Re: [R] prop.test or chisq.test ..?
Hi
Some comments are inside.
Dylan Beaudette writes:
Hi everyone,
Suppose I have a count the occurrences of positive results, and the total
number of occurrences:
pos - 14
total - 15
testing that the proportion of positive occurrences is greater than 0.5
gives
a p-value and confidence interval:
prop.test( pos, total, p=0.5, alternative='greater')
1-sample proportions test with continuity correction
data: 14 out of 15, null probability 0.5
X-squared = 9.6, df = 1, p-value = 0.0009729
alternative hypothesis: true p is greater than 0.5
95 percent confidence interval:
0.706632 1.00
sample estimates:
p
0.933
First of all by default there is a continuity correction in
prop.test(). If you use
prop.test(pos, total, p=0.5, alternative=greater, correct = FALSE)
1-sample proportions test without continuity correction
data: pos out of total, null probability 0.5
X-squared = 11.2667, df = 1, p-value = 0.0003946
alternative hypothesis: true p is greater than 0.5
95 percent confidence interval:
0.7492494 1.000
sample estimates:
p
0.933
Remark that know the X-squared is identical to your result from
chisq.test(), but the p-value is 0.0007891/2
The reason is that you are testing here the against the
alternative greater
If you use a two sided test
prop.test(pos, total, p=0.5, alternative=two.sided, correct = FALSE)
then you reporduce the results form chisq.test().
My question is how does the use of chisq.test() differ from the above
operation. For example:
chisq.test(table( c(rep('pos', 14), rep('neg', 1)) ))
Chi-squared test for given probabilities
data: table(c(rep(pos, 14), rep(neg, 1)))
X-squared = 11.2667, df = 1, p-value = 0.0007891
... gives slightly different results. I am corrent in interpreting that the
chisq.test() function in this case is giving me a p-value associated with
the
test that the probabilities of pos are *different* than the probabilities of
neg -- and thus a larger p-value than the prop.test(... , p=0.5,
alternative='greater') ?
Yes. In your example chisq.test() tests the null hypothesis if
all population probabilities are equal
?chisq.test says:
In this case, the hypothesis tested is whether the population
probabilities equal those in 'p', or are all equal if 'p' is not
given.
which means p1 = p2 = 0.5 in your two population case against
the alternative p1 != p2.
This is similar to the test in prop.test() p=0.5 against p!=0.5
and therefore you get identical results if you choose
alternative=two.sided in prop.test().
I realize that this is a rather elementary question, and references to a
text
would be just as helpful. Ideally, I would like a measure of how much I
can 'trust' that a larger proportion is also statistically meaningful. Thus
far the results from prop.test() match my intuition, but
affirmation would be
Your intuition was correct. Nevertheless in your original
results (with the continuity correction), the p-value of
prop.test() (0.0009729) was larger than the p-value of
chisq.test() (0.0007891) and therefore against your intuition.
great.
Cheers,
--
Dylan Beaudette
Soils and Biogeochemistry Graduate Group
University of California at Davis
530.754.7341
__
R-help@stat.math.ethz.ch mailing list
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.
Hope this helps
Christoph Buser
--
Credit and Surety PML study: visit our web page www.cs-pml.org
--
Christoph Buser [EMAIL PROTECTED]
Seminar fuer Statistik, LEO C13
ETH Zurich 8092 Zurich SWITZERLAND
phone: x-41-44-632-4673 fax: 632-1228
http://stat.ethz.ch/~buser/
__
R-help@stat.math.ethz.ch mailing list
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.
Re: [R] prop.test or chisq.test ..?
On Wednesday 28 February 2007 01:07, Christoph Buser wrote:
Hi
Some comments are inside.
