

If my sample size is small is there a particular switch option that I need to use with t.test so that it calculates the t ratio correctly?
Here is a dummy example?
á =0.05
Mean pain reduction for A =27; B =31 and SD are SDA=9 SDB=12
drgA.p<rnorm(5,27,9);
drgB.p<rnorm(5,31,12)
t.test(drgA.p,drgB.p) # what do I need to give as additional parameter here?
I can do it manually but was looking for a switch option that I need to specify for t.test.
Thanks ../Murli
[[alternative HTML version deleted]]
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Hi Nair,
If the two populations are normal the ttest gives you
the exact result for whatever the sample size is (the
sample size will affect the number of degrees of
freedom).
When the populations are not normal and the sample
size is large it is still OK to use ttest (because of
the Central Limit Theorem) but this is not necessarily
true for the small sample size.
You could use simulation to find the relevant
probabilities.
 "Nair, Murlidharan T" < [hidden email]> wrote:
> If my sample size is small is there a particular
> switch option that I need to use with t.test so that
> it calculates the t ratio correctly?
>
> Here is a dummy example?
>
> á =0.05
>
> Mean pain reduction for A =27; B =31 and SD are
> SDA=9 SDB=12
>
> drgA.p<rnorm(5,27,9);
>
> drgB.p<rnorm(5,31,12)
>
> t.test(drgA.p,drgB.p) # what do I need to give as
> additional parameter here?
>
>
>
> I can do it manually but was looking for a switch
> option that I need to specify for t.test.
>
>
>
> Thanks ../Murli
>
>
>
>
> [[alternative HTML version deleted]]
>
> > ______________________________________________
> [hidden email] mailing list
> https://stat.ethz.ch/mailman/listinfo/rhelp> PLEASE do read the posting guide
> http://www.Rproject.org/postingguide.html> and provide commented, minimal, selfcontained,
> reproducible code.
>
______________________________________________
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Indeed, I understand what you say. The df of freedom for the dummy example is n1+n22 = 8. But when I run the t.test I get it as 5.08, am I missing something?
Original Message
From: Moshe Olshansky [mailto: [hidden email]]
Sent: Tuesday, August 07, 2007 9:05 PM
To: Nair, Murlidharan T; [hidden email]
Subject: Re: [R] small sample techniques
Hi Nair,
If the two populations are normal the ttest gives you
the exact result for whatever the sample size is (the
sample size will affect the number of degrees of
freedom).
When the populations are not normal and the sample
size is large it is still OK to use ttest (because of
the Central Limit Theorem) but this is not necessarily
true for the small sample size.
You could use simulation to find the relevant
probabilities.
 "Nair, Murlidharan T" < [hidden email]> wrote:
> If my sample size is small is there a particular
> switch option that I need to use with t.test so that
> it calculates the t ratio correctly?
>
> Here is a dummy example?
>
> á =0.05
>
> Mean pain reduction for A =27; B =31 and SD are
> SDA=9 SDB=12
>
> drgA.p<rnorm(5,27,9);
>
> drgB.p<rnorm(5,31,12)
>
> t.test(drgA.p,drgB.p) # what do I need to give as
> additional parameter here?
>
>
>
> I can do it manually but was looking for a switch
> option that I need to specify for t.test.
>
>
>
> Thanks ../Murli
>
>
>
>
> [[alternative HTML version deleted]]
>
> > ______________________________________________
> [hidden email] mailing list
> https://stat.ethz.ch/mailman/listinfo/rhelp> PLEASE do read the posting guide
> http://www.Rproject.org/postingguide.html> and provide commented, minimal, selfcontained,
> reproducible code.
>
______________________________________________
[hidden email] mailing list
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On Wed, 8 Aug 2007, Nair, Murlidharan T wrote:
> Indeed, I understand what you say. The df of freedom for the dummy example is n1+n22 = 8. But when I run the t.test I get it as 5.08, am I missing something?
>
Yes. You are probably looking for the version of the t.test that assumes equal variances (the original one), so you need var.equal=TRUE.
thomas
> Original Message
> From: Moshe Olshansky [mailto: [hidden email]]
> Sent: Tuesday, August 07, 2007 9:05 PM
> To: Nair, Murlidharan T; [hidden email]
> Subject: Re: [R] small sample techniques
>
> Hi Nair,
>
> If the two populations are normal the ttest gives you
> the exact result for whatever the sample size is (the
> sample size will affect the number of degrees of
> freedom).
> When the populations are not normal and the sample
> size is large it is still OK to use ttest (because of
> the Central Limit Theorem) but this is not necessarily
> true for the small sample size.
> You could use simulation to find the relevant
> probabilities.
>
>  "Nair, Murlidharan T" < [hidden email]> wrote:
>
>> If my sample size is small is there a particular
>> switch option that I need to use with t.test so that
>> it calculates the t ratio correctly?
>>
>> Here is a dummy example?
>>
>> á =0.05
>>
>> Mean pain reduction for A =27; B =31 and SD are
>> SDA=9 SDB=12
>>
>> drgA.p<rnorm(5,27,9);
>>
>> drgB.p<rnorm(5,31,12)
>>
>> t.test(drgA.p,drgB.p) # what do I need to give as
>> additional parameter here?
>>
>>
>>
>> I can do it manually but was looking for a switch
>> option that I need to specify for t.test.
>>
>>
>>
>> Thanks ../Murli
>>
>>
>>
>>
>> [[alternative HTML version deleted]]
>>
>>> ______________________________________________
>> [hidden email] mailing list
>> https://stat.ethz.ch/mailman/listinfo/rhelp>> PLEASE do read the posting guide
>> http://www.Rproject.org/postingguide.html>> and provide commented, minimal, selfcontained,
>> reproducible code.
>>
>
> ______________________________________________
> [hidden email] mailing list
> https://stat.ethz.ch/mailman/listinfo/rhelp> PLEASE do read the posting guide http://www.Rproject.org/postingguide.html> and provide commented, minimal, selfcontained, reproducible code.
>
Thomas Lumley Assoc. Professor, Biostatistics
[hidden email] University of Washington, Seattle
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About using t tests and confidence intervals for "large" samples 
"large" may need to be very large.
The old precomputerage rule of n >= 30 is inadequate.
For example, for an exponential distribution, the actual size
of a nominal 2.5% onesided ttest is not accurate to within 10%
(i.e. between 2.25% & 2.75%) until n is around 5000.
The error (actual  nominal size) decreases very slowly, at the rate 1/sqrt(n).
In practice, real distributions may be even more skewed than
the exponential distribution, even though they appear less skewed,
if they have long tails. In this case the sample size would need
to be even larger for t procedures to be reasonably accurate.
An alternative is to use bootstrapping. Bootstrap procedures that
decrease at the rate 1/n include bootstrap t, BCa, and bootstrap
tilting.
Moshe Olshansky < [hidden email]> wrote:
>If the two populations are normal the ttest gives you
>the exact result for whatever the sample size is (the
>sample size will affect the number of degrees of
>freedom).
>When the populations are not normal and the sample
>size is large it is still OK to use ttest (because of
>the Central Limit Theorem) but this is not necessarily
>true for the small sample size.
>You could use simulation to find the relevant
>probabilities.
>...
========================================================
 Tim Hesterberg Senior Research Scientist 
 [hidden email] Insightful Corp. 
 (206)8022319 1700 Westlake Ave. N, Suite 500 
 (206)2838691 (fax) Seattle, WA 981093044, U.S.A. 
 www.insightful.com/Hesterberg 
========================================================
Short course  Bootstrap Methods and Permutation Tests
Oct 1011 San Francisco, 34 Oct UK.
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As Thomas Lumley noted, there exist several versions
of ttest.
If you use t1 < t.test(x,y) then no assumption is
made of x and y having equal variance and of the two
sample sizes being equal and then an approximate
ttest is used with an approximate number of degrees
of freedom (and this is what you got).
If you use t2 < t.test(x,y,var.equal=TRUE) then equal
variance is assumed and you get 8 degrees of freedom.
If you use t3 < t.test(x,y,paired=TRUE) then equal
sample sizes are assumed and the number of degrees of
freedom is 4 (51).
