confidence level of 90% for true proportion voting?

Question

Out of 200 people sampled, 134 preferred Candidate A.

Based on this, find a 90% confidence level for the true proportion of the voting population () prefers Candidate A.

Give your answers as decimals, to three places.

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Answer 1

Well, the answer will depend on what estimation method one wants to use. Does the instructor want to use a normal approximation or use the binomial/beta distribution method, which is more complex but more accurate?

Using the simple, but inaccurate, normal approximation we guess the population proportion/probability p hat by dividing 134/200 to get 0.67. (which is biased - the unbiased estimator is 0.6683)

Calculating the 90% confidence interval using the normal approximation we determine the margin of error (moe) by the formulae: moe = Z_90% * sqrt(p*(1-p)/n).

moe = 1.645 * √(.67*.33/200) = 0.054690

The C.I. to 3 decimal places is therefore [ 0.615 , 0.725]

Using the exact calculations (Baysian)

The likely value (p hat) is .668. ; the C.I. is [0.626 , 0.710].

Which, while being more accurate, is certainly not the same as the normal approximation which is probably what your instructor is looking for.