search for: mu_1

Hello, For a simple problem with 1 predictor (x) and 2 classes (0 and 1), the linear discriminant function should be something like 2(mu_0 - mu_1)/var x + x-independent-terms where var is the common variance. Question 1: Why does lda() report only a single "Coefficients of linea...

Hello, Is there a built-in test in R for hypothesis testing with samples of known variance? For example, I've got a set of data, a mean to compare against, and a known variance, and I want to determine the p-value for which I can reject the null hypothesis (mu_1 = mu_0) and accept the alternative (mu_1 > mu_0). I've found that JMP and Minitab can both do this (in JMP, it's a one-sided confidence interval in the Distribution platform), but I haven't figured out how to do such a test in R. Thanks, Tom

...gative binomially distributed data, I need to estimate the parameter theta. I have been doing this by using a negative binomial glm of the same model (except that all the effects are fixed), and estimating mu as the fitted model like so: model_1 <-glm.nb(y~x1+x2+x3, data = datafilename) mu_1 <- fitted(model_1) theta_1 <- theta.ml(y, mu_1, length(data), limit = 10, eps = .Machine$double.eps^0.25, trace = FALSE) Then, I conduct the lmer, using the estimated theta: model_11 <-lmer(y~x1+x2+(1|x3), family = negative.binomial(theta = theta_1, link = “log”), method = “...

...ints out as k rows (one for each Xi) of 8 numbers M: N x (3+k) matrix of cases. The first 3 cols are values of (F1,F2,F3) as above (and the remainder are values of (X1,...,Xk) but this doesn't really matter). AIM: For each row of M, find from mu the set of k means mu_1,...,mu_k corresponding to the values of F1, F2, F3 in that row. E.g. if M[1,1:3] -> 2,1,1 I want mu[,2,1,1] BUT: Not to have to do it by mu[,M[1,1],M[1,2],M[1,3]] but more like mu[,M[1,1:3]] -- except that this kind of try does not work, leading to error: "incorrect number of d...

Hi you guys: I have been wondering if there is any way to change the labeling in plot.ts( ), for example , if I plot two sequences, i always got y labels as "series1", "series2", I tried to use ylab=c((expression(mu_1)),(expression(mu_2))), but it doesn't work. thanks in advance for any help best.

Dear R People: I have a few questions about multiple comparisons and contrasts for ANOVA, please. I've tried some things but with no success. Suppose I have a completely randomized design, and I want to have the contrast \mu_1 - 0.5 \mu_2 - 0.5 \mu_2 How do I set that up, please? I used the C command, and ran aov, but the results were identical to those with no contrasts. Also, is there a command for Dunne'ts multiple range test? How about Scheffe's simultaneous intervals? Thank you so much for your help, in...

...[[i]]]; n = length(x); y = log(x[-1]/x[-n]);#log price difference y2 = y^2 m = length(y); z999 = (qnorm(1 - .001)) if (m>=260) { ##raw moments rtn = sum(y) * 100 rv = sum(y*y);#realized variance [Y] bv = (m/(m-1))*(pi/2)*(sum(abs(y[-1]*y[-m])));#bi-power vairiation #mu_1^{-2}*{Y}^[1,1] rj = (rv-bv)/rv tmp = apply(embed(y, 4), 1, prod);#y{j}*y{j-1}*y{j-2}*y{j-3} tp1 = apply(embed(y, 3), 1, prod)#y{j}*y{j-1}*y{j-2} tpq = (mu43^-3)*((m^2)/(m-3))*(sum((abs(tp1))^(4/3))) qp = ((pi*pi)/4)*((m^2)/(m-3))*(sum(abs(tmp)));#quad-power #mu_1^{-4}*{Y}^[1,...

Hello, I have some parameters from Mclust function. The parameters are in the form *parametersDf * * mu_1 mu_2 var_mc1 var_mc2 c1 c2 * *2 1.357283 2.962736 0.466154 0.1320129 0.5258975 0.4741025 * *21 8.357283 9.962736 0.466154 0.1320129 0.5258975 0.4741025 * Each row in the above data frame represent the segment a...

...nging the desciple to true enlightenment. Point 3 --- what hypothesis is being tested by SSA? Let factor A correspond to index i, and B to index j. Let the cell means be mu_ij. (In the overparameterized notation, mu_ij = mu + alpha_i + beta_j + gamma_ij.) The hypothesis being tested is H_0: mu_1.-bar = mu_2.-bar = ... = mu_a.-bar where factor A has a levels, and ``mu_i.-bar'' means the average (arithmetic mean) of mu_i1, mu_i2, ..., mu_ib. (Note --- factor B has b levels.) I.e. the hypothesis is that there is no difference, on average, between the levels of A, the average being t...

...MC_iter=0;MCMC_iter<iter_2;MCMC_iter++) { for(MCMC_iter2=0;MCMC_iter2<thin;MCMC_iter2++) { indice_adapt += 1; /************ SIMULAZIONE MU **********/ F77_NAME(dcopy)(&incJ, mu_acc_P, &incOne, mu_app_P, &incOne); // mu_1 app1_2 = 1/((1+pow(phi_acc_P[0],2))/zeta2_acc_P[0]+Sum_Sigma_inv); app1_1 = phi_acc_P[0]*mu_app_P[1]/zeta2_acc_P[0]; for(t=0; t<n_j;t++) { app1_1 += Sigma_inv_one[t]*(x_P[t]+2*M_PI*k_acc_P[t]-alpha_acc_P[0]); } ap...

Hi R users! I am trying to learn some multilevel analysis, but unfortunately i am now very confused. The reason: http://www.ats.ucla.edu/stat/hlm/seminars/hlm_mlm/mlm_hlm_seminar.htm http://www.ats.ucla.edu/stat/sas/seminars/sas_mlm/mlm_sas_seminar.htm and MlmSoftRev. pdf from mlmRev package. >From what i see, the first two links seem to declare the level one variable as a random part (i

Here's a fairly minimal-case example in which the saved EPS does not match the screen. The error comes when using bquote(.(i)) instead of bquote(1), as demonstrated by the two minimally different cases below. Very strange. Any clues as to why? #---------------- begin ------------------- # Version A. X axis labels have subscripts as constants. EPS is correct. windows() layout( matrix( 1:2 ,