 "Nair, Murlidharan T" < [hidden email]> wrote:
> Indeed, I understand what you say. The df of freedom
> for the dummy example is n1+n22 = 8. But when I run
> the t.test I get it as 5.08, am I missing something?
>
>
> Original Message
> From: Moshe Olshansky [mailto: [hidden email]]
>
> Sent: Tuesday, August 07, 2007 9:05 PM
> To: Nair, Murlidharan T; [hidden email]
> Subject: Re: [R] small sample techniques
>
> Hi Nair,
>
> If the two populations are normal the ttest gives
> you
> the exact result for whatever the sample size is
> (the
> sample size will affect the number of degrees of
> freedom).
> When the populations are not normal and the sample
> size is large it is still OK to use ttest (because
> of
> the Central Limit Theorem) but this is not
> necessarily
> true for the small sample size.
> You could use simulation to find the relevant
> probabilities.
>
>  "Nair, Murlidharan T" < [hidden email]> wrote:
>
> > If my sample size is small is there a particular
> > switch option that I need to use with t.test so
> that
> > it calculates the t ratio correctly?
> >
> > Here is a dummy example?
> >
> > á =0.05
> >
> > Mean pain reduction for A =27; B =31 and SD are
> > SDA=9 SDB=12
> >
> > drgA.p<rnorm(5,27,9);
> >
> > drgB.p<rnorm(5,31,12)
> >
> > t.test(drgA.p,drgB.p) # what do I need to give as
> > additional parameter here?
> >
> >
> >
> > I can do it manually but was looking for a switch
> > option that I need to specify for t.test.
> >
> >
> >
> > Thanks ../Murli
> >
> >
> >
> >
> > [[alternative HTML version deleted]]
> >
> > > ______________________________________________
> > [hidden email] mailing list
> > https://stat.ethz.ch/mailman/listinfo/rhelp> > PLEASE do read the posting guide
> > http://www.Rproject.org/postingguide.html> > and provide commented, minimal, selfcontained,
> > reproducible code.
> >
>
>
______________________________________________
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https://stat.ethz.ch/mailman/listinfo/rhelpPLEASE do read the posting guide http://www.Rproject.org/postingguide.htmland provide commented, minimal, selfcontained, reproducible code.


On 9/08/2007, at 2:57 PM, Moshe Olshansky wrote:
> As Thomas Lumley noted, there exist several versions
> of ttest.
<snip>
> If you use t3 < t.test(x,y,paired=TRUE) then equal
> sample sizes are assumed and the number of degrees of
> freedom is 4 (51).
This is seriously misleading. The assumption is not that the sample
sizes
are equal, but rather that there is ***just one sample***, namely
the sample of differences.
More explicitly the assumptions are that
x_i  y_i
are i.i.d. Gaussian with mean mu and variance sigma^2.
One is trying to conduct inference about mu, of course.
It should also be noted that it is a crucial assumption for the
``nonpaired''
ttest that the two samples be ***independent*** of each other, as
well as
being Gaussian.
None of this is however germane to Nair's original question; it is
clear
that he is interested in a twoindependentsample ttest.
cheers,
Rolf Turner
######################################################################
Attention:\ This email message is privileged and confidenti...{{dropped}}
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Well, this an explanation of what is done in the
paired ttest (and why the number of df is as it is).
I was too lazy to write all this.
It is nice that some list members are less lazy!
 Rolf Turner < [hidden email]> wrote:
>
> On 9/08/2007, at 2:57 PM, Moshe Olshansky wrote:
>
> > As Thomas Lumley noted, there exist several
> versions
> > of ttest.
>
> <snip>
>
> > If you use t3 < t.test(x,y,paired=TRUE) then
> equal
> > sample sizes are assumed and the number of degrees
> of
> > freedom is 4 (51).
>
> This is seriously misleading. The assumption is
> not that the sample
> sizes
> are equal, but rather that there is ***just one
> sample***, namely
> the sample of differences.
>
> More explicitly the assumptions are that
>
> x_i  y_i
>
> are i.i.d. Gaussian with mean mu and variance
> sigma^2.
>
> One is trying to conduct inference about mu, of
> course.
