Bonferroni Correction in Constructing Confidence Intervals

Hi, hope all is well. I wanted to use the Bonferroni correction when constructing my confidence intervals. I am very confused. Say my linear regression model is as follows:

y=-499.166 + 61.780 (x1) + -5.554 (x2) + -29.431 (x3) + 28.443 (x4) + 22.143 (x5)

Say I am told to construct 95% confidence intervals. Would the alpha for my critical t be 0.05/(2*6) or would it be something else? To elaborate, for x1, would the confidence interval be

[61.780-(t0.0042, df)*Error, 61.780+(t0.0042, df)*Error]

or no?


Less is more. Stay pure. Stay poor.
Can we step back, since you provided your model. Why are you applying a correction?

Say if I had 3 categories and was comparing 1 v 2, 1 v 3, and 2 v3, and wanted a Bonferroni corrected 95% CI, I would use 0.05 alpha. But since I apply it to both sides of the estimate and have 3 comparisons - I would use (0.05/3)/2 when calculating: estimate +/- (SE * critical value). And the alpha would come into play for the critical value.
Thank you so much for responding. To be honest, I am not really sure. In the problem statement, it simply says “Carry out a full analysis of the model, including tests for each coefficient (with the significance level adjusted for multiple tests) and confidence intervals for each coefficient.” I just assumed that “with the significance level adjusted for multiple tests” meant that a bonferroni correction should be used.

So is my answer correct? Since there are 6 comparisons? I am not sure how many comparisons/categories there are in this model if that makes any sense.


Less is more. Stay pure. Stay poor.
When I think of pairwise corrections - my mind goes to corrections. So does this model have any categorical variables with > 2 groups?
Oh. I am not sure. In the problem statement, I am simply asked to fit multiple linear regression model that relates
weight (y) to length1 (x1), length2 (x2), length3 (x3), height (x4), and width (x5). I am not sure how the term pairwise correction applies to this problem.