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I made a model using repeated measures univariate ANOVA in R.

> g <- aov(bis ~ x1 + x2 + bg.sol + x1:x2:I(bg.sol * k1) + Error(subject), coded)
> summary.lm(g$Within)
Call:
NULL

Residuals:
     Min       1Q   Median       3Q      Max 
-24.7459  -4.8055  -0.1518   5.1696  17.6015 

Coefficients:
                     Estimate Std. Error t value Pr(>|t|)    
x1                     3.1170     0.8444   3.691 0.000275 ***
x2                    -1.0906     0.1230  -8.864  < 2e-16 ***
I(bg.sol * k1)         2.0522     1.0216   2.009 0.045645 *  
x1:x2:I(bg.sol * k1)  -0.3191     0.1254  -2.545 0.011543 *  
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 

Residual standard error: 7.256 on 246 degrees of freedom
Multiple R-squared: 0.2743, Adjusted R-squared: 0.2654 
F-statistic: 30.99 on 3 and 246 DF,  p-value: < 2.2e-16 

I calculated confidence limit for each estimates. I thought SE * critical value would work. In case of x1 (continuous variable) 95% confidence limit was,

> 0.8444 * qt(0.975, df = 1)
[1] 10.72912

I'm wondering whether the calculated value is real confidence limit for x1. The estimates for x1 is 3.1170, and the limit is 10.72912. Plus-minus it includes zero value. But P-value showed value less than 0.05!

I want to know where I made an error!

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  • 9
    $\begingroup$ The confidence interval for x1 is 3.1170+c(-1,1)*qt(0.975, df=246)*0.8444, or $(1.45,4.78)$. Your only issue is the degrees of freedom in your qt call. $\endgroup$ – assumednormal Aug 25 '12 at 10:13
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    $\begingroup$ Hi, @Max, that seems to be the answer. Would you mind making it an official answer (& perhaps elaborating it a little) so we can upvote it & the OP can accept it? $\endgroup$ – gung Aug 25 '12 at 13:06
  • $\begingroup$ Hi @Max, please do consider posting this as an answer! (Almost four years later, but still.) $\endgroup$ – amoeba May 12 '16 at 12:55
  • $\begingroup$ @Max Would you mind writing this up into an official answer to close this thread? $\endgroup$ – Charlotte R Sep 20 '16 at 14:08
  • $\begingroup$ @Max, do you want to turn your comment into an answer? $\endgroup$ – Stephan Kolassa Nov 6 '17 at 12:25

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