Dear Md Kamruzzaman,
I've copied this response to the r-help list, where you originally asked
your question. That way, other people can follow the conversation, if
they're interested and there will be a record of the solution. Please
keep r-help in the loop
See below:
On 2024-01-17 9:47 p.m., Md. Kamruzzaman wrote:
Caution: External email.
Dear John
Thank you so much for your reply.
I have calculated the 95%CI of the separate two proportions by using the
survey package. The code is given below.
svyby(~Diabetes_Cate, ~Year, nhc, svymean, na=TRUE)
Here: nhc is the weighted survey data.
I understand your point that it is possible to calculate the 95%CI of
the proportional difference manually. It is time consuming, that's why
I was looking for a function with a design effect to calculate this
easily. I couldn't find this kind of function.
However, it will be okay for me to calculate this manually, if there are
no functions like this.
If you intend to do this computation once, it's not terribly time
consuming. If you intend to do it repeatedly, you can write a simple
function to do the calculation, probably in less time than it takes to
search for one.
For manual calculation, could you please share the formula? to calculate
the 95%CI of proportional difference.
Here's a simple function to compute the confidence interval, assuming
that the normal distribution is used. The formula is based on the
elementary result that the variance of the difference of two independent
random variables is the sum of their variances, plus the observation
that the width of the confidence interval is 2*z*SE, where z is the
normal quantile corresponding to the confidence level (e.g., 1.96 for a
95% CI).
ciDiff <- function(ci1, ci2, level=0.95){
p1 <- mean(ci1)
p2 <- mean(ci2)
z <- qnorm((1 - level)/2, lower.tail=FALSE)
se1 <- (ci1[2] - ci1[1])/(2*z)
se2 <- (ci2[2] - ci2[1])/(2*z)
seDiff <- sqrt(se1^2 + se2^2)
(p1 - p2) + c(-z, z)*seDiff
}
Example: Prevalence of Diabetes:
2011: 11.0 (95%CI
10.1-11.9)
2017: 10.1 (95%CI
9.4-10.9)
Diff: 0.9% (95%CI: ??)
These are percentages, not proportions, but you can use either:
> ciDiff(c(10.1, 11.9), c(9.4, 10.9))
[1] -0.3215375 2.0215375
> ciDiff(c(.101, .119), c(.094, .109))
[1] -0.003215375 0.020215375
You'll want more significant digits in the inputs to get sufficiently
precise results.
Since I did this quickly, if I were you I'd check the results manually.
Best,
John
With Kind Regards
-------------------------
*/Md Kamruzzaman/*
On Thu, Jan 18, 2024 at 12:44 AM John Fox <j...@mcmaster.ca
<mailto:j...@mcmaster.ca>> wrote:
Dear Md Kamruzzaman,
To answer your second question first, you could just use the svychisq()
function. The difference-of-proportion test is equivalent to a
chisquare
test for the 2-by-2 table.
You don't say how you computed the confidence intervals for the two
separate proportions, but if you have their standard errors (and if
not,
you should be able to infer them from the confidence intervals) you can
compute the variance of the difference as the sum of the variances
(squared standard errors), because the two proportions are independent,
and from that the confidence interval for their difference.
I hope this helps,
John
--
John Fox, Professor Emeritus
McMaster University
Hamilton, Ontario, Canada
web: https://www.john-fox.ca/ <https://www.john-fox.ca/>
On 2024-01-16 10:21 p.m., Md. Kamruzzaman wrote:
> [You don't often get email from mkzama...@gmail.com
<mailto:mkzama...@gmail.com>. Learn why this is important at
https://aka.ms/LearnAboutSenderIdentification
<https://aka.ms/LearnAboutSenderIdentification> ]
>
> Caution: External email.
>
>
> Hello Everyone,
> I was analysing big survey data using survey packages on RStudio.
Survey
> package allows survey data analysis with the design effect.The survey
> package included functions for all other statistical analysis except
> two-proportion z tests.
>
> I was trying to calculate the difference in prevalence of
Diabetes and
> Prediabetes between the year 2011 and 2017 (with 95%CI). I was
able to
> calculate the weighted prevalence of diabetes and prediabetes in
the Year
> 2011 and 2017 and just subtracted the prevalence of 2011 from the
> prevalence of 2017 to get the difference in prevalence. But I
could not
> calculate the 95%CI of the difference in prevalence considering
the weight
> of the survey data.
>
> I was also trying to see if this difference in prevalence is
statistically
> significant. I could do it using the simple two-proportion z test
without
> considering the weight of the sample. But I want to do it
considering the
> weight of the sample.
>
>
> Example: Prevalence of Diabetes:
> 2011: 11.0
(95%CI
> 10.1-11.9)
> 2017: 10.1
(95%CI
> 9.4-10.9)
> Diff: 0.9%
(95%CI: ??)
> Proportion
Z test P
> Value: ??
> Your cooperation will be highly appreciated.
>
> Thanks in advance.
>
> With Regards
>
> *--------------------------------*
>
> *Md Kamruzzaman*
>
> *PhD **Research Fellow (**Medicine**)*
> Discipline of Medicine and Centre of Research Excellence in
Translating
> Nutritional Science to Good Health
> Adelaide Medical School | Faculty of Health and Medical Sciences
> The University of Adelaide
> Adelaide SA 5005
>
> [[alternative HTML version deleted]]
>
> ______________________________________________
> R-help@r-project.org <mailto:R-help@r-project.org> mailing list
-- To UNSUBSCRIBE and more, see
> https://stat.ethz.ch/mailman/listinfo/r-help
<https://stat.ethz.ch/mailman/listinfo/r-help>
> PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
<http://www.R-project.org/posting-guide.html>
> and provide commented, minimal, self-contained, reproducible code.
______________________________________________
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