>
> It should also be noted that it is a crucial
> assumption for the
> ``nonpaired''
> ttest that the two samples be ***independent*** of
> each other, as
> well as
> being Gaussian.
>
> None of this is however germane to Nair's original
> question; it is
> clear
> that he is interested in a twoindependentsample
> ttest.
>
> cheers,
>
> Rolf Turner
>
>
######################################################################
> Attention:
> This email message is privileged and confidential.
> If you are not the
> intended recipient please delete the message and
> notify the sender.
> Any views or opinions presented are solely those of
> the author.
>
> This email has been scanned and cleared by
> MailMarshal
> www.marshalsoftware.com
>
######################################################################
>
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Thanks, that discussion was helpful. Well, I have another question
I am comparing two proportions for its deviation from the hypothesized
difference of zero. My manually calculated z ratio is 1.94.
But, when I calculate it using prop.test, it uses Pearson's chisquared
test and the Xsquared value that it gives it 0.74. Is there a function
in R where I can calculate the z ratio? Which is
('p1'p2)(p1p2)
Z= 
S
('p1'p2)
Where S is the standard error estimate of the difference between two
independent proportions
Dummy example
This is how I use it
prop.test(c(30,23),c(300,300))
Cheers../Murli
Original Message
From: Moshe Olshansky [mailto: [hidden email]]
Sent: Thursday, August 09, 2007 12:01 AM
To: Rolf Turner; [hidden email]
Cc: Nair, Murlidharan T; Moshe Olshansky
Subject: Re: [R] small sample techniques
Well, this an explanation of what is done in the
paired ttest (and why the number of df is as it is).
I was too lazy to write all this.
It is nice that some list members are less lazy!
 Rolf Turner < [hidden email]> wrote:
>
> On 9/08/2007, at 2:57 PM, Moshe Olshansky wrote:
>
> > As Thomas Lumley noted, there exist several
> versions
> > of ttest.
>
> <snip>
>
> > If you use t3 < t.test(x,y,paired=TRUE) then
> equal
> > sample sizes are assumed and the number of degrees
> of
> > freedom is 4 (51).
>
> This is seriously misleading. The assumption is
> not that the sample
> sizes
> are equal, but rather that there is ***just one
> sample***, namely
> the sample of differences.
>
> More explicitly the assumptions are that
>
> x_i  y_i
>
> are i.i.d. Gaussian with mean mu and variance
> sigma^2.
>
> One is trying to conduct inference about mu, of
> course.
>
> It should also be noted that it is a crucial
> assumption for the
> ``nonpaired''
> ttest that the two samples be ***independent*** of
> each other, as
> well as
> being Gaussian.
>
> None of this is however germane to Nair's original
> question; it is
> clear
> that he is interested in a twoindependentsample
> ttest.
>
> cheers,
>
> Rolf Turner
>
>
######################################################################
> Attention:
> This email message is privileged and confidential.
> If you are not the
> intended recipient please delete the message and
> notify the sender.
> Any views or opinions presented are solely those of
> the author.
>
> This email has been scanned and cleared by
> MailMarshal
> www.marshalsoftware.com
>
######################################################################
>
______________________________________________
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> Original Message
> From: [hidden email]
> [mailto: [hidden email]] On Behalf Of Nair,
> Murlidharan T
> Sent: Thursday, August 09, 2007 9:19 AM
> To: Moshe Olshansky; Rolf Turner; [hidden email]
> Subject: Re: [R] small sample techniques
>
> Thanks, that discussion was helpful. Well, I have another question
> I am comparing two proportions for its deviation from the hypothesized
> difference of zero. My manually calculated z ratio is 1.94.
