Answer:
The sample of students required to estimate the mean weekly earnings of students at one college is of size, 3458.
Step-by-step explanation:
The (1 - α)% confidence interval for population mean (μ) is:
[tex]CI=\bar x\pm z_{\alpha/2}\times \frac{\sigma}{\sqrt{n}}[/tex]
The margin of error of a (1 - α)% confidence interval for population mean (μ) is:
[tex]MOE=z_{\alpha/2}\times \frac{\sigma}{\sqrt{n}}[/tex]
The information provided is:
σ = $60
MOE = $2
The critical value of z for 95% confidence level is:
[tex]z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96[/tex]
Compute the sample size as follows:
[tex]MOE=z_{\alpha/2}\times \frac{\sigma}{\sqrt{n}}[/tex]
[tex]n=[\frac{z_{\alpha/2}\times \sigma }{MOE}]^{2}[/tex]
[tex]=[\frac{1.96\times 60}{2}]^{2}[/tex]
[tex]=3457.44\\\approx 3458[/tex]
Thus, the sample of students required to estimate the mean weekly earnings of students at one college is of size, 3458.
