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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.

Respuesta :

Answer:

The margin of error  = 2.13

Step-by-step 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

[tex]M.E = Z_{\alpha } \frac{S.D}{\sqrt{n} }[/tex]

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

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