Question

A random sample of 432 voters revealed that 100 are in favor of a certain bond issue. A 95 percent confidence interval for the proportion of the population of voters who are in favor of the bond issue is A 100 + 1.96 0.5(0.5) 432 100 + 1.645 0.5(0.5) 432 100 + 1.96 0.231(0.769) 432 0.231 +1.96 0.231(0.769) 432 0.231(0.709) 0.231 +1.6451 432​

Answer 1

Answer:

The 95% confidence interval is $0.231 \pm 1.96\sqrt{\frac{0.231*0.769}{432}}$

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$, and a confidence level of $1-\alpha$, we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the z-score that has a p-value of $1 - \frac{\alpha}{2}$.

A random sample of 432 voters revealed that 100 are in favor of a certain bond issue.

This means that $n = 432, \pi = \frac{100}{432} = 0.231$

95% confidence level

So $\alpha = 0.05$, z is the value of Z that has a p-value of $1 - \frac{0.05}{2} = 0.975$, so $Z = 1.96$.

Confidence interval:

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

$0.231 \pm 1.96\sqrt{\frac{0.231*0.769}{432}}$

The 95% confidence interval is $0.231 \pm 1.96\sqrt{\frac{0.231*0.769}{432}}$