Question

A travel agent wants to estimate the proportion of vacationers who plan to travel outside the United States in the next 12 months. A random sample of 130 vacationers revealed that 40 had plans for foreign travel in that time frame. Construct a 95% confidence interval estimate of the population proportion.

Answer 1

Answer:

The 95% confidence interval estimate of the population proportion is (0.2284, 0.3870).

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:

A random sample of 130 vacationers revealed that 40 had plans for foreign travel in that time frame. This means that $n = 130, p = \frac{40}{130} = 0.3077$

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:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.3077 - 1.96\sqrt{\frac{0.3077*0.6923}{130}} = 0.2284$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.3077 + 1.96\sqrt{\frac{0.3077*0.6923}{130}} = 0.3870$

The 95% confidence interval estimate of the population proportion is (0.2284, 0.3870).