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.

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

Cross multiplication , we get

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

the answer is 510

step-by-step explanation:

15 • 13 = 195

15 • 12 = 180

15 • 5 = 75

12 • 5 = 60

195 + 180 = 375

375 + 75 = 450

450 + 60 = 510