Answer:
The sample size must be approximately 120.
Step-by-step explanation:
We are given the following in the question:
Margin of error,E = 9
Standard Deviation, σ = 60
90% confidence level
[tex]\mu \pm z_{critical}\frac{\sigma}{\sqrt{n}}[/tex]
[tex]z_{critical}\text{ at}~\alpha_{0.05} = 1.645[/tex]
Sample size is given by:
[tex]n = (\dfrac{z\times s}{E})^2[/tex]
Putting values, we get,
[tex]n = (\dfrac{1.645\times 60}{9})^2\\\\n = 120.26\\n \approx 120[/tex]
Thus, the sample size must be approximately 120.
