Question

A sport analyst wants to determine the mean salary of a Baseball player for 2015. He believes an estimate of this average salary using a confidence interval is sufficient. How large a sample should he take to be within $497,000 of the actual average with 80% confidence? He calculates the standard deviation of salary for all baseball players for 2015 is about $5,478,384.55. Round your answer to whole number.

Answer 1

Answer:

The large sample size 'n' =  199.6569≅ 200

Step-by-step explanation:

Step(i):-

Given Standard deviation of salary for all baseball players for 2015 is about $5,478,384.55

Standard deviation of  of salary for all baseball players for 2015

  (S.D ) σ =  $5,478,384.55

Given estimate of this average salary for all baseball players for 2015

                          =  $497,000

Given Margin of error of error is  =  $497,000

Level of significance ∝ = 80%

The critical value Z₀.₂₀ = 1.282  

Step(ii):-

Margin of error of error is  determined by

          $M.E = \frac{Z_{0.20} S.D}{\sqrt{n} }$

         $497,000 = \frac{1.282 X 5,478,384.55}{\sqrt{n} }$

Cross multiplication , we get

 $\sqrt{n} = \frac{1.282 X 5,478,384.55}{497,000 }$

On calculation , we get

√n = 14.13

Squaring on both sides, we get

n = 199.6569

Conclusion:-

The large sample size 'n' =  199.6569≅ 200