Question

The number of job applications submitted before landing an interview are normally distributed with a population standard deviation of 4 applications and an unknown population mean. A random sample of 19 job seekers is taken and results in a sample mean of 55 applications. The confidence intervalis (52.87.57.14). What is the margin of error? Round to two decimal places.

Answer 1

The margin of error = 2.13

Explanation:

Explanation:-

Given random sample size 'n' =19

mean of the sample(x⁻) = 55 applicants

Given standard deviation of the Population(S.D) = 4

Given confidence intervals are

((52.87.57.14)

we know that The Margin of error is determined by

`M.E = Z_(\alpha ) (S.D)/(√(n) )`

The confidence intervals are determined by

(x⁻ - M.E , x⁻+ M.E)

Step(ii):-

Given confidence intervals are

((52.87.57.14)

Now equating

(x⁻ - M.E , x⁻+ M.E) = ((52.87 , 57.14)

Given mean of the sample x⁻ = 55

( 55 - M.E , 55 + M.E) =((52.87.57.14)

Equating

55 - M.E = 52.87

M.E = 55 - 52.87

M.E = 2.13

Final answer:-

The margin of error = 2.13