Question

A survey of 865 voters in one state reveals that 408 favor approval of an issue before the legislature. Construct the 95% confidence interval for the true proportion of all voters in the state who favor approval.

Answer 1

Answer:

The 95% confidence interval for the true proportion of all voters in the state who favor approval is (0.4384, 0.5050).

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 zscore that has a pvalue of $1 - \frac{\alpha}{2}$.

For this problem, we have that:

$n = 865, \pi = \frac{408}{865} = 0.4717$

95% confidence level

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

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.4717 - 1.96\sqrt{\frac{0.4717*0.5283}{865}} = 0.4384$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.4717 + 1.96\sqrt{\frac{0.4717*0.5283}{865}} = 0.5050$

The 95% confidence interval for the true proportion of all voters in the state who favor approval is (0.4384, 0.5050).