> But, when I calculate it using prop.test, it uses Pearson's
> chisquared
> test and the Xsquared value that it gives it 0.74. Is there
> a function
> in R where I can calculate the z ratio? Which is
>
>
> ('p1'p2)(p1p2)
> Z= 
> S
> ('p1'p2)
>
> Where S is the standard error estimate of the difference between two
> independent proportions
>
> Dummy example
> This is how I use it
> prop.test(c(30,23),c(300,300))
>
>
> Cheers../Murli
>
>
Murli,
I think you need to recheck you computations. You can run a ttest on your data in a variety of ways. Here is one:
> x<c(rep(1,30),rep(0,270))
> y<c(rep(1,23),rep(0,277))
> t.test(x,y)
Welch Two Sample ttest
data: x and y
t = 1.0062, df = 589.583, pvalue = 0.3147
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
0.02221086 0.06887752
sample estimates:
mean of x mean of y
0.10000000 0.07666667
Hope this is helpful,
Dan
Daniel J. Nordlund
Research and Data Analysis
Washington State Department of Social and Health Services
Olympia, WA 985045204
______________________________________________
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n=300
30% taking A relief from pain
23% taking B relief from pain
Question; If there is no difference are we likely to get a 7% difference?
Hypothesis
H0: p1p2=0
H1: p1p2!=0 (not equal to)
1>Weighed average of two sample proportion
300(0.30)+300(0.23)
 = 0.265
300+300
2>Std Error estimate of the difference between two independent proportions
sqrt((0.265 *0.735)*((1/300)+(1/300))) = 0.03603
3>Evaluation of the difference between sample proportion as a deviation from the hypothesized difference of zero
((0.300.23)(0))/0.03603 = 1.94
z did not approach 1.96 hence H0 is not rejected.
This is what I was trying to do using prop.test.
prop.test(c(30,23),c(300,300))
What function should I use?
Original Message
From: [hidden email] on behalf of Nordlund, Dan (DSHS/RDA)
Sent: Thu 8/9/2007 1:26 PM
To: [hidden email]
Subject: Re: [R] small sample techniques
> Original Message
> From: [hidden email]
> [mailto: [hidden email]] On Behalf Of Nair,
> Murlidharan T
> Sent: Thursday, August 09, 2007 9:19 AM
> To: Moshe Olshansky; Rolf Turner; [hidden email]
> Subject: Re: [R] small sample techniques
>
> Thanks, that discussion was helpful. Well, I have another question
> I am comparing two proportions for its deviation from the hypothesized
> difference of zero. My manually calculated z ratio is 1.94.
> But, when I calculate it using prop.test, it uses Pearson's
> chisquared
> test and the Xsquared value that it gives it 0.74. Is there
> a function
> in R where I can calculate the z ratio? Which is
>
>
> ('p1'p2)(p1p2)
> Z= 
> S
> ('p1'p2)
>
> Where S is the standard error estimate of the difference between two
> independent proportions
>
> Dummy example
> This is how I use it
> prop.test(c(30,23),c(300,300))
>
>
> Cheers../Murli
>
>
Murli,
I think you need to recheck you computations. You can run a ttest on your data in a variety of ways. Here is one:
> x<c(rep(1,30),rep(0,270))
> y<c(rep(1,23),rep(0,277))
> t.test(x,y)
Welch Two Sample ttest
data: x and y
t = 1.0062, df = 589.583, pvalue = 0.3147
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
0.02221086 0.06887752
sample estimates:
mean of x mean of y
0.10000000 0.07666667
Hope this is helpful,
Dan
Daniel J. Nordlund
Research and Data Analysis
Washington State Department of Social and Health Services
Olympia, WA 985045204
______________________________________________
[hidden email] mailing list
https://stat.ethz.ch/mailman/listinfo/rhelpPLEASE do read the posting guide http://www.Rproject.org/postingguide.htmland provide commented, minimal, selfcontained, reproducible code.
______________________________________________
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> Original Message
> From: [hidden email]
> [mailto: [hidden email]] On Behalf Of Nair,
> Murlidharan T
> Sent: Thursday, August 09, 2007 12:02 PM
> To: Nordlund, Dan (DSHS/RDA); [hidden email]
> Subject: Re: [R] small sample techniques
>
> n=300
> 30% taking A relief from pain
> 23% taking B relief from pain
> Question; If there is no difference are we likely to get a 7%
> difference?
>
> Hypothesis
> H0: p1p2=0
> H1: p1p2!=0 (not equal to)
>
> 1>Weighed average of two sample proportion
> 300(0.30)+300(0.23)
>  = 0.265
> 300+300
> 2>Std Error estimate of the difference between two
> independent proportions
> sqrt((0.265 *0.735)*((1/300)+(1/300))) = 0.03603
>
> 3>Evaluation of the difference between sample proportion as a
> deviation from the hypothesized difference of zero
> ((0.300.23)(0))/0.03603 = 1.94
>
>
> z did not approach 1.96 hence H0 is not rejected.
