Plotting polynomial regression

[James Canning]

Hi there,

I am trying to plot a polynomial regression between the variables hiscam and my third order polynomial time variable below:

the intention is for the third order polynomial to fit the variation in the variable hiscam. How would I go about plotting this fitted relationship?

Best wishes,

James

42.87512 -1000
43.74237 -1000
43.49274 -729
43.85648 -729
44.71942 -512
45.68772 -512
43.78705 -343
44.99989 -343
43.08729 -216
44.15181 -216
43.06003 -125
43.11882 -125
43.62434 -64
44.06446 -64
44.13048 -27
45.65981 -27
43.56574 -8
45.83962 -8
43.35312 -1
44.99914 -1
47.03517 0
48.04344 0
47.05454 1
48.23038 1
47.78739 8
48.77758 8
47.53606 27
48.73902 27
47.94943 64
48.08339 64
48.14172 125
49.26078 125
48.65994 216
49.85739 216
48.77392 343
49.07422 343
48.99594 512
49.09679 512
49.44331 729
50.71514 729
48.81572 1000
49.97142 1000

[Clyde Schechter]

* Example generated by -dataex-. For more info, type help dataex
clear
input float hiscam int time3
42.87512 -1000
43.74237 -1000
43.49274  -729
43.85648  -729
44.71942  -512
45.68772  -512
43.78705  -343
44.99989  -343
43.08729  -216
44.15181  -216
43.06003  -125
43.11882  -125
43.62434   -64
44.06446   -64
44.13048   -27
45.65981   -27
43.56574    -8
45.83962    -8
43.35312    -1
44.99914    -1
47.03517     0
48.04344     0
47.05454     1
48.23038     1
47.78739     8
48.77758     8
47.53606    27
48.73902    27
47.94943    64
48.08339    64
48.14172   125
49.26078   125
48.65994   216
49.85739   216
48.77392   343
49.07422   343
48.99594   512
49.09679   512
49.44331   729
50.71514   729
48.81572  1000
49.97142  1000
end


graph twoway scatter hiscam time3 || lfit hiscam time3, name(cubic, replace)

I could not help noticing, however, that, at least in your example data, this is a really badly fitting model. A simple linear model against time describes the data much better:

gen time = sign(time3)*abs(time3)^(1/3)
graph twoway scatter hiscam time || lfit hiscam time, name(linear, replace)

In the future, when showing data examples, please use the -dataex- command to do so, as I have here. If you are running version 17, 16 or a fully updated version 15.1 or 14.2, -dataex- is already part of your official Stata installation. If not, run -ssc install dataex- to get it. Either way, run -help dataex- to read the simple instructions for using it. -dataex- will save you time; it is easier and quicker than typing out tables. It includes complete information about aspects of the data that are often critical to answering your question but cannot be seen from tabular displays or screenshots. It also makes it possible for those who want to help you to create a faithful representation of your example to try out their code, which in turn makes it more likely that their answer will actually work in your data.

When asking for help with code, always show example data. When showing example data, always use -dataex-.

[James Canning]

Hi Clyde,

Thank you for your help.

You are right! The linear model is a much better estimator.

I just have a few further questions.

I am trying to investigate the treatment effect of an intervention called the "crusade" that begins at the cutoff point. However I am struggling to graph what it is that I my regression results are showing.

the dataset is below:

Example generated by -dataex-. To install: ssc install dataex
clear
input float(hiscam YEAR crusade) byte summer
42.87512 -10 0 1
43.74237 -10 0 0
43.49274 -9 0 1
43.85648 -9 0 0
44.71942 -8 0 1
45.68772 -8 0 0
43.78705 -7 0 1
44.99989 -7 0 0
43.08729 -6 0 1
44.15181 -6 0 0
43.06003 -5 0 1
43.11882 -5 0 0
43.62434 -4 0 1
44.06446 -4 0 0
44.13048 -3 0 1
45.65981 -3 0 0
43.56574 -2 0 1
45.83962 -2 0 0
43.35312 -1 0 1
44.99914 -1 0 0
47.03517 0 1 1
48.04344 0 1 0
47.05454 1 1 1
48.23038 1 1 0
47.78739 2 1 1
48.77758 2 1 0
47.53606 3 1 1
48.73902 3 1 0
47.94943 4 1 1
48.08339 4 1 0
48.14172 5 1 1
49.26078 5 1 0
48.65994 6 1 1
49.85739 6 1 0
48.77392 7 1 1
49.07422 7 1 0
48.99594 8 1 1
49.09679 8 1 0
49.44331 9 1 1
50.71514 9 1 0
48.81572 10 1 1
49.97142 10 1 0
end
[/CODE]

The years have been normalised so that they are equal to 0 at the cutoff point. The crusade intervention is a binary variable and there is a summer seasonal dummy as there are two observations in each year, a winter and summer one.

My question is how can I graph this result showing a discontinuity at the cutoff whilst still maintaining the linear model?


Many thanks

[Clyde Schechter]

regress hiscam i.crusade##c.YEAR i.summer
predict xb, xb
by YEAR, sort: egen model = mean(xb)
graph twoway scatter hiscam YEAR || line model YEAR, xline(0)