Question

The weekly earnings of students in one age group are normally
distributed with a standard deviation of 47 dollars. A researcher
wishes to estimate the mean weekly earnings of students in this age
group. Find the sample size needed to assure with 95 percent
confidence that the sample mean will not differ from the population
mean by more than 5 dollars.

Answer #1

*Solution:*

standard deviation = =47

Margin of error = E = 5

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z/2 = Z0.025 = 1.96

sample size = n = [Z/2* / E] 2

n = ( 1.96*47 /5 )2

n =339.44

Sample size = n =340

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The weekly earnings of students in one age group are
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