calculating sample sizes for independent groups
 

Can someone give me the spreadsheet formulas for calculating a sample size
for independent samples?

Thanks

Ariel

calculating sample sizes for independent groups
 

I'm sure someone can ... but you need to learn to state the requirements
clearly first.
__________________________________________________ ______________________

"Ariel" wrote in message
...
Can someone give me the spreadsheet formulas for calculating a sample size
for independent samples?

Thanks

Ariel

calculating sample sizes for independent groups
 

I had to look this up, its ages since did this.

It is based on the standard error of the mean. Say a sample of 100 items has
a mean of 50 and a standard deviation of 20. The interval estimate with 95%
confidence for the population mean is:

=1.96*std_Dev/SQRT(sample_size) which gives four.
The table below shows the figures and formulas. (I created Range names for
the numbers to make it more clear. Select labels and figures and choose
Insert, Names, Create - make sure the left labels checkbox is ticked and
click OK

sample size 100 100
mean 50 50
std Dev 20 20

Error of mean 3.92 =1.96*std_Dev/SQRT(sample_size)
Mean + Err 53.92 =mean+b5
Mean - Err 46.08 =mean-b5
new N 384.16 =(std_Dev*1.96/2)^2

so to reduce the error of the mean by half we increase the sample size by:
=(std_Dev*1.96/2)^2

of course you have to get you mean and stdev first or make an estimate of
them first.

Hope this helps
Peter

"Ariel" wrote:

Can someone give me the spreadsheet formulas for calculating a sample size
for independent samples?

Thanks

Ariel

calculating sample sizes for independent groups
 

Hi Billy,

Thank you for your effort, but this calculation is not for two independent
groups.

Assume that the mean and SD of the first group is 0.50 and 0.224,
respectively. We estimate that the second group will have a mean value of
0.045 (difference between the two groups is 10%). I can even throw in that we
estimate that the second group's SD will be 0.212.

What will the sample size needed to acheive this in the second group,
assuming a power of 80% and an alpha of 0.05.?

An online calculator gave me a sample size of 29,864 if the two SDs are
known (using the z-test) and a sample size of 31,506 when only the first SD
is known and the second estimated.

Thanks for your help!

"Billy Liddel" wrote:

I had to look this up, its ages since did this.

It is based on the standard error of the mean. Say a sample of 100 items has
a mean of 50 and a standard deviation of 20. The interval estimate with 95%
confidence for the population mean is:

=1.96*std_Dev/SQRT(sample_size) which gives four.
The table below shows the figures and formulas. (I created Range names for
the numbers to make it more clear. Select labels and figures and choose
Insert, Names, Create - make sure the left labels checkbox is ticked and
click OK

sample size 100 100
mean 50 50
std Dev 20 20

Error of mean 3.92 =1.96*std_Dev/SQRT(sample_size)
Mean + Err 53.92 =mean+b5
Mean - Err 46.08 =mean-b5
new N 384.16 =(std_Dev*1.96/2)^2

so to reduce the error of the mean by half we increase the sample size by:
=(std_Dev*1.96/2)^2

of course you have to get you mean and stdev first or make an estimate of
them first.

Hope this helps
Peter

"Ariel" wrote:

Can someone give me the spreadsheet formulas for calculating a sample size
for independent samples?

Thanks

Ariel

calculating sample sizes for independent groups
 

Hi Ariel

I think you lost me somewhere there!

"Ariel" wrote:

Hi Billy,

Thank you for your effort, but this calculation is not for two independent
groups.

Assume that the mean and SD of the first group is 0.50 and 0.224,
respectively. We estimate that the second group will have a mean value of
0.045 (difference between the two groups is 10%). I can even throw in that we
estimate that the second group's SD will be 0.212.

What will the sample size needed to acheive this in the second group,
assuming a power of 80% and an alpha of 0.05.?

An online calculator gave me a sample size of 29,864 if the two SDs are
known (using the z-test) and a sample size of 31,506 when only the first SD
is known and the second estimated.

Thanks for your help!

"Billy Liddel" wrote:

I had to look this up, its ages since did this.

It is based on the standard error of the mean. Say a sample of 100 items has
a mean of 50 and a standard deviation of 20. The interval estimate with 95%
confidence for the population mean is:

=1.96*std_Dev/SQRT(sample_size) which gives four.
The table below shows the figures and formulas. (I created Range names for
the numbers to make it more clear. Select labels and figures and choose
Insert, Names, Create - make sure the left labels checkbox is ticked and
click OK

sample size 100 100
mean 50 50
std Dev 20 20

Error of mean 3.92 =1.96*std_Dev/SQRT(sample_size)
Mean + Err 53.92 =mean+b5
Mean - Err 46.08 =mean-b5
new N 384.16 =(std_Dev*1.96/2)^2

so to reduce the error of the mean by half we increase the sample size by:
=(std_Dev*1.96/2)^2

of course you have to get you mean and stdev first or make an estimate of
them first.