>
> This is what I was trying to do using prop.test.
>
> prop.test(c(30,23),c(300,300))
>
> What function should I use?
>
>
The proportion test above indicates that p1=0.1 and p2=0.07666667. But in your ttest you specify p1=0.3 and p2=0.23. Which is correct? If p1=0.3 and p2=0.23, then use
prop.test(c(.30*300,.23*300),c(300,300))
Hope this is helpful,
Dan
Daniel J. Nordlund
Research and Data Analysis
Washington State Department of Social and Health Services
Olympia, WA 985045204
______________________________________________
[hidden email] mailing list
https://stat.ethz.ch/mailman/listinfo/rhelpPLEASE do read the posting guide http://www.Rproject.org/postingguide.htmland provide commented, minimal, selfcontained, reproducible code.


30 is not 30% of 300 (it is 10%), so your prop.test below is testing
something different from your hand calculations. Try:
> prop.test(c(.30,.23)*300,c(300,300), correct=FALSE)
2sample test for equality of proportions without continuity
correction
data: c(0.3, 0.23) * 300 out of c(300, 300)
Xsquared = 3.7736, df = 1, pvalue = 0.05207
alternative hypothesis: two.sided
95 percent confidence interval:
0.000404278 0.140404278
sample estimates:
prop 1 prop 2
0.30 0.23
> sqrt(3.7736)
[1] 1.942576
Notice that the square root of the Xsquared value matches your hand
calculations (with rounding error). This is true if Yates continuty
correction is not used (the correct=FALSE in the call to prop.test).
Hope this helps,

Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
[hidden email]
(801) 4088111
> Original Message
> From: [hidden email]
> [mailto: [hidden email]] On Behalf Of Nair,
> Murlidharan T
> Sent: Thursday, August 09, 2007 1:02 PM
> To: Nordlund, Dan (DSHS/RDA); [hidden email]
> Subject: Re: [R] small sample techniques
>
> n=300
> 30% taking A relief from pain
> 23% taking B relief from pain
> Question; If there is no difference are we likely to get a 7%
> difference?
>
> Hypothesis
> H0: p1p2=0
> H1: p1p2!=0 (not equal to)
>
> 1>Weighed average of two sample proportion
> 300(0.30)+300(0.23)
>  = 0.265
> 300+300
> 2>Std Error estimate of the difference between two independent
> 2>proportions
> sqrt((0.265 *0.735)*((1/300)+(1/300))) = 0.03603
>
> 3>Evaluation of the difference between sample proportion as a
> deviation
> 3>from the hypothesized difference of zero
> ((0.300.23)(0))/0.03603 = 1.94
>
>
> z did not approach 1.96 hence H0 is not rejected.
>
> This is what I was trying to do using prop.test.
>
> prop.test(c(30,23),c(300,300))
>
> What function should I use?
>
>
> Original Message
> From: [hidden email] on behalf of Nordlund,
> Dan (DSHS/RDA)
> Sent: Thu 8/9/2007 1:26 PM
> To: [hidden email]
> Subject: Re: [R] small sample techniques
>
> > Original Message
> > From: [hidden email]
> > [mailto: [hidden email]] On Behalf Of Nair,
> > Murlidharan T
> > Sent: Thursday, August 09, 2007 9:19 AM
> > To: Moshe Olshansky; Rolf Turner; [hidden email]
> > Subject: Re: [R] small sample techniques
> >
> > Thanks, that discussion was helpful. Well, I have another
> question I
> > am comparing two proportions for its deviation from the
> hypothesized
> > difference of zero. My manually calculated z ratio is 1.94.
