topic Re: Logistic analysis of cumulative percentages in Statistical Procedures

Logistic analysis of cumulative percentages

spirto — Fri, 08 Jul 2016 18:18:58 GMT

Hi everyone I am trying to analyze cumulative percentages and get a fitted curve. Here is what my data looks like:

n	n_cum	pct	time
0	0	0	15
2	2	0.05	45.5
10	12	0.3	75.5
17	29	0.725	120.5
4	33	0.825	165.5
5	38	0.95	225.5
2	40	1	318
0	40	1	410

data have;
   input n n_cum pct time;
datalines;
0   0    0    15
2   2    0.05    45.5
10   12    0.3    75.5
17   29    0.725    120.5
4   33    0.825    165.5
5   38    0.95    225.5
2   40    1    318
0   40    1    410
;

What I would like to do is fit a curve to the pct (defined as n_cum/40) over time. My initial attempt was to run a logistic regression (where trials is a variable equal to 40)

data have; set have; trials=40; run;

proc logistic data=have plots(only)=effect;
model n_cum/trials=time / rsquare;
run;

Here is my output.

My question to everyone is given that the number of trials is not independent at each time point (40 trials were not performed at each time point) is this still a valid way to run the analysis. My initial thoughts are that the model fit is correct but the standard errors and corresponding confidence intervals and inferential tests are not.

Thoughts?

Re: Logistic analysis of cumulative percentages

Ksharp — Sat, 09 Jul 2016 03:13:36 GMT

Why would you do that ? LOGISTIC is only for sparse data ,not continuous data.
If it is not time series analysis , you can use PROC LOESS, PROC ADPDTIVE .....   
some non-parameter regression model.

Re: Logistic analysis of cumulative percentages

PGStats — Sat, 09 Jul 2016 04:28:08 GMT

You are right. What you are trying to estimate is the distribution of time. One way to do such estimation is survival (or reliability) analysis.


data have;
    input n n_cum pct time;
datalines;
0    0    0    15
2    2    0.05    45.5
10    12    0.3    75.5
17    29    0.725    120.5
4    33    0.825    165.5
5    38    0.95    225.5
2    40    1    318
0    40    1    410
;

proc lifereg data=have;
model time = / distribution=llogistic;
weight n;
probplot / pupper=99.5 plower=0.5 ppout;
inset scale;
run;

You would probably get a better fit if your times were not so coarsely binned.