COMPGROUPS.NET | Search | Post Question | Groups | Stream | About | Register

### ANOVA and F-statistic

• Email
• Follow

```Hi there,

When using the anova1() and then looking at the F-statistic, I don't get the same value when determining it separately in workspace. I'm using this code in the test

X = [-1.4969 1.7449 2.7270; 1.4413 2.1644 4.5763;  -0.3981 2.7477 2.5191];
anova1(X)

providing an F=7.41. Then trying with this code

F = finv(1-0.05,2,6);

provides F=5.14. The degrees of freedom are of course determined on the basis of X, but the significance level does not seem to fit the stadard 0.05 or 0.01. It F in the ANOVA table not the same F-statistic as usual?

Thanks
```
 0

See related articles to this posting

```I mixed up the F statistic in the ANOVA table with the comparation level for accepting the H0 hypothesis, which are not equal.
```
 0

1 Replies
446 Views

Similar Articles

12/6/2013 3:07:11 PM
[PageSpeed]

Similar Artilces:

The monthly statistic
This month, Tiobe rates Lisp/Scheme as having a rating of .770% giving them an `A' (mainstream language) rating. August .688 September .718 (aggregated with Dylan on this month only) October .665 November .770 Joe Marshall wrote: > This month, Tiobe rates Lisp/Scheme as having a rating of .770% giving > them an `A' (mainstream language) rating. Darn, guess I'll have to switch to a less popular language :/ -- Alex On Fri, 04 Nov 2005 03:04:01 +0100, Alex Shinn <alexshinn@gmail.com> wrote: > Joe Marshall wrote: >> This month, Tiobe rates L

CMH test statistic
Hi all, I have a study, in which i would like to compare treatment A with B, C & D. I'd like to do a CMH adjusting for the study and prior use of medication. The response is a binary response(Yes/No). Can someone tell me how can i do this using Proc freq? The data contains more than two studies. Regards,

p-value for resampled statistic
A function I wrote runs a bootstrapped regression with nlinfit 2000 times and returns 2000 estimates of each parameter in the model. I want to test the hypothesis that the second parameter ~= 0. The apparent sample size is artificially large because of the resampling procedure. Therefore, I can't use ttest and instead have to calculate the probability using the standard deviation rather than the standard error in the calculations. I bastardized the code in ztest (lines 89 & 94) substituting the std for stderr. Here's what I've come up with: mu=0; % null hypothesis is that th

How can I get the mode (a kind of statistic parameter) correctly?
--6968417.1058450811568.JavaMail.quiq.tekken Content-Type: text/plain; charset=us-ascii Content-Transfer-Encoding: 7bit I tried to use the 'mode' vi under LabVIEW6.1 to get the mode of an array. But the result obviously wrong. It returned a number that never appeared in the array. And the vi ask me to provide interval, but how can an array of numbers has an interval? Who can tell me how can I get the mode correctly? My array is attached with this question. I'm working on my thesis, the problem is urgent to me. Thanks! --6968417.1058450811568.JavaMail.quiq.tekken Content-Type: tex

deriving likelihood ratio chi-square statistic from proc logistic
Is there a way to derive the likelihood ratio chi-square statistic from proc logistic for each parameter? My output only contains the Wald chi-square statistic. I know that the proc genmod will produce the likelihood, but just wondering if proc logistic has some option that will allow you to obtain it. Thanks. On 11 Dez., 22:56, anwesto...@SBCGLOBAL.NET (Arthur Westover) wrote: > Is there a way to derive the likelihood ratio > chi-square statistic from proc logistic for each > parameter? My output only contains the Wald > chi-square statistic. > > I know that the proc genmod will produce the > likelihood, but just wondering if proc logistic has > some option that will allow you to obtain it. > > Thanks. Dear Arthur, I am not aware of PROC LOGISTIC having this option. An option might be to compare the -2LogLikelihoods of the models with/ without the respective covariate. See the example below where the LR test for the covariate "treatment" is computed: data temp1; input study response treatment number; cards; 0 1 0 20 0 0 0 80 0 1 1 30 0 0 1 70 1 1 0 10 1 0 0 90 1 1 1 25 1 0 1 75 ; run; proc