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).

solution:

as, mean is defined as sum of all the observations divided by total number of observations.

and , median is mid value of the data whether arranged in ascending order or descending order.

consider, roadsters:

2,3,4,5,6,7,8,9

mean =

also, median= , because number of observations is even, so median is mean of two mid values.

now, coming to bandits

2,3,4,5,6

mean =  

median = 4, as number of observations is odd.

so, roadsters has greater median as well mean than bandits.

→option a is true about the repairs performed on 2 types of cars:

bandits have lower median and mean than roadsters.

109 is the nearest whole number/final answer! : )