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.)

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.

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

98% confidence level

So , z is the value of Z that has a pvalue of , so .

The lower limit of this interval is:

The upper limit of this interval is:

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

do you mean add both of the equations together

because if you do the answer is 10x-5

step-by-step explanation: