Question

A pollster plans to survey a random sample of voters in a certain city to ask whether they support an increase in property taxes to fund the construction of a new elementary school. A preliminary poll of 40 voters finds that 29 of them support the proposal. (a) Find a 95% confidence lower bound for the proportion of voters who favor the proposal.

Answer 1

Answer:

0.5867 is the lower bound for the proportion of voters who favor the proposal.  

Step-by-step explanation:

We are given the following in the question:

Sample size, n = 40

Number of voters who support the proposal, x = 29

$\hat{p} = \dfrac{x}{n} = \dfrac{29}{40} = 0.725$

95% Confidence interval:

$\hat{p}\pm z_{stat}\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}$

$z_{critical}\text{ at}~\alpha_{0.05} = 1.96$

Putting the values, we get:

$0.725\pm 1.96(\sqrt{\dfrac{0.725(1-0.725)}{40}})\\\\ = 0.725\pm 0.1383\\\\=(0.5867,0.8633)$

0.5867 is the lower bound for the proportion of voters who favor the proposal.