estimated standard error

**emmalou264** · Jun 10th 2009, 01:26 PM

SAMPLE 1 SAMPLE 2

Sample size 56 Sample size 55
mean 3.2 mean 2.8
s. deviation 1.8 s deviation 1.5

How do i find the Estimated Standard Error of the difference between the 2 sample mean numbers.

i know that ESE is S / Square root of N

But dont know how to progress

**pickslides** · Jun 10th 2009, 03:28 PM

$SE_1 = \frac{s_1}{\sqrt{n_1}}$

$SE_1 = \frac{1.8}{\sqrt{56}}=\cdots$

$SE_2 = \frac{s_2}{\sqrt{n_2}}$

$SE_2 = \frac{1.5}{\sqrt{55}}=\cdots$

Now check out this link Confidence Interval

**matheagle** · Jun 10th 2009, 04:03 PM

The st. error is the estimate of the st. deviation.
NOW there are two cases here.
If you assume that the population variances are equal, you pool the sample variances.

The st. deviation of $\bar X_1-\bar X_2$ is the square root of the variance $\sqrt{ {\sigma^2_1\over n_1} + {\sigma^2_2\over n_2} }$ .

Since we usually do not know $\sigma^2_1$ or $\sigma^2_2$ we use the sample variances

and that's what we call the st. error, $\sqrt{ {s^2_1\over n_1} + {s^2_2\over n_2} }$ .

BUT in may cases we assume that $\sigma^2_1=\sigma^2_2$ and in that case we pool the two estimators,

here the st. error is $s_p\sqrt{{1\over n_1} + {1\over n_2} }$ which is the same as $\sqrt{ {s^2_p\over n_1} + {s^2_p\over n_2} }$ .

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