Hope this helps
Peter

"Ariel" wrote:

Can someone give me the spreadsheet formulas for calculating a sample size
for independent samples?

Thanks

Ariel

calculating sample sizes for independent groups
 

Does that mean you can't help me, or that I need to clarify it further?

"Billy Liddel" wrote:

Hi Ariel

I think you lost me somewhere there!

"Ariel" wrote:

Hi Billy,

Thank you for your effort, but this calculation is not for two independent
groups.

Assume that the mean and SD of the first group is 0.50 and 0.224,
respectively. We estimate that the second group will have a mean value of
0.045 (difference between the two groups is 10%). I can even throw in that we
estimate that the second group's SD will be 0.212.

What will the sample size needed to acheive this in the second group,
assuming a power of 80% and an alpha of 0.05.?

An online calculator gave me a sample size of 29,864 if the two SDs are
known (using the z-test) and a sample size of 31,506 when only the first SD
is known and the second estimated.

Thanks for your help!

"Billy Liddel" wrote:

I had to look this up, its ages since did this.

It is based on the standard error of the mean. Say a sample of 100 items has
a mean of 50 and a standard deviation of 20. The interval estimate with 95%
confidence for the population mean is:

=1.96*std_Dev/SQRT(sample_size) which gives four.
The table below shows the figures and formulas. (I created Range names for
the numbers to make it more clear. Select labels and figures and choose
Insert, Names, Create - make sure the left labels checkbox is ticked and
click OK

sample size 100 100
mean 50 50
std Dev 20 20

Error of mean 3.92 =1.96*std_Dev/SQRT(sample_size)
Mean + Err 53.92 =mean+b5
Mean - Err 46.08 =mean-b5
new N 384.16 =(std_Dev*1.96/2)^2

so to reduce the error of the mean by half we increase the sample size by:
=(std_Dev*1.96/2)^2

of course you have to get you mean and stdev first or make an estimate of
them first.

Hope this helps
Peter

"Ariel" wrote:

Can someone give me the spreadsheet formulas for calculating a sample size
for independent samples?

Thanks

Ariel

calculating sample sizes for independent groups
 

Hi
Sorry, I can't help you

Peter

"Ariel" wrote:

Does that mean you can't help me, or that I need to clarify it further?

"Billy Liddel" wrote:

Hi Ariel

I think you lost me somewhere there!

"Ariel" wrote:

Hi Billy,

Thank you for your effort, but this calculation is not for two independent
groups.

Assume that the mean and SD of the first group is 0.50 and 0.224,
respectively. We estimate that the second group will have a mean value of
0.045 (difference between the two groups is 10%). I can even throw in that we
estimate that the second group's SD will be 0.212.

What will the sample size needed to acheive this in the second group,
assuming a power of 80% and an alpha of 0.05.?

An online calculator gave me a sample size of 29,864 if the two SDs are
known (using the z-test) and a sample size of 31,506 when only the first SD
is known and the second estimated.

Thanks for your help!

"Billy Liddel" wrote:

I had to look this up, its ages since did this.

It is based on the standard error of the mean. Say a sample of 100 items has
a mean of 50 and a standard deviation of 20. The interval estimate with 95%
confidence for the population mean is:

=1.96*std_Dev/SQRT(sample_size) which gives four.
The table below shows the figures and formulas. (I created Range names for
the numbers to make it more clear. Select labels and figures and choose
Insert, Names, Create - make sure the left labels checkbox is ticked and
click OK

sample size 100 100
mean 50 50
std Dev 20 20

Error of mean 3.92 =1.96*std_Dev/SQRT(sample_size)
Mean + Err 53.92 =mean+b5
Mean - Err 46.08 =mean-b5
new N 384.16 =(std_Dev*1.96/2)^2

so to reduce the error of the mean by half we increase the sample size by:
=(std_Dev*1.96/2)^2

of course you have to get you mean and stdev first or make an estimate of
them first.

Hope this helps
Peter

"Ariel" wrote:

Can someone give me the spreadsheet formulas for calculating a sample size
for independent samples?

Thanks

Ariel

calculating sample sizes for independent groups
 

Fri, 29 Jun 2007 08:38:01 -0700 from Ariel
:
Assume that the mean and SD of the first group is 0.50 and 0.224,
respectively. We estimate that the second group will have a mean value of
0.045 (difference between the two groups is 10%). I can even throw in that we
estimate that the second group's SD will be 0.212.

What will the sample size needed to acheive this in the second group,
assuming a power of 80% and an alpha of 0.05.?

Ask in a statistics group. The formula is pretty trivial, once you
know what the calculation should be.

--
Stan Brown, Oak Road Systems, Tompkins County, New York, USA
http://OakRoadSystems.com/