Question

It is said that happy and healthy workers are efficient and productive. A company that manufactures exercising machines wanted to know the percentage of large companies that provide on-site health club facilities. A sample of 240 such companies showed that 96 of them provide such facilities. Construct a 97% confidence interval for the percentage of all such companies that provide such facilities on-site. What is the margin of error for this estimate

Answer 1

Answer:

The 97% confidence interval for the percentage of all such companies that provide such facilities on-site is (0.3314, 0.4686). The margin of error is of 0.0686 = 6.86 percentage points.

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:

The absolute value of the subtraction of one of the bounds by the estimate $\pi$

For this problem, we have that:

$n = 240, \pi = \frac{96}{240} = 0.4$

97% confidence level

So $\alpha = 0.03$, z is the value of Z that has a pvalue of $1 - \frac{0.03}{2} = 0.985$, so $Z = 2.17$.

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.4 - 2.17\sqrt{\frac{0.4*0.6}{240}} = 0.3314$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.4 + 2.17\sqrt{\frac{0.4*0.6}{240}} = 0.4686$

0.4686 - 0.4 = 0.0686

The 97% confidence interval for the percentage of all such companies that provide such facilities on-site is (0.3314, 0.4686). The margin of error is of 0.0686 = 6.86 percentage points.