Question

The national vote in the presidential election had 127,676,926 voters. The proportion of voters who

voted for Donald Trump was 0.473. Let us consider this to be the population proportion. Fox News ran a

poll a few days before the election with a sample size of 1,295. Describe the distribution for the sampling

proportion (for Trump) of the Fox News poll (including the mean and the standard deviation). (2 points)

The Fox News poll found that the sample proportion of people planning to vote for Trump was 0.443.

Based on your description of the sampling distribution above, what is the probability that a poll of 1295

individuals would find a sample proportion of 0.443 or less?

Answer 1

Sample size, n = 1295

Population proportion, p = 0.473

Distribution of proportion: Normal

E(p) = Population proportion = 0.473

sd(p) = sqrt(p*(1-p)/n) = sqrt(0.473*(1-0.473)/1295)

sd(p) = sqrt(0.473*0.527/1295)

sd(p) = 0.0139

Sample proportion, p^ = 0.443

P( p < 0.443 ) = P( ( p - E(p) )/sd(p) < ( 0.443 - E(p) )/sd(p) )

P( p < 0.443 ) = P( ( p - 0.473 )/0.0139 < ( 0.443 - 0.473 )/0.0139 )

P( p < 0.443 ) = P( Z< ( 0.443 - 0.473 )/0.0139 )

P( p < 0.443 ) = P( Z < -2.16 )

P( p < 0.443 ) = 0.0154