Question

Eliana wants to estimate the number of out of state visitors at a national park. She surveys 200 visitors and finds that 83 of them are from out of state. Identify the values needed to calculate a confidence interval at the 90% confidence level. Then find the confidence interval.

Answer 1

Answer:

The 90% confidence interval for the proportion of out of state visitors at a national park is (0.3577, 0.4723).

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:

$n = 200, \pi = \frac{83}{200} = 0.415$

90% confidence level

So $\alpha = 0.1$, z is the value of Z that has a pvalue of $1 - \frac{0.1}{2} = 0.95$, so $Z = 1.645$.

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.415 - 1.645\sqrt{\frac{0.415*0.585}{200}} = 0.3577$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.415 + 1.645\sqrt{\frac{0.415*0.585}{200}} = 0.4723$

The 90% confidence interval for the proportion of out of state visitors at a national park is (0.3577, 0.4723).