Just before a referendum on a school budget a local newspape
Just before a referendum on a school budget, a local newspaper 427 voters to predict whether the budget will pass. Suppose the budget has the support of 52% of the voters. What is the probability that the newspapers sample will lead it to predict defeat?
Solution
The standard deviation of the sampling distirbution of p^ is
s(p^) = sqrt(p(1-p)/n) = sqrt(0.52*(1-0.52)/427) = 0.024177331
Predicting defeat means less than 0.50 of voters.
We first get the z score for the critical value. As z = (x - u) / s, then as
x = critical value = 0.5
u = mean = 0.52
s = standard deviation = 0.024177331
Thus,
z = (x - u) / s = -0.827221168
Thus, using a table/technology, the left tailed area of this is
P(z < -0.827221168 ) = 0.204055858 [ANSWER]
