Question

A large trucking company wants to estimate the proportion of its tracker truck population with refrigerated carrier capacity. A random sample of 200 tracker trucks is taken and 30% of the sample have refrigerated carrier capacity. The 95% confidence interval to estimate the population proportion is _______.

Answer 1

The 95% confidence interval to estimate the population proportion is (0.2365, 0.3635).

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 = 200, \pi = 0.3`

95% confidence level

So
`\alpha = 0.05`, z is the value of Z that has a pvalue of
`1 - (0.05)/(2) = 0.975`, so
`Z = 1.96`.

The lower limit of this interval is:

`\pi - z\sqrt{(\pi(1-\pi))/(n)} = 0.3 - 1.96\sqrt{(0.3*0.7)/(200)} = 0.2365`

The upper limit of this interval is:

`\pi + z\sqrt{(\pi(1-\pi))/(n)} = 0.3 + 1.96\sqrt{(0.3*0.7)/(200)} = 0.3635`

The 95% confidence interval to estimate the population proportion is (0.2365, 0.3635).