Question

Hiring In preparing a report on the economy, we need to estimate the percentage of businesses that plan to hire additional employees in the next 60 days.
a) How many randomly selected employers must we contact in order to create an estimate in which we are 98% confident with a margin of error of 5%

Answer 1

Answer:

We need to contact 542 employees.

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 $1 - \frac{\alpha}{2}$.

The margin of error is:

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

98% confidence level

So $\alpha = 0.02$, z is the value of Z that has a pvalue of $1 - \frac{0.02}{2} = 0.99$, so $Z = 2.327$.

a) How many randomly selected employers must we contact in order to create an estimate in which we are 98% confident with a margin of error of 5%

We dont know the proportion, so we use $\pi = 0.5$, which is when we are going to need the largest sample size.

We have to find n when M = 0.05. So

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

$0.05 = 2.327\sqrt{\frac{0.5*0.5)}{n}}$

$0.05\sqrt{n} = 2.327*0.5$

$\sqrt{n} = \frac{2.327*0.5}{0.05}$

$(\sqrt{n})^{2} = (\frac{2.327*0.5}{0.05})^{2}$

$n = 541.49$

Rounding up

We need to contact 542 employees.