Question

Random sample of 330 voters, finding 144 who says that will vote “yes” on the upcoming school budget

e chronicle polls a random sample of 330 voters, finding 144 who says that will vote "yes" on the upcoming school budget. What is the value of the sample proportion? What is the value of the standard error? Have we met the conditions to construct a 1-proportion z interval? Construct a 95% confidence interval for the actual sentiment of all the voters.

Answer 1

n=330

p=144/330 = 0.436

q = 186/330 =0.564

Se =√(pq/n)

=√(0.436 *0.564 / 330)

=0.027

Margin of error = 1.96*0.027

=0.053

CI =0.436+/- 0.053

=(0.383, 0.489)

We are 95% confident that between 38.3% and 48.9% of voters will vote on upcoming polls