend.test(smokers, patients)
Output:
Chi-squared Test for Trend inProportions
data: smokers out of patients ,
using scores: 1 2 3 4
X-squared = 8.2249, df = 1, p-value = 0.004132
# trend test for proportions indicates proportions aretrending.
How does one identify the direction of trending?
# prop.test indicates that the proportions are unequal but doeslittle to
indicate trend direction.
All the best,
Thomas Subia
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Hello,
By visual inspection it seems that there is a decreasing trend.
Note that the sample estimates of prop.test and smokers/patients are equal.
smokers <- c( 83, 90, 129, 70 )
patients <- c( 86, 93, 136, 82 )
prop.test(smokers, patients)$estimate
#> prop 1 prop 2 prop 3 prop 4
#> 0.9651163 0.9677419 0.9485294 0.8536585
smokers/patients
#> [1] 0.9651163 0.9677419 0.9485294 0.8536585
plot(smokers/patients, type = "b")
Hope this helps,
Rui Barradas
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Hello,
Actually, the t-statistic is not the signed square root of the X-squared
test statistic. I have edited the function, assigned the lm fit and
returned it as is. (print.htest won't print this new list member so the
output is not cluttered with irrelevant noise.)
smokers <- c( 83, 90, 129, 70 )
patients <- c( 86, 93, 136, 82 )
edit(prop.trend.test, file = "ptt.R")
source("ptt.R")
# stats::prop.trend.test edited to include the results
# of the lm fit and saved under a new name
ptt <- function (x, n, score = seq_along(x))
{
method <- "Chi-squared Test for Trend in Proportions"
dname <- paste(deparse1(substitute(x)), "out of",
deparse1(substitute(n)),
",\n using scores:", paste(score, collapse = " "))
x <- as.vector(x)
n <- as.vector(n)
p <- sum(x)/sum(n)
w <- n/p/(1 - p)
a <- anova(fit <- lm(freq ~ score, data = list(freq = x/n, score =
as.vector(score)),
weights = w))
chisq <- c(`X-squared` = a["score", "Sum Sq"])
structure(list(statistic = chisq, parameter = c(df = 1),
p.value = pchisq(as.numeric(chisq), 1, lower.tail =
FALSE),
method = method, data.name = dname
, lmfit = fit
), class = "htest")
}
ht <- ptt(smokers, patients)
ht$statistic |> sqrt()
#> X-squared
#> 2.867913
ht$lmfit |> summary() |> coef()
#> Estimate Std. Error t value Pr(>|t|)
#> (Intercept) 1.02187141 0.04732740 21.591539 0.00213815
#> score -0.03341563 0.01723384 -1.938954 0.19207036
Hope this helps,
Rui Barradas
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