Confidence interval for percent change in rate

[Andrew Adkins Second]

Building off of the example shown here: https://www.statalist.org/forums/new-content/51

I'm attempting to do something similar in reporting percent change with confidence intervals compared to some baseline period.

* Example generated by -dataex-. For more info, type help dataex
clear
input float ac_drug_oral double volume_per10000 byte period float(monthly ln_volume_per10000)
2 1260.6831300832241 1 732 7.139409
2 1121.1703693867348 1 733 7.022129
2 1297.5034939189118 1 734 7.168197
2  1244.835560987853 1 735 7.126759
2 1209.8619960138883 1 736 7.098261
2 1276.7094465389555 1 737 7.152041
2 1257.8345016583257 1 738 7.137147
2 1250.1367395163436 1 739 7.131008
2 1194.3472777250283 1 740 7.085355
2  1217.227766272708 1 741 7.104331
2 1188.4083496816286 1 742  7.08037
2 1220.8647452373323 1 743 7.107315
2 1195.1871343100497 2 744 7.086058
2  1093.872239363099 2 745 6.997479
2 1217.4365842951077 2 746 7.104503
2 1155.7952853628883 2 747 7.052544
2 1183.4685905249155 2 748 7.076205
2 1180.5756265808982 2 749 7.073758
2  1148.722932051039 3 750 7.046406
2  1206.729594179629 3 751 7.095669
2 1119.5237418002291 3 752 7.020658
2 1132.7168821291311 3 753 7.032374
2 1107.2286632215928 3 754 7.009615
2 1137.4805390050224 3 755 7.036571
end
format %tm monthly

reg ln_volume_per10000 i.period contrast r.period nlcom 100*(1-exp(_b[2.period])) nlcom 100*(1-exp(_b[3.period]))

The issue that I encounter is trying to calculate a percent change compared to the baseline period (period of interest/baseline-1)*100). Is there a way to generate a percent change in the right direction? For example, one is positive and one is negative while the formula has both as positive.

Thanks,

[Clyde Schechter]

If you are talking about getting a percentage change in ln_volume, rather than figure out how to do it, I will ask you why you want to do it? Percentage changes in log-transformed outcomes are not really meaningful.

If you are talking about getting a percentage change in volume itself, on the assumption that your "baseline" period is period == 1, then all you need is to look at the coefficients of the other two baseline periods in this regression of ln_volume, and multiply those, and their confidence limits, by 100. That will give you a reasonable approximation to the percentage difference in the expected value of volume between those periods and the baseline period, as long as those coefficients are < 0.1 in magnitude.

If your coefficients of those periods are more than 0.1 in magnitude, then the approximation is not as good, and instead you should calculate 100*(exp(coefficient)-1), and similarly for the confidence limits. This formula is not an approximation--it gives the exact value. You could also use it in the smaller coefficient scenario if you need more than one decimal place worth of accuracy in the percentage.

[Andrew Adkins Second]

I am trying to determine percent change between periods for volume per 10,000. Instead of doing this via regression to obtain expected value of volume between periods, is there a way in Stata to obtain percent changes in the mean values between periods? In this example, and others, I still face the issue where one is positive and one is negative though the formula estimates both as negative values.

mean mean_volume_per10000, over(ac_drug_oral period)

Is there something either before or after this that I can use in order to estimate to obtain percent changes with confidence intervals?

[Clyde Schechter]

In this example, and others, I still face the issue where one is positive and one is negative though the formula estimates both as negative values.

I don't know what you're talking about here.

. * Example generated by -dataex-. For more info, type help dataex
. clear

. input float ac_drug_oral double volume_per10000 byte period float(monthly ln_volume_per10000)

     ac_drug~l  volu~10000    period    monthly  ln_~10000
  1. 2 1260.6831300832241 1 732 7.139409
  2. 2 1121.1703693867348 1 733 7.022129
  3. 2 1297.5034939189118 1 734 7.168197
  4. 2  1244.835560987853 1 735 7.126759
  5. 2 1209.8619960138883 1 736 7.098261
  6. 2 1276.7094465389555 1 737 7.152041
  7. 2 1257.8345016583257 1 738 7.137147
  8. 2 1250.1367395163436 1 739 7.131008
  9. 2 1194.3472777250283 1 740 7.085355
 10. 2  1217.227766272708 1 741 7.104331
 11. 2 1188.4083496816286 1 742  7.08037
 12. 2 1220.8647452373323 1 743 7.107315
 13. 2 1195.1871343100497 2 744 7.086058
 14. 2  1093.872239363099 2 745 6.997479
 15. 2 1217.4365842951077 2 746 7.104503
 16. 2 1155.7952853628883 2 747 7.052544
 17. 2 1183.4685905249155 2 748 7.076205
 18. 2 1180.5756265808982 2 749 7.073758
 19. 2  1148.722932051039 3 750 7.046406
 20. 2  1206.729594179629 3 751 7.095669
 21. 2 1119.5237418002291 3 752 7.020658
 22. 2 1132.7168821291311 3 753 7.032374
 23. 2 1107.2286632215928 3 754 7.009615
 24. 2 1137.4805390050224 3 755 7.036571
 25. end

. format %tm monthly

.
. mean volume_per10000, over(ac_drug_oral period)

Mean estimation                                                      Number of obs = 24

---------------------------------------------------------------------------------------
                                      |       Mean   Std. err.     [95% conf. interval]
--------------------------------------+------------------------------------------------
c.volume_per10000@ac_drug_oral#period |
                                 2 1  |   1228.299   13.63556      1200.091    1256.506
                                 2 2  |   1171.056   17.48179      1134.892     1207.22
                                 2 3  |   1142.067   14.20789      1112.676    1171.458
---------------------------------------------------------------------------------------

.
end of do-file

So all three means are positive. And both differences from baseline are negative. Moreover, the means of volume_per100000 must be positive, as much all the values of that variable, or you would not have been able to log-transform them: only positive numbers have logarithms.

Also, I don't know what "formula" you are talking about.

Anyway, after that those means, you can follow with

. nlcom 100*(_b[[email protected]_drug_oral#2.period] - _b[[email protected]_drug_oral#1.period]) ///
>     / _b[[email protected]_drug_oral#1.period]

       _nl_1: 100*(_b[[email protected]_drug_oral#2.period] - _b[[email protected]_drug_oral#1.period]) / _b[[email protected]_
> drug_oral#1.period]

------------------------------------------------------------------------------
        Mean | Coefficient  Std. err.      z    P>|z|     [95% conf. interval]
-------------+----------------------------------------------------------------
       _nl_1 |  -4.660325   1.773646    -2.63   0.009    -8.136607   -1.184042
------------------------------------------------------------------------------

. nlcom 100*(_b[[email protected]_drug_oral#3.period] - _b[[email protected]_drug_oral#1.period]) ///
>     / _b[[email protected]_drug_oral#1.period]

       _nl_1: 100*(_b[[email protected]_drug_oral#3.period] - _b[[email protected]_drug_oral#1.period]) / _b[[email protected]_
> drug_oral#1.period]

------------------------------------------------------------------------------
        Mean | Coefficient  Std. err.      z    P>|z|     [95% conf. interval]
-------------+----------------------------------------------------------------
       _nl_1 |  -7.020407   1.550286    -4.53   0.000    -10.05891   -3.981903
------------------------------------------------------------------------------

to get your percentage changes.

Added: By the way, this gets you the same results as the regression based approach.

[Andrew Adkins Second]

Got it; thank you.

I was looking at a different example than the one I shared which is why I thought one percent change was negative and one was positive. Apologies for the confusion. This is exactly what I was looking to accomplish.