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

Answer:

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

Step-by-step 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.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$.

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

This means that $n = 11, \pi = \frac{2}{11} = 0.182$

80% confidence level

So $\alpha = 0.2$, z is the value of Z that has a pvalue of $1 - \frac{0.2}{2} = 0.9$, so $Z = 1.28$.

The lower limit of this interval is:

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

The upper limit of this interval is:

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

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