Question

in a small town of 5,832 people, the mayor wants to determine the proportion of voters who would support an increase to the food tax. an assistant to the mayor surveys 500 randomly chosen people, and finds that 240 support the increase. assuming the conditions for inference have been met, what is the 95% confidence interval for the true proportion of people who would support the increase in food tax?

0.48 ± 1.65√(0.48(1 - 0.48)/500)
0.48 ± 1.96√(0.48(1 - 0.48)/500)
0.52 ± 1.65√(0.52(1 - 0.52)/500)
0.52 ± 1.96√(0.52(1 - 0.52)/500)

Explanation:

Step1: Calculate the sample proportion

The sample proportion $\hat{p}=\frac{240}{500} = 0.48$.

Step2: Determine the critical value

For a 95% confidence interval, the critical value $z$ is 1.96.

Step3: Write the formula for the confidence interval

The formula for a confidence interval for a proportion is $\hat{p}\pm z\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}$. Substituting $\hat{p}=0.48$, $z = 1.96$, and $n = 500$ into the formula gives $0.48\pm1.96\sqrt{\frac{0.48(1 - 0.48)}{500}}$.

Answer:

B. $0.48\pm1.96\sqrt{\frac{0.48(1 - 0.48)}{500}}$