prop.trend.test issue

I have the clinical study data.

                                           Year 0       Year 3
Retinol (nmol/L)        N       Mean +-sd       Mean +-sd
Vitamin A group         73      1.89+-0.36      2.06+-0.53
Trace group                57      1.83+-0.31     1.78+-0.30

where N is the number of male for the clinical study.

I want to test if the mean serum retinol has increased over 3 years
among subjects in the vitamin A group.

 1.89+0.36

[1] 2.25

1.89-0.36

[1] 1.53

2.06+0.53

[1] 2.59

2.06-0.53

[1] 1.53

prop.trend.test(c(2.25, 1.53),c( 2.59, 1.53))

       Chi-squared Test for Trend in Proportions

data:  c(2.25, 1.53) out of c(2.59, 1.53) ,
using scores: 1 2
X-squared = 0.2189, df = 1, p-value = 0.6399

The issue I am seeing that N of Vitamin A group = 73 seems not reflected.
This leads me to think that I can't implement the test based on the
data just presented.
Nor a two-tailed test is possible.

       2-sample test for equality of proportions with continuity correction

data:  c(2.25, 1.53) out of c(2.59, 1.53)
X-squared = 0, df = 1, p-value = 1
alternative hypothesis: two.sided
95 percent confidence interval:
-0.6738203  0.4112720
sample estimates:
  prop 1    prop 2
0.8687259 1.0000000

Warning message:
Chi-squared approximation may be incorrect in: prop.test(c(2.25,
1.53), c(2.59, 1.53))

Any ideas, please?

thx

ej

At 05:55 03/12/2006, Ethan Johnsons wrote:

I have the clinical study data.

                                           Year 0       Year 3
Retinol (nmol/L)        N       Mean +-sd       Mean +-sd
Vitamin A group         73      1.89+-0.36      2.06+-0.53
Trace group                57      1.83+-0.31     1.78+-0.30

where N is the number of male for the clinical study.

I want to test if the mean serum retinol has increased over 3 years
among subjects in the vitamin A group.

If you want to test means why did you think a test for proportions
was a good idea?

 1.89+0.36

[1] 2.25

1.89-0.36

[1] 1.53

2.06+0.53

[1] 2.59

2.06-0.53

[1] 1.53

prop.trend.test(c(2.25, 1.53),c( 2.59, 1.53))

       Chi-squared Test for Trend in Proportions

data:  c(2.25, 1.53) out of c(2.59, 1.53) ,
using scores: 1 2
X-squared = 0.2189, df = 1, p-value = 0.6399

The issue I am seeing that N of Vitamin A group = 73 seems not reflected.
This leads me to think that I can't implement the test based on the
data just presented.
Nor a two-tailed test is possible.

       2-sample test for equality of proportions with continuity correction

data:  c(2.25, 1.53) out of c(2.59, 1.53)
X-squared = 0, df = 1, p-value = 1
alternative hypothesis: two.sided
95 percent confidence interval:
-0.6738203  0.4112720
sample estimates:
  prop 1    prop 2
0.8687259 1.0000000

Warning message:
Chi-squared approximation may be incorrect in: prop.test(c(2.25,
1.53), c(2.59, 1.53))

Any ideas, please?

thx

ej

Michael Dewey
http://www.aghmed.fsnet.co.uk

I don't find any other test avail for this?
Am I missing something?

thx

ej

On 12/3/06, Michael Dewey <info at aghmed.fsnet.co.uk> wrote:

At 05:55 03/12/2006, Ethan Johnsons wrote:

I have the clinical study data.

                                           Year 0       Year 3
Retinol (nmol/L)        N       Mean +-sd       Mean +-sd
Vitamin A group         73      1.89+-0.36      2.06+-0.53
Trace group                57      1.83+-0.31     1.78+-0.30

where N is the number of male for the clinical study.

I want to test if the mean serum retinol has increased over 3 years
among subjects in the vitamin A group.

If you want to test means why did you think a test for proportions
was a good idea?

 1.89+0.36

[1] 2.25

1.89-0.36

[1] 1.53

2.06+0.53

[1] 2.59

2.06-0.53

[1] 1.53

prop.trend.test(c(2.25, 1.53),c( 2.59, 1.53))

       Chi-squared Test for Trend in Proportions

data:  c(2.25, 1.53) out of c(2.59, 1.53) ,
using scores: 1 2
X-squared = 0.2189, df = 1, p-value = 0.6399

The issue I am seeing that N of Vitamin A group = 73 seems not reflected.
This leads me to think that I can't implement the test based on the
data just presented.
Nor a two-tailed test is possible.

       2-sample test for equality of proportions with continuity 

correction

data:  c(2.25, 1.53) out of c(2.59, 1.53)
X-squared = 0, df = 1, p-value = 1
alternative hypothesis: two.sided
95 percent confidence interval:
-0.6738203  0.4112720
sample estimates:
  prop 1    prop 2
0.8687259 1.0000000

Warning message:
Chi-squared approximation may be incorrect in: prop.test(c(2.25,
1.53), c(2.59, 1.53))

Any ideas, please?

thx

ej

Michael Dewey
http://www.aghmed.fsnet.co.uk

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01/12/2006 13:12

I have the clinical study data.

                                           Year 0       Year 3
Retinol (nmol/L)        N       Mean +-sd       Mean +-sd
Vitamin A group         73      1.89+-0.36      2.06+-0.53
Trace group                57      1.83+-0.31     1.78+-0.30

where N is the number of male for the clinical study.

I want to test if the mean serum retinol has increased over 3 years
among subjects in the vitamin A group.

At 03:55 AM 12/3/2006, Ethan Johnsons wrote:

I have the clinical study data.

                                           Year 0       Year 3
Retinol (nmol/L)        N       Mean +-sd       Mean +-sd
Vitamin A group         73      1.89+-0.36      2.06+-0.53
Trace group                57      1.83+-0.31     1.78+-0.30

where N is the number of male for the clinical study.

I want to test if the mean serum retinol has increased over 3 years
among subjects in the vitamin A group.

If  You desire check mean serum retinol has increased over 3 years in
vitamin A group.
You may use t.test
Look this example:

#Generate random Data

set.seed(123)
VitA1<-rnorm(73,1.89,.36)
Trace1<-rnorm(57,1.83,0.31)
VitA2<-rnorm(73,2.06,.53)
Trace2<-rnorm(57,1.78,0.30)

# Calculate diference Year 3 - Year 0

dVitA<-VitA2-VitA1
dTrace<-Trace2-Trace1

# Testing diference
t.test(dVitA,dTrace)


         Welch Two Sample t-test

data:  dVitA and dTrace
t = 2.2762, df = 117.746, p-value = 0.02464
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
  0.02756874 0.39659494
sample estimates:
   mean of x   mean of y
  0.15905162 -0.05303022





Bernardo Rangel Tura, MD, Phd
National Institute of Cardiology Laranjeiras
Rio de Janeiro Brazil