pooled standard error

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pooled standard error
Message posted by Lisa Appeddu (via 164.58.150.32) on October 5, 2001 at 11:19 AM (ET)

I have three treatments with uneven experimental units. How do I calculate a pooled standard error of the mean?

READERS RESPOND:
(In chronological order. Most recent at the bottom.)

Re: pooled standard error
Message posted by Tomi (via 154.32.142.220) on October 5, 2001 at 4:27 PM (ET)

Let's refer to your three sets as x_i, y_i and z_i, with differing numbers of measurements being n_x, n_y and n_z.

Then you can calculate the mean values by: x_bar = sum {from i=1 to i=n_x} (x_i) etc.

For each set calculate the sum of the square deviations: ssd_x = sum ((x_i - x_bar)^2) etc.

The pooled standard deviation will be: psd= (ssd_x + ssd_y + ssd_z)/(n_x + n_y + n_z - 3).

The standard error of the mean can then be calculated by: t*psd/(square root (n_x + n_y + n_z)), where t is taken from t tables and depends on the confidence level.

Hope this helps.

Re: pooled standard error
Message posted by Jack Tomsky (via 208.249.113.130) on October 5, 2001 at 6:24 PM (ET)

In Tomi's notation, the standard error of the mean is sqrt(psd).

Re: pooled standard error
Message posted by Tomi (via 154.32.143.67) on October 6, 2001 at 2:32 AM (ET)

As Jack pointed out, there was an error in what I wrote. I should have said psd = sqrt( (ssd_x + ssd_y + ssd_z)/(n_x + n_y + n_z - 3)).

I still think that this is the pooled standard deviation, not the pooled standard error of the mean, and that the next step is still necessary. Am I wrong, Jack?

Re: pooled standard error
Message posted by Jack Tomsky (via 208.249.113.130) on October 8, 2001 at 2:03 PM (ET)

If xbar is an unbiased estimate of the population mean, then the standard error and standard deviation are equivalent. As a separate issue, the confidence interval on the population mean is

xbar +/- t*psd/sqrt(n_x + n_y + n_z)

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