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I am developing a new psychometric scale consisting of a several dichotomous items taken from a larger item bank. The p values (item means) are known. Now I need to calculate the expected mean and standard deviation of the scale after having picked the items. is there a formula, or well known procedure for it? I am looking for the best predicted mean and SD. I know there will allways be uncertainty about the true mean and SD, but for now that's not my main concern.

Many thanks in advance.

Edit. The scale is a lineair combination (just the sum) of the dichotomous items.

edit. This is an example of all the information I have of the picked items (=variables). Unfortenately I don't have the covariances.

   M_item   SD_item
0.8622312   0.344657159720439
0.3368913   0.472647386520131
0.5976337   0.490375020391853
0.6673509   0.471162048842211
0.8394329   0.367131184180246
0.7136281   0.452065299365468
0.7948207   0.403832582701681
0.7589461   0.427722944550313
0.8300466   0.375591855380864
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  • $\begingroup$ Is "digotomous" intended to be "dichotomous"? Exactly what is the formula for your scale? (Presumably it's a linear combination of responses, but it doesn't have to be linear.) $\endgroup$ – whuber Apr 13 '15 at 17:52
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If you are developing a scale, you should collect a new sample using just the questions in your new smaller scale. There are a lot of problems and considerations when estimating mean and variance (especially variance) of a linear combination of means and variances.

However, if you need to get an estimate, the variance of the sum of any 2 random variables is $$\text{Var}(X+Y) =\text{Var}(X)+\text{Var}(Y)+2\text{Cov}(X,Y)$$

You have more than 2 so the equation becomes $$\text{Var}\Bigl(\,\sum_{i=1}^n X_i\,\Bigr)= \sum_{i=1}^n\text{Var}( X_i)+ 2\sum_{i< j} \text{Cov}(X_i,X_j)$$

The mean of the sum of random variables is easier. Just sum the means. $$\mu_T=\sum_{i=1}^n \mu_i$$

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  • $\begingroup$ Thanks for your answer! This works for dichotomous items as well ofcourse. But I don't think I have the covariances of the items... let me think on this. Do you happen to know an r function that does exactly the same as your formula? $\endgroup$ – rdatasculptor Apr 13 '15 at 22:44
  • $\begingroup$ If all you have is the mean, you cannot calculate the variance of items or covariance between items. You need the original data. Alternatively, you said you had p values. I don't know what the p value means in this context (what is being compared?). However, if you have F or T statistics, mean squared error, and degrees of freedom, you can back out what the variances are. Again, though this depends on what the tests were that generated the p value. $\endgroup$ – le_andrew Apr 13 '15 at 22:49
  • $\begingroup$ Regarding dichotomous items, the variance is p(1-p), so you can actually get that from means (since the mean is p). $\endgroup$ – le_andrew Apr 13 '15 at 22:55
  • $\begingroup$ by p value I mean variable mean. So I only have means of dichotomous variables. Variance of the variables shouldn't be so hard to caluculate because there are only two possible values of each variable. Covariates are not possible to calculate with as less information as I have, I guess. $\endgroup$ – rdatasculptor Apr 13 '15 at 22:59
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    $\begingroup$ Yes, you cant get a covariance with just means. You can estimate a range though if you know n. I cant figure out the equation right now, but if you have n, you can figure out the maximum and minimum value that the covar can take, and after adding it to your other estimates you know you overall variance should be within that range. It'll probably be a rather large range however. Edit: Accidentally deleted previous comment, for future readers, here is the question with a shortcut for calculating covariance $\endgroup$ – le_andrew Apr 14 '15 at 19:14

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