A researcher wishes to estimate within $5 the average repair cost of for a refrigerator. If she wishes to be 90% confident, how large of a sample would be necessary if the population standard deviation is $15.50?

Respuesta :

Answer:

The sample size 'n' = 26

Step-by-step explanation:

Explanation:-

Given a researcher wishes to estimate within $5 the average repair cost of for a refrigerator

Given estimate "E" = $5

The  estimate within  the average is determined by

                [tex]E = \frac{Z_{0.10}S.D }{\sqrt{n} }[/tex]

 Level of significance    ∝ = 90 % or 10%

                                            = 0.90 or o.10

Z₀.₁₀ = 1.645

                                 [tex]E = \frac{Z_{0.10}S.D }{\sqrt{n} }[/tex]

                             [tex]5 = \frac{1.645 X15.50 }{\sqrt{n} }[/tex]

                        √n =  [tex]\frac{25.49}{5} = 5.099[/tex]

Squaring on both sides , we get

                        n = 26

Conclusion:-

The sample size 'n' = 26