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

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

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{(\pi(1-\pi))/(n)}`

In which

z is the zscore that has a pvalue of
`1 - (\alpha)/(2)`.

For this problem, we have that:

`n = 94, \pi = (27)/(94) = 0.2872`

98% confidence level

So
`\alpha = 0.02`, z is the value of Z that has a pvalue of
`1 - (0.02)/(2) = 0.99`, so
`Z = 2.327`.

The lower limit of this interval is:

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

The upper limit of this interval is:

`\pi + z\sqrt{(\pi(1-\pi))/(n)} = 0.2872 + 2.327\sqrt{(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).