The standard error of a proportion p is estimated by the square root of pq/
N, where q=1-p and N is sample size (if the sample is a simple random
sample). The 95% confidence interval is 1.96 times the standard error. There
you are. If your N is, say, 400 cases, the square root of N is 20, and then
e=the square root of (0.65 x 0.35)/400=0.02385. The confidence interval
would be 0.65 +/- (1.96 x 0.02385), ranging approx from 0.6261 to 0.6738.
Any proportion higher than 0.6738 would be regarded as significantly
different from 0.65, if your sample is N=400 and the chosen level of
significance is 95%.
Now, this refers to STATISTICAL significance. It means that with such sample
size you are reasonably certain that the observed difference is not a fluke;
that is, reasonably certain that the 'real' difference is not zero (you'd
still have a [lower than 5%] chance of the difference being actually zero).
But it does not mean that the difference is of any 'real' or 'substantive'
significance. For instance, if you are using the proportion to rank classes
of students, or for any other similar practical purpose, perhaps a reading
of 0.6738 makes no real difference compared to another of 0.65, even if you
are pretty certain that the difference is not due to sampling fluke.
Hector
-----Original Message-----
From: SPSSX(r) Discussion [mailto:
[hidden email]] On Behalf Of
Mark Webb
Sent: 03 August 2009 22:42
To:
[hidden email]
Subject: What to increase a % to to make it sig different ?
We track top-2-box % scores over time & want treat the current reading
as a benchmark and add some increment to each to make the new % a
target. The thinking being that the new % will be significantly
different from the benchmark at 95% sig level.
How can I calculate the incremental increase?
Eg - what must I increase 65% to so that the new % is sig different to 65% ?
Regards
Mark Webb
+27 21 786 4379
+27 72 199 1000
Skype - webbmark
[hidden email]
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