> > But, when I calculate it using prop.test, it uses Pearson's
> > chisquared test and the Xsquared value that it gives it 0.74. Is
> > there a function in R where I can calculate the z ratio? Which is
> >
> >
> > ('p1'p2)(p1p2)
> > Z= 
> > S
> > ('p1'p2)
> >
> > Where S is the standard error estimate of the difference
> between two
> > independent proportions
> >
> > Dummy example
> > This is how I use it
> > prop.test(c(30,23),c(300,300))
> >
> >
> > Cheers../Murli
> >
> >
>
> Murli,
>
> I think you need to recheck you computations. You can run a
> ttest on your data in a variety of ways. Here is one:
>
> > x<c(rep(1,30),rep(0,270))
> > y<c(rep(1,23),rep(0,277))
> > t.test(x,y)
>
> Welch Two Sample ttest
>
> data: x and y
> t = 1.0062, df = 589.583, pvalue = 0.3147 alternative
> hypothesis: true difference in means is not equal to 0
> 95 percent confidence interval:
> 0.02221086 0.06887752
> sample estimates:
> mean of x mean of y
> 0.10000000 0.07666667
>
> Hope this is helpful,
>
> Dan
>
> Daniel J. Nordlund
> Research and Data Analysis
> Washington State Department of Social and Health Services
> Olympia, WA 985045204
>
> ______________________________________________
> [hidden email] mailing list
> https://stat.ethz.ch/mailman/listinfo/rhelp> PLEASE do read the posting guide
> http://www.Rproject.org/postingguide.html> and provide commented, minimal, selfcontained, reproducible code.
>
> ______________________________________________
> [hidden email] mailing list
> https://stat.ethz.ch/mailman/listinfo/rhelp> PLEASE do read the posting guide
> http://www.Rproject.org/postingguide.html> and provide commented, minimal, selfcontained, reproducible code.
>
______________________________________________
[hidden email] mailing list
https://stat.ethz.ch/mailman/listinfo/rhelpPLEASE do read the posting guide http://www.Rproject.org/postingguide.htmland provide commented, minimal, selfcontained, reproducible code.


Hi Murli,
First of all, regarding prop.test, you made a typo:
you should have used prop.test(c(69,90),c(300,300))
which gives you the squared value of 3.4228, and it's
square root is 1.85 which is not too far from 1.94.
I would use Fisher Exact Test (fisher.test). Two
sided test has a pvalue of 0.06411 so you do not
reject H0, One sided test (i.e. H1 is that the first
probability of success is smaller than the second) has
a pvalue of 0.03206, so you reject H0 (with 95%
confidence level).
You get similar results with twosided and onesided
ttest.
Moshe.
P.S. if you use paired ttest you get nonsense since
it uses pairwise differences, and in your case only 21
of 300 differences are nonzero!
 "Nair, Murlidharan T" < [hidden email]> wrote:
> n=300
> 30% taking A relief from pain
> 23% taking B relief from pain
> Question; If there is no difference are we likely to
> get a 7% difference?
>
> Hypothesis
> H0: p1p2=0
> H1: p1p2!=0 (not equal to)
>
> 1>Weighed average of two sample proportion
> 300(0.30)+300(0.23)
>  = 0.265
> 300+300
> 2>Std Error estimate of the difference between two
> independent proportions
> sqrt((0.265 *0.735)*((1/300)+(1/300))) =
> 0.03603
>
> 3>Evaluation of the difference between sample
> proportion as a deviation from the hypothesized
> difference of zero
> ((0.300.23)(0))/0.03603 = 1.94
>
>
> z did not approach 1.96 hence H0 is not rejected.
>
> This is what I was trying to do using prop.test.
>
> prop.test(c(30,23),c(300,300))
>
> What function should I use?
>
>
> Original Message
> From: [hidden email] on behalf of
> Nordlund, Dan (DSHS/RDA)
> Sent: Thu 8/9/2007 1:26 PM
> To: [hidden email]
> Subject: Re: [R] small sample techniques
>
> > Original Message
> > From: [hidden email]
> > [mailto: [hidden email]] On
> Behalf Of Nair,
> > Murlidharan T
> > Sent: Thursday, August 09, 2007 9:19 AM
> > To: Moshe Olshansky; Rolf Turner;
> [hidden email]
> > Subject: Re: [R] small sample techniques
> >
> > Thanks, that discussion was helpful. Well, I have
> another question
> > I am comparing two proportions for its deviation
> from the hypothesized
> > difference of zero. My manually calculated z ratio
> is 1.94.
