Question

A national survey of companies included a question that asked whether the customers like the new flavor of a cola from company A. The sample results of 1000 customers, and 850 of them indicated that they liked the new flavor. The 98% confidence interval on the population proportion of people who like the new flavor is _______________.

Answer 1

The 98% confidence interval on the population proportion of people who like the new flavor is (0.8237, 0.8763).

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 = 1000, \pi = (850)/(1000) = 0.85`

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)} = 0.85 - 2.327\sqrt{(0.85*0.15)/(1000)} = 0.8237`

The upper limit of this interval is:

`\pi + z\sqrt{(\pi(1-\pi))/(n)} = 0.85 + 2.327\sqrt{(0.85*0.15)/(1000)} = 0.8763`

The 98% confidence interval on the population proportion of people who like the new flavor is (0.8237, 0.8763).