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A vote transfer poll asks the electorate about how they voted in the last election and how they would vote if the next election was now. With the result from both questions one could build a matrix of vote transfer, say: 77% of Party A voters admit maintaining their vote, whereas 12% admit voting for Party B. How confident can we be in these estimates? How do they relate to sample size? How big does the sample need to be for required degree of confidence? Does it depend on the number of parties?

EDIT 1: To give you an example, here is what such a matrix would look like (columns for previous vote, rows for future vote intentions):

vote transfer poll

All the columns sum up to 100%, since every voter of every party must choose the same or another party, choose not to answer (Nr) or say they don't know (Ns).

How confident can we be in the values of each cell? Does it depend on the sample represented on each column?

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Your question is a bit vague for a precise sample size computation.

Exactly which proportions do you want to compare. Just 'stay' vs 'change' for Party A? Some sort of comparison between Party A and Party B? [In your one numerical example, $77\% + 12\% = 89\%;$ what about the rest?]

General rule for margin of error in a CI: In cases where proportions are 30-70%, a rough rule is that the margin of error $M$ (half CI width) in a 95% Wald CI is $1/\sqrt{n}.$ So you'd need $n\approx 2500$ to get within $\pm M = \pm 2\%.$

Notes: (1) Remember that such computations address only random sampling error. Systematic errors due to (a) subjects not feeling as free to express an opinion is an opinion poll as on a secret ballot, (b) subjects of one party being easier to reach than those of another party, and so on, have to be addressed separately.

(2) Wald CIs are not accurate for small $n,$ but are OK for moderately large $n$ (high hundreds and larger) and for proportions not too far from 1/2.$

(3) I see your edit showing data. Now, which counts (not percentages) do you want to compare?

if you just wanted validation that 12% is significantly different from 72%, then it wouldn't take a large random sample: Output from a relevant 'power and sample size' computation from a recent version of Minitab is shown below.

Power and Sample Size 

Test for Two Proportions

Testing comparison p = baseline p (versus >)
Calculating power for baseline p = 0.12
α = 0.05

              Sample  Target
Comparison p    Size   Power  Actual Power
        0.72       8     0.8      0.839019
        0.72      10     0.9      0.911786

The sample size is for each group.
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  • $\begingroup$ Thanks for your input so far. I want to know what is the 95% confidence interval in the value of each cell. $\endgroup$ May 13, 2021 at 20:37

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