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Marina CMI [18]
3 years ago
15

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.
Mathematics
1 answer:
djyliett [7]3 years ago
7 0

Answer:

<em>The large sample size 'n' =  199.6569≅ 200</em>

Step-by-step explanation:

<u><em>Step(i)</em></u>:-

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

<em>Standard deviation of  of salary for all baseball players for 2015 </em>

<em>   (S.D ) σ =  $5,478,384.55</em>

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

                          =  $497,000

<em>Given Margin of error of error is  =  $497,000 </em>

<em>Level of significance ∝ = 80% </em>

<em>The critical value Z₀.₂₀ = 1.282  </em>

<u><em>Step(ii):</em></u><em>-</em>

<em>Margin of error of error is  determined by</em>

<em>           </em>M.E = \frac{Z_{0.20} S.D}{\sqrt{n} }<em></em>

<em>          </em>497,000 = \frac{1.282 X 5,478,384.55}{\sqrt{n} }<em></em>

Cross multiplication , we get

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

On calculation , we get

√n = 14.13

<em>Squaring on both sides, we get</em>

n = 199.6569

<u><em>Conclusion</em></u><em>:-</em>

<em>The large sample size 'n' =  199.6569≅ 200</em>

<em>    </em>

<em></em>

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