Respuesta :
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 [tex]\pi[/tex], and a confidence level of [tex]1-\alpha[/tex], we have the following confidence interval of proportions.
[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
In which
z is the zscore that has a pvalue of [tex]1 - \frac{\alpha}{2}[/tex].
The margin of error is:
The absolute value of the subtraction of one of the bounds by the estimate [tex]\pi[/tex]
For this problem, we have that:
[tex]n = 240, \pi = \frac{96}{240} = 0.4[/tex]
97% confidence level
So [tex]\alpha = 0.03[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.03}{2} = 0.985[/tex], so [tex]Z = 2.17[/tex].
The lower limit of this interval is:
[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.4 - 2.17\sqrt{\frac{0.4*0.6}{240}} = 0.3314[/tex]
The upper limit of this interval is:
[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.4 + 2.17\sqrt{\frac{0.4*0.6}{240}} = 0.4686[/tex]
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.
