Question

A survey of 542 consumers reveals that 301 favor the new design for a product. Construct a 90% confidence interval for the true proportion of all consumers who favor the design.

Answer 1

The 90% confidence interval for the true proportion of all consumers who favor the design is (0.52, 0.59).

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 = 542, \pi = (301)/(542) = 0.555`

90% confidence level

So
`\alpha = 0.1`, z is the value of Z that has a pvalue of
`1 - (0.1)/(2) = 0.95`, so
`Z = 1.645`.

The lower limit of this interval is:

`\pi - z\sqrt{(\pi(1-\pi))/(n)} = 0.555 - 1.645\sqrt{(0.555*0.445)/(542)} = 0.52`

The upper limit of this interval is:

`\pi + z\sqrt{(\pi(1-\pi))/(n)} = 0.555 + 1.645\sqrt{(0.555*0.445)/(542)} = 0.59`

The 90% confidence interval for the true proportion of all consumers who favor the design is (0.52, 0.59).