Question

answer please

The Chamber of Commerce of a mid-sized city has supported a It will change the zoning laws for a particular part of the town. Worried that the proposal does not have the required support of at least 50% of the population, the Mayor has commissioned a survey that asks prospective voters whether or not they support the proposal. A random sample of 450 voters is selected, and 243 said they would likely support the proposal. How wide would be the 95% confidence interval for the true proportion of voters who are likely to support the proposal? O A. 0.046 B. 0.108 C. 0.092 D. 0.024 E. 0.070

Answer 1

Solution :

Given that,

n = 450

x = 243

Point estimate = sample proportion = $\hat p$ = x / n = 243 / 450 = 0.54

1 - $\hat p$ = 1 - 0.54 = 0.46

a) At 95% confidence level

$\alpha$ = 1 - 95%

$\alpha$ =1 - 0.95=0.05

$\alpha$/2 = 0.025

Z$\alpha$/2 = Z_{0.025 = 1.96}

Margin of error = E = Z_{$\alpha$ / 2} * $\sqrt$ (($\hat p$ * (1 - $\hat p$ )) / n)

= 1.96($\sqrt$((0.54* 0.46) / 450)

= 0.046

A 95% confidence interval for population proportion p is ,

$\hat p$ ± E   

= 0.54 ± 0.046

= (0.494, 0.586)

width of the confidence interval is = 0.586 - 0.494 = 0.092

Ans: option c) 0.092 is correct