Re: unequal n's: quadratic weights
On Tue, 30 Jan 2001, Kathleen Bloom wrote: If you have unequal n's, and want to determine linear parameters, you can develop new coefficients by taking the normal unweighted coefficients (e.g., -1, 0, +1, for three group design) and the formula: n1(X1) + n2(X2) + n3(X3)/ n1+n2+n3 where the X's are 1, 2, and 3 because you have 3 groups. This gives you a new mean of the Xs... (i.e., no longer 1+2+3/3 = 2), and from there you calculate the new coefficients (e.g., 1 - ?, 2 - ?, 3 - ?, gives you the new linear coefficients) for the 3-group design with unequal n's. You get the same results if you use X's corresponding to the unweighted coefficients (-1, 0, +1). I should suppose that for quadratic estimates you'd play the same game with the quadratic unweighted coefficients (+1, -2, +1). However, I've never played around much with weighted trend analyses, so my supposition may possibly be incorrect. It may be better to retain the grand mean calculated from (-1,0,+1), equal to your "?" minus 2 (let's call that ""), and generate coefficients from the unweighted quadratic coefficients as 1-, -2-, 1-. I note in passing that your decision to pursue weighting-by-sample-size implies that you have decided to assign equal weight to individual cases and NOT to assign equal weight to each subgroup. (Had you chosen equal weight for each subgroup, you'd use the "unweighted" (they're not really UNweighted, they're _equally_ weighted) coefficients directly.) I've not encountered situations where it seemed necessary to give equal importance to each individual case, enough to make it worth the extra effort to weight the coefficients -- and think about what it means, that the "grand mean" for the data depends on which trend you're currently pursuing, and that the several trends (linear, quadratic, cubic, ...) are explicitly NOT orthogonal. From there you can do things like determine the the weighted linear estimated parameters. They are given in the spss oneway printout... as I understand it... i.e., the weighted (for sample size) beta for the linear contrast. Notice that all this does is change the distance between each group mean and the grand mean; it does not change the relative distances between groups, which are still equally spaced. It has never been very clear to me what advantage one gets from the weighted parameters, especially as those estimates depend on the accident of how many observations you were able to find for each group. For this reason (among others) I am inclined to favor equal weighting in general. If it turns out that the choice of weighting influences the conclusion(s) to be drawn, one has compelling arguments for repeating the experiment, this time with a proper (equal-numbers-of-cases) design and carefully random selection of cases. snip ... My means are 2.05, 6.38, and 12.08 for the three groups respectively. In other words.. what does one calculate and how? You might reasonably try using the regression module (rather than one-way anova) to compare output: predict Y from X1 (X1 = 1,2,3 for the three groups, and X1 = -1,0,+1 if you want to confirm that the results are the same for this coding); and for an alternate (quadratic) model, predict Y from X1 and X2 (X1 = -1,0,+1; X2 = +1,-2,+1). You have not, by the way, said what you're doing this analysis FOR, so it's a bit difficult to know whether one is offering useful advice. Or not. -- Don. -- Donald F. Burrill[EMAIL PROTECTED] 348 Hyde Hall, Plymouth State College, [EMAIL PROTECTED] MSC #29, Plymouth, NH 03264 (603) 535-2597 Department of Mathematics, Boston University[EMAIL PROTECTED] 111 Cummington Street, room 261, Boston, MA 02215 (617) 353-5288 184 Nashua Road, Bedford, NH 03110 (603) 471-7128 = Instructions for joining and leaving this list and remarks about the problem of INAPPROPRIATE MESSAGES are available at http://jse.stat.ncsu.edu/ =
Re: unequal n's: quadratic weights
Serve me right for not checking my work. Here's a correction: On Wed, 31 Jan 2001 12:37:13 GMT, [EMAIL PROTECTED] (Duncan Murdoch) wrote: To find it, do this. Suppose the quadratic curve is A x^2 + B x + C. Then you've got three equations: A (-1)^2 + B (-1) + C = 2.05 A (0)^2 + B (0) + C = 6.38 A (1)^2 + B (1) + C = 12.08 There's a unique solution to these equations; it is C = 6.38, A = (12.08+2.05)/2-C, B = (12.08-2.05)/2-C.Those are the least-squares parameter estimates. Duncan Murdoch = Instructions for joining and leaving this list and remarks about the problem of INAPPROPRIATE MESSAGES are available at http://jse.stat.ncsu.edu/ =
Re: unequal n's: quadratic weights
