Let p1 = 0.54, p2 = 0.52 and let N1 = 660 and N2 = 1010.
Now let p = the weighted average of p1 and p2: that is, p = (p1 * N1 + p2 * N2) / (N1 + N2) = 0.528
The standard error of p is SE(p) = sqrt(p(1-p) * (1/N1 + 1/N2)) = sqrt(6.244)
Since N1 and N2 are large, p1, p2 (and, therefore p) can be treated as if it came from a normal distribution. The associated z-statistic is
z = (p1 - p2)/SE(p) = 8.004 > 3 so the probability of a z-score this large and no change is the proportion approving is extremely small.
Hope this fits the bill. It first appeared as if a McNemar test was needed, but since the original sample is not resampled, this is the way to go (IMHO).
Thx. and best wishes.
BTW, the tail end should read "and the probability of no change in the proportion approving is extremely small". Whoops. :-)
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