Dylan Beaudette writes:
Hi everyone,
Suppose I have a count the occurrences of positive results, and the
total number of occurrences:
pos - 14
total - 15
testing that the proportion of positive occurrences is greater than 0.5
gives a p-value and confidence interval:
prop.test( pos, total, p=0.5, alternative='greater')
1-sample proportions test with continuity correction
data: 14 out of 15, null probability 0.5
X-squared = 9.6, df = 1, p-value = 0.0009729
alternative hypothesis: true p is greater than 0.5
95 percent confidence interval:
0.706632 1.00
sample estimates:
p
0.933
First of all by default there is a continuity correction in
prop.test(). If you use
prop.test(pos, total, p=0.5, alternative=greater, correct = FALSE)
1-sample proportions test without continuity correction
data: pos out of total, null probability 0.5
X-squared = 11.2667, df = 1, p-value = 0.0003946
alternative hypothesis: true p is greater than 0.5
95 percent confidence interval:
0.7492494 1.000
sample estimates:
p
0.933
Remark that know the X-squared is identical to your result from
chisq.test(), but the p-value is 0.0007891/2
The reason is that you are testing here the against the
alternative greater
If you use a two sided test
prop.test(pos, total, p=0.5, alternative=two.sided, correct = FALSE)
then you reporduce the results form chisq.test().
My question is how does the use of chisq.test() differ from the above
operation. For example:
chisq.test(table( c(rep('pos', 14), rep('neg', 1)) ))
Chi-squared test for given probabilities
data: table(c(rep(pos, 14), rep(neg, 1)))
X-squared = 11.2667, df = 1, p-value = 0.0007891
... gives slightly different results. I am corrent in interpreting that
the chisq.test() function in this case is giving me a p-value associated
with the test that the probabilities of pos are *different* than the
probabilities of neg -- and thus a larger p-value than the prop.test(...
, p=0.5, alternative='greater') ?
Yes. In your example chisq.test() tests the null hypothesis if
all population probabilities are equal
?chisq.test says:
In this case, the hypothesis tested is whether the population
probabilities equal those in 'p', or are all equal if 'p' is not
given.
which means p1 = p2 = 0.5 in your two population case against
the alternative p1 != p2.
This is similar to the test in prop.test() p=0.5 against p!=0.5
and therefore you get identical results if you choose
alternative=two.sided in prop.test().
I realize that this is a rather elementary question, and references to a
text would be just as helpful. Ideally, I would like a measure of how
much I can 'trust' that a larger proportion is also statistically
meaningful. Thus far the results from prop.test() match my intuition,
but
affirmation would be
Your intuition was correct. Nevertheless in your original
results (with the continuity correction), the p-value of
prop.test() (0.0009729) was larger than the p-value of
chisq.test() (0.0007891) and therefore against your intuition.
great.
Cheers,
--
Dylan Beaudette
Soils and Biogeochemistry Graduate Group
University of California at Davis
530.754.7341
__
R-help@stat.math.ethz.ch mailing list
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.
Hope this helps
Christoph Buser
Thanks for the tips Christoph, this was the help that I was looking for.
cheers,
--
Dylan Beaudette
Soils and Biogeochemistry Graduate Group
University of California at Davis
530.754.7341
__
R-help@stat.math.ethz.ch mailing list
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.
[R] prop.test or chisq.test ..?
Hi everyone,
Suppose I have a count the occurrences of positive results, and the total
number of occurrences:
pos - 14
total - 15
testing that the proportion of positive occurrences is greater than 0.5 gives
a p-value and confidence interval:
prop.test( pos, total, p=0.5, alternative='greater')
1-sample proportions test with continuity correction
data: 14 out of 15, null probability 0.5
X-squared = 9.6, df = 1, p-value = 0.0009729
alternative hypothesis: true p is greater than 0.5
95 percent confidence interval:
0.706632 1.00
sample estimates:
p
0.933
My question is how does the use of chisq.test() differ from the above
operation. For example:
chisq.test(table( c(rep('pos', 14), rep('neg', 1)) ))
Chi-squared test for given probabilities
data: table(c(rep(pos, 14), rep(neg, 1)))
X-squared = 11.2667, df = 1, p-value = 0.0007891
... gives slightly different results. I am corrent in interpreting that the
chisq.test() function in this case is giving me a p-value associated with the
test that the probabilities of pos are *different* than the probabilities of
neg -- and thus a larger p-value than the prop.test(... , p=0.5,
alternative='greater') ?
I realize that this is a rather elementary question, and references to a text
would be just as helpful. Ideally, I would like a measure of how much I
can 'trust' that a larger proportion is also statistically meaningful. Thus
far the results from prop.test() match my intuition, but affirmation would be
great.
Cheers,
--
Dylan Beaudette
Soils and Biogeochemistry Graduate Group
University of California at Davis
530.754.7341
__
R-help@stat.math.ethz.ch mailing list
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.
