Unequal p-values when using tcdf and regstats

Hi,
I noticed some inconsistencies with tcdf and regstats and wonder if I am missing something or if this is a bug of some kind. This relates to the statisticsl toolbox.
To illustrate:
a_predicted = [92.9536;93.0658;93.0670;93.2834;93.2858;93.4251;93.4267;93.5773; 93.5789;93.7254;93.7271;93.8718;93.8734;94.0277;94.0293;94.2216];
a_predictor1 = [84.2436;88.1282;88.1718;92.1000;92.0696;89.3000;89.3359;92.6000;92.6352;95.8000;95.7978;95.6000;95.5848;94.2000;94.2315;97.1000];
a_predictor2 = [133.9558;130.0221;129.9779;126.0000;125.7500;103.0000;102.9457;98.0000;98.0220;100.0000;100.0330;103.0000;103.6304;161.0000;161.5978;216.0000];
a_regstat = regstats(a_predicted, [a_predictor1 a_predictor2], 'linear');
Now, a_regstat.tstat.pval(2) results in 5.1918e-06 But 1-tcdf(a_regstat.tstat.t(2),13) results in 2.5959e-06 which seems to be half the regstats pval.
Why is that?

 采纳的回答

That is because the pvalue 5.1918e-06 is for a two-tailed hypothesis test.
1-tcdf(a_regstat.tstat.t(2),13)
is only giving you the result for 1 tail of the t-distribution. If you double that result (the t-distribution is symmetric), you'll see the agreement.

1 个评论

Thanks, it wasn't obvious from regstats documentation, or I just missed it.

请先登录,再进行评论。

更多回答(0 个)

类别

帮助中心File Exchange 中查找有关 Electrophysiology 的更多信息

标签

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by