Question

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?

Answer 1

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

                $E = \frac{Z_{0.10}S.D }{\sqrt{n} }$

 Level of significance    ∝ = 90 % or 10%

                                            = 0.90 or o.10

Z₀.₁₀ = 1.645

                                 $E = \frac{Z_{0.10}S.D }{\sqrt{n} }$

                             $5 = \frac{1.645 X15.50 }{\sqrt{n} }$

                        √n =  $\frac{25.49}{5} = 5.099$

Squaring on both sides , we get

                        n = 26

Conclusion:-

The sample size 'n' = 26