Variance Components Output

[constant survivor]

Hello again,
sorry for bomb texting the forum, I hope that at least some other people can learn from my questions too.

I am referring to the 19th Edition of the pdf-MARK book:

On page 6-97 is the output of a Variance Components estimation shown.
Below that in the first paragraph is written: "We use this estimated SE [ SE(Beta-hat) ] in the usual way to derive 95% CL to the estimated mean". A quite straight sentence and without the next two paragraphs I could easily accept that. It is also explained, that SE(beta-hat) contains sampling AND process variance.

In the next two paragraphs, it is explained that sigma^2 is the pure process variance. Ok. But nothing said about if this is now the basis for calculating the CL of the mean...

So my question: I guess that sigma (meaning square root of sigma^2) is the pure process SE (if one can say so), right?! So do I have to take this SE to calculate the 95% CL of the mean or, as it is clearly written in the first paragraph, do I have to take SE(beta-hat)?

Sorry if this is obvious to some of you. To me it's not. And it doesn't matter how often I read the section. I still remain unsure.

Thanks a lot
Hannes

[cooch]

Hello again,
sorry for bomb texting the forum, I hope that at least some other people can learn from my questions too.

I am referring to the 19th Edition of the pdf-MARK book:

Please only refer to the current version (i.e., the online version), which is the 20th edition (21st edition coming soon).

On page 6-97 is the output of a Variance Components estimation shown.
Below that in the first paragraph is written: "We use this estimated SE [ SE(Beta-hat) ] in the usual way to derive 95% CL to the estimated mean". A quite straight sentence and without the next two paragraphs I could easily accept that. It is also explained, that SE(beta-hat) contains sampling AND process variance.

In the next two paragraphs, it is explained that sigma^2 is the pure process variance. Ok. But nothing said about if this is now the basis for calculating the CL of the mean...

So my question: I guess that sigma (meaning square root of sigma^2) is the pure process SE (if one can say so), right?! So do I have to take this SE to calculate the 95% CL of the mean or, as it is clearly written in the first paragraph, do I have to take SE(beta-hat)?

Sorry if this is obvious to some of you. To me it's not.

What is presumed to be obvious is that you want your 95% CI to reflect overall uncertainty -- meaning, the sum of process and sampling. Meaning, use the SE(beta-hat).

[cooch]

Before you get too invested in 'comparing CI's', have a read through viewtopic.php?f=1&t=3987#p13189

[B.K. Sandercock]

On variance components

Take a look at the example on page D-11 of Appendix D

Code: Select all

Beta-hat SE(Beta-hat)
0.482526   0.033946

S-hat SE(S-hat)   s-tilde  SE(S-tilde) RMSE(S-tilde)
0.640000   0.096000    0.548337   0.048955    0.103917
0.360000   0.096000    0.431320    0.048955   0.086505
0.480000   0.099920    0.481505    0.049351   0.049374
0.440000   0.099277    0.465242    0.049289   0.055377
0.480000   0.099920    0.481505    0.049351   0.049374
0.320000   0.093295    0.412990    0.048656   0.104951
0.600000   0.097980    0.530797    0.049160   0.084887
0.400000   0.097980    0.448615    0.049160   0.069139
0.560000   0.099277    0.514014    0.049289   0.067410
0.560000   0.099277    0.514014    0.049289   0.067410
Naive estimate of sigma^2 = 0.0015950 with 95% CI ( -0.0042991 to 0.0277050)
Estimate of sigma^2 = 0.0019503 with 95% CI ( -0.0039312 to 0.0280522)
Estimate of sigma = 0.0441616 with 95% CI (0.0000000 to 0.1674878)
Trace of G matrix = 4.7017092

Compare the three estimates in the output:
SE(Beta-hat) = SE for both process and sampling variance
Estimate of sigma^2 = variance for process variance only
Estimate of sigma = standard deviation for process variance only

From page D-12 of the appendix, it's possible to calculate 95% confidence intervals with total variance
0.482526 ± (1.96 × 0.033946) = [0.4160, 0.5491]

If sigma was a SE for process variance, it would be puzzling because it is greater than the SE for total variance: 0.04416 > 0.03395

For standard error, need to divide by the square root of the number of estimates in the table (n = 10)
sigma/sqrt(n) = standard error for process variance only
0.0441616/sqrt(10) = 0.01396

Beta-hat ± 1.96SE = 95%CI with process variance only
0.482526 ± (1.96 * 0.01396) = [0.45515, 0.50989]

percentage of total variance due to process variance
0.01396/0.033946 = 41.1%

[constant survivor]

Hi cooch, Hi Brett,
thank you for your distributed answers! Also for the hint on CI's ...

@Brett

If sigma was a SE for process variance, it would be puzzling because it is greater than the SE for total variance: 0.04416 > 0.03395

Of course youre right. I also meant that sigma = SD (not SE) and I already did the following calculations you described. But as cooch said, its obviously more reasonable to report overall uncertainty...

Makes perfect sense to me now. Thank you so much!

[constant survivor]

...question to new thread "t-test for means from VC estimation"

[cooch]

Hi cooch, Hi Brett,
thank you for your distributed answers! Also for the hint on CI's ...

@Brett

If sigma was a SE for process variance, it would be puzzling because it is greater than the SE for total variance: 0.04416 > 0.03395

Of course youre right. I also meant that sigma = SD (not SE) and I already did the following calculations you described. But as cooch said, its obviously more reasonable to report overall uncertainty...

Thatwould be my preference,philosophically, but I could also make an argument for using the SE for the process variance. Brett outlined the steps.