Question

The weekly earnings of students in one age group are normally distributed with a standard deviation of $110. A researcher wishes to estimate the mean weekly earnings of students in this age group. Find the sample size needed to assure with 99 percent confidence that the sample mean will not differ from the population mean by more than $10.

Answer 1

Answer: 806

Explanation:

Given that:

Standard deviation (s) = $110

Error = $10

Zcritical ; Zα/2 at 99% confidence interval = 2.58

Using the relation :

[(Zα/2 * s) / sqrt(n)] < 10

[(2.58 * 110) /sqrt(n)] = 10

283.8 / sqrt(n) = 10

283.8 = 10 * sqrt(n)

Sqrt(n) = 283.8/ 10

Sqrt(n) = 28.38

Square both sides

n = 28.38^2

n = 805.4244

n = 806

Hence, sample size = 806