> > But, when I calculate it using prop.test, it uses
> Pearson's
> > chisquared
> > test and the Xsquared value that it gives it
> 0.74. Is there
> > a function
> > in R where I can calculate the z ratio? Which is
> >
> >
> > ('p1'p2)(p1p2)
> > Z= 
> > S
> > ('p1'p2)
> >
> > Where S is the standard error estimate of the
> difference between two
> > independent proportions
> >
> > Dummy example
> > This is how I use it
> > prop.test(c(30,23),c(300,300))
> >
> >
> > Cheers../Murli
> >
> >
>
> Murli,
>
> I think you need to recheck you computations. You
> can run a ttest on your data in a variety of ways.
> Here is one:
>
> > x<c(rep(1,30),rep(0,270))
> > y<c(rep(1,23),rep(0,277))
> > t.test(x,y)
>
> Welch Two Sample ttest
>
> data: x and y
> t = 1.0062, df = 589.583, pvalue = 0.3147
> alternative hypothesis: true difference in means is
> not equal to 0
> 95 percent confidence interval:
> 0.02221086 0.06887752
> sample estimates:
> mean of x mean of y
> 0.10000000 0.07666667
>
> Hope this is helpful,
>
> Dan
>
> Daniel J. Nordlund
> Research and Data Analysis
> Washington State Department of Social and Health
> Services
> Olympia, WA 985045204
>
> ______________________________________________
> [hidden email] mailing list
> https://stat.ethz.ch/mailman/listinfo/rhelp> PLEASE do read the posting guide
> http://www.Rproject.org/postingguide.html> and provide commented, minimal, selfcontained,
> reproducible code.
>
> ______________________________________________
> [hidden email] mailing list
> https://stat.ethz.ch/mailman/listinfo/rhelp> PLEASE do read the posting guide
> http://www.Rproject.org/postingguide.html> and provide commented, minimal, selfcontained,
> reproducible code.
>
______________________________________________
[hidden email] mailing list
https://stat.ethz.ch/mailman/listinfo/rhelpPLEASE do read the posting guide http://www.Rproject.org/postingguide.htmland provide commented, minimal, selfcontained, reproducible code.


> Original Message
> From: [hidden email]
> [mailto: [hidden email]] On Behalf Of Nair,
> Murlidharan T
> Sent: Thursday, August 09, 2007 12:02 PM
> To: Nordlund, Dan (DSHS/RDA); [hidden email]
> Subject: Re: [R] small sample techniques
>
> n=300
> 30% taking A relief from pain
> 23% taking B relief from pain
> Question; If there is no difference are we likely to get a 7%
> difference?
>
> Hypothesis
> H0: p1p2=0
> H1: p1p2!=0 (not equal to)
>
> 1>Weighed average of two sample proportion
> 300(0.30)+300(0.23)
>  = 0.265
> 300+300
> 2>Std Error estimate of the difference between two
> independent proportions
> sqrt((0.265 *0.735)*((1/300)+(1/300))) = 0.03603
>
> 3>Evaluation of the difference between sample proportion as a
> deviation from the hypothesized difference of zero
> ((0.300.23)(0))/0.03603 = 1.94
>
>
> z did not approach 1.96 hence H0 is not rejected.
>
> This is what I was trying to do using prop.test.
>
> prop.test(c(30,23),c(300,300))
>
> What function should I use?
>
>
I sent this from work but it seems to have disappeared into the luminiferous ether.
The proportion test above indicates that p1=0.1 and p2=0.07666667. But in your ttest you specify p1=0.3 and p2=0.23. Which is correct? If p1=0.3 and p2=0.23, then use
prop.test(c(.30*300,.23*300),c(300,300))
Hope this is helpful,
Dan
Daniel J. Nordlund
Research and Data Analysis
Washington State Department of Social and Health Services
Olympia, WA 985045204
______________________________________________
[hidden email] mailing list
https://stat.ethz.ch/mailman/listinfo/rhelpPLEASE do read the posting guide http://www.Rproject.org/postingguide.htmland provide commented, minimal, selfcontained, reproducible code.

