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

The sample size 'n' = 26

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 = (Z_(0.10)S.D )/(√(n) )`

Level of significance ∝ = 90 % or 10%

= 0.90 or o.10

Z₀.₁₀ = 1.645

`E = (Z_(0.10)S.D )/(√(n) )`

`5 = (1.645 X15.50 )/(√(n) )`

√n =
`(25.49)/(5) = 5.099`

Squaring on both sides , we get

n = 26

Conclusion:-

The sample size 'n' = 26