Answer:
The 98% confidence interval for the proportion of applicants that fail the test is (0.025, 0.067).
Step-by-step explanation:
In a sample with a number n of people surveyed with a probability of a success of
, and a confidence level of
, we have the following confidence interval of proportions.
![\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}](https://tex.z-dn.net/?f=%5Cpi%20%5Cpm%20z%5Csqrt%7B%5Cfrac%7B%5Cpi%281-%5Cpi%29%7D%7Bn%7D%7D)
In which
z is the zscore that has a pvalue of
.
For this problem, we have that:
560 random tests conducted, 26 employees failed the test. This means that ![n = 560, \pi = \frac{26}{560} = 0.046](https://tex.z-dn.net/?f=n%20%3D%20560%2C%20%5Cpi%20%3D%20%5Cfrac%7B26%7D%7B560%7D%20%3D%200.046)
98% confidence level
So
, z is the value of Z that has a pvalue of
, so
.
The lower limit of this interval is:
![\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.046 - 2.33\sqrt{\frac{0.046*0.954}{560}} = 0.025](https://tex.z-dn.net/?f=%5Cpi%20-%20z%5Csqrt%7B%5Cfrac%7B%5Cpi%281-%5Cpi%29%7D%7Bn%7D%7D%20%3D%200.046%20-%202.33%5Csqrt%7B%5Cfrac%7B0.046%2A0.954%7D%7B560%7D%7D%20%3D%200.025)
The upper limit of this interval is:
![\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.046 + 2.33\sqrt{\frac{0.046*0.954}{560}} = 0.067](https://tex.z-dn.net/?f=%5Cpi%20%2B%20z%5Csqrt%7B%5Cfrac%7B%5Cpi%281-%5Cpi%29%7D%7Bn%7D%7D%20%3D%200.046%20%2B%202.33%5Csqrt%7B%5Cfrac%7B0.046%2A0.954%7D%7B560%7D%7D%20%3D%200.067)
The 98% confidence interval for the proportion of applicants that fail the test is (0.025, 0.067).