Question

During an economic downturn, 11 companies were sampled and asked whether they were planning to increase their workforce. Only 2 of the 11 companies were planning to increase their workforce. Use the small-sample method to construct an 80% confidence interval for the proportion of companies that are planning to increase their workforce. Round the answers to at least three decimal places.

An 80% confidence interval for the proportion of companies that are planning to increase their workforce is _______ < p< ________

Answer 1

An 80% confidence interval for the proportion of companies that are planning to increase their workforce is 0.033 < p < 0.331.

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)`.

Only 2 of the 11 companies were planning to increase their workforce

This means that
`n = 11, \pi = (2)/(11) = 0.182`

80% confidence level

So
`\alpha = 0.2`, z is the value of Z that has a pvalue of
`1 - (0.2)/(2) = 0.9`, so
`Z = 1.28`.

The lower limit of this interval is:

`\pi - z\sqrt{(\pi(1-\pi))/(n)} = 0.182 - 1.28\sqrt{(0.182*0.818)/(11)} = 0.033`

The upper limit of this interval is:

`\pi + z\sqrt{(\pi(1-\pi))/(n)} = 0.182 + 1.28\sqrt{(0.182*0.818)/(11)} = 0.331`

An 80% confidence interval for the proportion of companies that are planning to increase their workforce is 0.033 < p < 0.331.