Comments are embedded in Roy's post: On 13 Nov 2002, Roy Carambula wrote:
> I have conflicting web-based references for the following problem: The conflicting information, whatever it is, is not clear to me. > If a response variable y is normally distributed, The usual assumption it not that y be normally distributed (after all, it would not ordinarily be expected to be), but that e is, where e is the deviation of y from the value predicted by the model (for each observation) and the model (y = beta_0 + beta_1*x + e) is correct. > and so: > Beta(be) ~ N(Beta,sigma^2/Sxx) > > Beta(be)-Beta / [sigma/sq(Sxx)] ~ N(0,1) > > Where: > Beta = slope of least square fitted line > (be) = best estimate "best" is ambiguous; "least-squares estimate" is preferable. > Sxx = sum of squares of x > > With: > sigma^2=var(y) will be unknown rather, sigma^2 = var(e) > can estimate sigma^2 using sigma(be)^2=RSS/(n-2) If the model is correct. > Then > Beta(be)-Beta / se[Beta(be)] = Beta(be)-Beta / [sigma(be)/sq(Sxx)] > > Has "Students" t-distribution with n-2 df > > Where: > se = standard error OK so far. > For t-test: > x1(avg)-x2(avg)/se(x)[pooled] Relevance unclear. What do averages of x1 and x2 have to do with the slope beta (or its estimate b) of the regression line? And you have omitted the factor sqrt(2) in the denominator, unless by "se(x)" you meant the standard error of the difference in means, which is not what the equation appears to state. > (Q1) is [sigma(be)/sq(Sxx)] the standard error or the standard > deviation? A standard error IS a standard deviation. The answer to this question is indeterminate unless both terms are explicitly defined (or explicitly implied by context). The expression you cite [...] is the standard error of the estimate of the slope of the regressin line; it is also, and equivalently, the standard deviation of the sampling distribution of the estimate of the slope. > (Q2) if [sigma(be)/sq(Sxx)] is the standard deviation then the > standard error should be [sigma(be)/sq(Sxx)]/sqrt(1/n)? I believe you may be confusing the standard error of a slope with the standard error of a sample mean. (And you appear also to be using both "sq" and "sqrt" as abbreviations for "square root of".) The idea of 1/n (by which the variance sigma^2 would ordinarily be multiplied, not divided) appears already in [1/Sxx] since (if x were a random variable) Sxx = Var(x)*n approximately. > Any comment would be greatly appreciated. > Roy Carambula I hope this may have been helpful. -- DFB. ----------------------------------------------------------------------- Donald F. Burrill [EMAIL PROTECTED] 56 Sebbins Pond Drive, Bedford, NH 03110 (603) 626-0816 [was: 184 Nashua Road, Bedford, NH 03110 (603) 471-7128] . . ================================================================= Instructions for joining and leaving this list, remarks about the problem of INAPPROPRIATE MESSAGES, and archives are available at: . http://jse.stat.ncsu.edu/ . =================================================================
