Answer:
The minimum sample size required is 71.
Step-by-step explanation:
The (1 - α)% confidence interval for population mean is:
[tex]CI=\bar x\pm z_{\alpha/2}\times \frac{\sigma}{\sqrt{n}}[/tex]
The margin of error for this interval is:
[tex]MOE=z_{\alpha/2}\times \frac{\sigma}{\sqrt{n}}[/tex]
It is provided that the population standard deviation σ can be estimated by the sample standard deviation s.
The given information is:
s = 18000,
Margin of error = $4200.
Confidence level = 95%
The z-value for 95% confidence level is,
[tex]z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96[/tex]
*Use a z-table for the critical value.
Compute the sample size required 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{18000\times 1.96}{4200}]^{2}[/tex]
[tex]=70.56\\\approx71[/tex]
Thus, the minimum sample size required is 71.
