Question

HighTech Inc. randomly tests its employees about company policies. Last year in the 560 random tests conducted, 26 employees failed the test. Develop a 98% confidence interval for the proportion of applicants that fail the test. (Round your answers to 3 decimal places.)

Answer 1

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

For this problem, we have that:

560 random tests conducted, 26 employees failed the test. This means that $n = 560, \pi = \frac{26}{560} = 0.046$

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 $Z = 2.33$.

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.046 - 2.33\sqrt{\frac{0.046*0.954}{560}} = 0.025$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.046 + 2.33\sqrt{\frac{0.046*0.954}{560}} = 0.067$

The 98% confidence interval for the proportion of applicants that fail the test is (0.025, 0.067).