One month before an election in a metropolitan area, a poll of 660 randomly selected voters showed 54% planning to vote for a certain candidate. A week later it became known that he had shared inappropriate pictures of himself, and a new poll showed only 52% of 1010 voters supporting him. Do these results indicate a decrease in voter support for his candidacy. The z is 0.80. Does anyone understand what x1 and x2 are?

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.

Apr 19th, 2015

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).

Apr 19th, 2015

Thx. and best wishes.

Apr 20th, 2015

BTW, the tail end should read "and the probability of no change in the proportion approving is extremely small". Whoops. :-)

Apr 20th, 2015

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