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. 6 8 37 340

Answer 1

Answer:

Option D) 340

Step-by-step explanation:

We are given the following in the question:

Alpha, α = 0.05

Population standard deviation, σ = $47

Margin of error = 5

95% Confidence Interval:

$\mu \pm z_{critical}\frac{\sigma}{\sqrt{n}}$

$\text{Margin of error} = z_{critical}\frac{\sigma}{\sqrt{n}}$

Putting the values, we get,

$z_{critical}\text{ at}~\alpha_{0.05} = 1.96$

$5 = 1.96(\dfrac{47}{\sqrt{n}} )\\\\\sqrt{n} = \dfrac{47\times 1.96}{5}\\\\n = 339.443776\\\Rightarrow n \approx 340$

Thus, the correct answer is

Option D) 340