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.

806

Step-by-step 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

none of the above.

for one of the equations to be used in the process, there has to be a term in the equation since squared is .

since there is only a term in the first equation, we cannot derive a from completing the square.

solving with completing the square

i think it is 70

step-by-step explanation: