How To Find Standard Error Of Two Samples
randomly drawn from the same normally distributed source population, belongs to a normally distributed sampling distribution whose overall mean is equal to zero and whose standard deviation ("standard error") is equal to square.root[(sd2/na) + (sd2/nb)] where sd2 = the variance of the source population (i.e., the square of the standard deviation); na = the size of sample A; and nb = the size of sample B. To calculate the standard error of any particular sampling distribution of sample-mean differences, enter the mean and standard deviation (sd) of the source population, along with the values of na andnb, and then click the "Calculate" button. -1sd mean +1sd <== sourcepopulation <== samplingdistribution standard error of sample-mean differences = ± sd of source population sd = ± size of sample A = size of sample B = Home Click this link only if you did not arrive here via the VassarStats main page. ©Richard Lowry 2001- All rights reserved.
say 10, years ago? Suppose a random sample of 100 student records from 10 years ago yields a sample average GPA of 2.90 with a standard deviation of .40. A random sample of 100 current students today yields a sample average of 2.98 with a standard deviation of .45. The difference between the two sample means is 2.98-2.90 = .08. Is this proof that GPA's are higher today than 10 years ago? Well....first we need to account for the fact that 2.98 and 2.90 are not the true averages, but are computed from random samples. Therefore, .08 is not the true difference, but simply an estimate of the true difference. Can this estimate miss by much? Fortunately, http://vassarstats.net/dist2.html statistics has a way of measuring the expected size of the ``miss'' (or error of estimation) . For our example, it is .06 (we show how to calculate this later). Therefore, we can state the bottom line of the study as follows: "The average GPA of WMU students today is .08 higher than 10 years ago, give or take .06 or so." We now show how to calculate the .06, the standard error of the estimate. But http://www.stat.wmich.edu/s216/book/node81.html first, a note on terminology. The estimate .08=2.98-2.90 is a difference between averages (or means) of two independent random samples. "Independent" refers to the sampling luck-of-the-draw: the luck of the second sample is unaffected by the first sample. In other words, there were two independent chances to have gotten lucky or unlucky with the sampling. The likely size of the error of estimation in the .08 is called the standard error of the difference between independent means. We calculate it using the following formula: (7.4) where and . Note that and are the SE's of and , respectively. The formula looks easier without the notation and the subscripts. 2.98 is a sample mean, and has standard error (since SE= ). Similarly, 2.90 is a sample mean and has standard error . Summarizing, we write the two mean estimates (and their SE's in parentheses) as 2.98 (SE=.045) 2.90 (SE=.040) If two independent estimates are subtracted, the formula (7.6) shows how to compute the SE of the difference : 2.98 - 2.90 (SE= ) or .08 .06. Remember the Pythagorean Theorem in geometry? Think of the two SE's as the length of the two sides of the triangle (call them a and b). The SE of the difference then equals the length of the hypotenuse (SE of difference = ). We are now ready to state a confidence interval for the differenc
Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site About Us Learn more about Stack Overflow the company Business http://stats.stackexchange.com/questions/55999/is-it-possible-to-find-the-combined-standard-deviation Learn more about hiring developers or posting ads with us Cross Validated Questions Tags Users Badges Unanswered Ask Question _ Cross Validated is a question and answer site for people interested in statistics, machine learning, data analysis, data mining, and data visualization. Join them; it only takes a minute: Sign up Here's how it works: Anybody can ask a question Anybody can answer The best answers are voted up and rise to the top Is it how to possible to find the combined standard deviation? up vote 12 down vote favorite 7 Suppose I have 2 sets: Set A: number of items $n= 10$, $\mu = 2.4$ , $\sigma = 0.8$ Set B: number of items $n= 5$, $\mu = 2$, $\sigma = 1.2$ I can find the combined mean ($\mu$) easily, but how am I supposed to find the combined standard deviation? standard-deviation share|improve this question edited Apr 13 '13 at 18:23 gpoo how to find 1951311 asked Apr 13 '13 at 9:04 kype 70115 add a comment| 2 Answers 2 active oldest votes up vote 15 down vote accepted So, if you just want to have two of these samples brought together into one you have: $s_1 = \sqrt{\frac{1}{n_1}\Sigma_{i = 1}^{n_1} (x_i - \bar{y}_1)^2}$ $s_2 = \sqrt{\frac{1}{n_2}\Sigma_{i = 1}^{n_2} (y_i - \bar{y}_2)^2}$ where $\bar{y}_1$ and $\bar{y}_2$ are sample means and $s_1$ and $s_2$ are sample standard deviations. To add them up you have: $s = \sqrt{\frac{1}{n_1 + n_2}\Sigma_{i = 1}^{n_1 + n_2} (z_i - \bar{y})^2}$ which is not that straightforward since the new mean $\bar{y}$ is different from $\bar{y}_1$ and $\bar{y}_2$: $\bar{y} = \frac{1}{n_1 + n_2}\Sigma_{i = 1}^{n_1 + n_2} z_i = \frac{n_1 \bar{y}_1 + n_2 \bar{y}_2}{n_1 + n_2}$ The final formula is: $s = \sqrt{\frac{n_1 s_1^2 + n_2 s_2^2+ n_1(\bar{y}_1-\bar{y})^2 +n_2(\bar{y}_2-\bar{y})^2}{n_1 + n_2 }}$ For the commonly-used Bessel-corrected ("$n-1$-denominator") version of standard deviation, the results for the means are as before, but $s = \sqrt{\frac{(n_1 - 1)s_1^2 + (n_2 - 1)s_2^2 + n_1(\bar{y}_1-\bar{y})^2 +n_2(\bar{y}_2-\bar{y})^2}{n_1+n_2 - 1} }$ You can read more info here: http://en.wikipedia.org/wiki/Standard_deviation share|improve this answer edited Apr 27 at 11:13 Glen_b♦ 150k19246514 answered Apr 13 '13 at 9:07 sashkello 1,35111124 1 If the OP is using the Bessel-corrected ($n-1$-denominator for the variance) version of sample standard deviation (as almost everyone who asks here will be doing), this answ