Question

The Royal Haciendas Resort in Playa del Carmen, Mexico, is thinking of starting a new promotion. When a customer checks out of the resort after spending 5 or more days, the customer would be given a voucher that is good for 2 free nights on the next stay of 5 or more nights at the resort. The marketing manager is interested in estimating the proportion of customers who return after getting a voucher. From a simple random sample of 100 customers, 62 returned within 1 year after receiving the voucher. Please determine the 95% confidence interval estimate of the true population proportion.

Answer 1

Answer:

The 95% confidence interval estimate of the true population proportion is (0.5249, 0.7151).

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 $1 - \frac{\alpha}{2}$.

From a simple random sample of 100 customers, 62 returned within 1 year after receiving the voucher.

This means that $n = 100, \pi = \frac{62}{100} = 0.62$

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:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.62 - 1.96\sqrt{\frac{0.62*0.38}{100}} = 0.5249$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.62 + 1.96\sqrt{\frac{0.62*0.38}{100}} = 0.7151$

The 95% confidence interval estimate of the true population proportion is (0.5249, 0.7151).