Question

Out of a sample of 94 purchases at the drive-up window of a fast-food establishment, 27 were made with a major credit card. Find a 98% confidence interval for the population proportion of purchases paid with a major credit card.

Answer 1

Answer:

The 98% confidence interval for the population proportion of purchases paid with a major credit card is (0.1786, 0.3958).

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:

$n = 94, \pi = \frac{27}{94} = 0.2872$

98% confidence level

So $\alpha = 0.02$, z is the value of Z that has a pvalue of $1 - \frac{0.02}{2} = 0.99$, so $Z = 2.327$.

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.2872 - 2.327\sqrt{\frac{0.2872*0.7128}{94}} = 0.1786$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.2872 + 2.327\sqrt{\frac{0.2872*0.7128}{94}} = 0.3958$

The 98% confidence interval for the population proportion of purchases paid with a major credit card is (0.1786, 0.3958).