Theorem 7.12(b) gives rise to statements we hear in the news, such as, Based on

Question

Theorem 7.12(b) gives rise to statements we hear in the news, such as, Based on a sample of 1103 potential voters, the percentage of people supporting Candidate Jones is 58% with an accuracy of plus or minus 3 percentage points. The experiment is to observe a voter at random and determine whether the voter supports Candidate Jones. We assign the value X = 1 if the voter supports Candidate Jones and X = 0 otherwise. The probability that a random voter supports Jones is E[X] = p. In this case, the data provides an estimate M_n(X) = 0.58 as an estimate of p. What is the confidence coefficient 1 - alpha corresponding to this statement? Since X is a Bernoulli (p) random variable, E[X] = p and Var[X] = p(1 - p). For c = 0.03, Theorem 7.12(b) says P[|M_n(X) - p| = 0.75. If the accuracy is +/-1% instead, recalculate the confidence coefficient. Can you conclude that when accuracy improves (to a smaller percentage, or the confidence interval shrinks), does the confidence coefficient increase or decrease? (b) In the same polling example as above, if we have accuracy +/-3% but increase the sample size to 2,000 voters. What is the new confidence coefficient? Is it increased or decreased from bigger sample size?

Explanation / Answer

a) if 1 % accuracy

alpha < 0.25/(0.01)^2 /n

= 2500/n

here n = 1103

hence alpha < 2.2665

or 1 - alpha >= -1.2665

confidence level decrease

b)

if n = 2000

alpha < = 0.1388888

1 - alpha >= 0.861111

hence 86.11 %

cofidence coefficient increase as sample size increases

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