On Tue, 30 Jan 2001 23:22:51 -0500, "K. Bloom" [EMAIL PROTECTED] wrote: If your answer IS the correct answer to my question, perhaps you would kindly explain it to me in a more concrete way suitable to my simple knowledge of the problem. My means are 2.05, 6.38, and 12.08 for the three groups respectively. In other words.. what does one calculate and how? I'm still not sure I understand your question completely, but what I think you're asking is this: You have 3 groups of observations. You'd graph them with x-coordinates equal to -1, 0 and +1. The y-coordinates would be the observed data. The means for those three groups are 2.05, 6.38, and 12.08, based on different numbers of observations in each group. You want the quadratic curve that provides the least-squares fit to your data. If that's the case, then the numbers of observations in each group doesn't matter. There's a quadratic curve that goes exactly through each group mean, and you can't find a better fit than that. To find it, do this. Suppose the quadratic curve is A x^2 + B x + C. Then you've got three equations: A (-1)^2 + B (-1) + C = 2.05 A (0)^2 + B (0) + C = 6.38 A (1)^2 + B (1) + C = 12.08 There's a unique solution to these equations; it is C = 6.38, A = (12.08+2.05)/2, B = (12.08-2.05)/2.Those are the least-squares parameter estimates. If you have more than three groups, then you won't be able to find an exact solution like this (you've only got three parameters to play with), and then the least-squares solution *does* depend on the group sizes. In general, you should solve linear least-squares problems using "multiple linear regression"; there are a ton of texts on that and I'd suggest you use one of those. In particular, with unequal group sizes you probably want to use "weighted least squares", with the weights equal to the group sizes. Duncan Murdoch = Instructions for joining and leaving this list and remarks about the problem of INAPPROPRIATE MESSAGES are available at http://jse.stat.ncsu.edu/ =
unequal n's: quadratic weights
Could someone please tell me how to calculate quadratic coefficients for unequal sample sizes (equal intervals, three groups)? The formula is not in the SPSS algorithm notes. Thank you, Kathleen Bloom = Instructions for joining and leaving this list and remarks about the problem of INAPPROPRIATE MESSAGES are available at http://jse.stat.ncsu.edu/ =
Re: unequal n's: quadratic weights
On Tue, 30 Jan 2001 19:53:03 -0500, "K. Bloom" [EMAIL PROTECTED] wrote: Could someone please tell me how to calculate quadratic coefficients for unequal sample sizes (equal intervals, three groups)? The formula is not in the SPSS algorithm notes. Maybe I'm misunderstanding the question, but if you only have 3 groups, your estimates will interpolate the group means. Just solve the 3 linear equations in 3 unknowns, using whatever equation solving method you like. Duncan Murdoch = Instructions for joining and leaving this list and remarks about the problem of INAPPROPRIATE MESSAGES are available at http://jse.stat.ncsu.edu/ =
Re: unequal n's: quadratic weights
I don't understand your reply... sorry. I'll try again. This may be a self-evident solution for you, but I'll need a simple and concrete example. Here is where I am right now: If you have unequal n's, and want to determine linear parameters, you can develop new coefficients by taking the normal unweighted coefficients (e.g., -1, 0, +1, for three group design) and the formula: n1(X1) + n2(X2) + n3(X3)/ n1+n2+n3 where the X's are 1, 2, and 3 because you have 3 groups. This gives you a new mean of the Xs... (i.e., no longer 1+2+3/3 = 2), and from there you calculate the new coefficients (e.g., 1 - ?, 2 - ?, 3 - ?, gives you the new linear coefficients) for the 3-group design with unequal n's. From there you can do things like determine the the weighted linear estimated parameters. They are given in the spss oneway printout... as I understand it... i.e., the weighted (for sample size) beta for the linear contrast. I calculated and checked this value with a data set in which the n's were 247, 29, and 37, for X1, X2, and X3 respectively. It worked. Now I want to calculate the weighted quadratic beta for this data set. To do this I need a formula.. for weighting... as above... but that formula above does not work for the quadratic parameter because it is for a linear not a quadratic trend. One text referred to a 1965 article by Gaito...for determining the coefficients, but I don't have access to that paper at this moment. So I asked, does anyone know this formula? If your answer IS the correct answer to my question, perhaps you would kindly explain it to me in a more concrete way suitable to my simple knowledge of the problem. My means are 2.05, 6.38, and 12.08 for the three groups respectively. In other words.. what does one calculate and how? Many thanks. Duncan Murdoch [EMAIL PROTECTED] wrote in message [EMAIL PROTECTED]">news:[EMAIL PROTECTED]... On Tue, 30 Jan 2001 19:53:03 -0500, "K. Bloom" [EMAIL PROTECTED] wrote: Could someone please tell me how to calculate quadratic coefficients for unequal sample sizes (equal intervals, three groups)? The formula is not in the SPSS algorithm notes. Maybe I'm misunderstanding the question, but if you only have 3 groups, your estimates will interpolate the group means. Just solve the 3 linear equations in 3 unknowns, using whatever equation solving method you like. Duncan Murdoch = Instructions for joining and leaving this list and remarks about the problem of INAPPROPRIATE MESSAGES are available at http://jse.stat.ncsu.edu/ =
