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).

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step-by-step explanation:

answer: i wpould go with c

step-by-step explanation: