Question

Assuming that s = 18,000 is a reasonable estimate for what sample size would be needed to ensure that we could estimate the true mean salary of all production managers with more than 15 years experience within $4200 if we wish to be 95% confident?

Answer 1

Answer:

The minimum sample size required is 71.

Step-by-step explanation:

The (1 - α)% confidence interval for population mean is:

$CI=\bar x\pm z_{\alpha/2}\times \frac{\sigma}{\sqrt{n}}$

The margin of error for this interval is:

$MOE=z_{\alpha/2}\times \frac{\sigma}{\sqrt{n}}$

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,

$z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96$

*Use a z-table for the critical value.

Compute the sample size required as follows:

$MOE=z_{\alpha/2}\times \frac{\sigma}{\sqrt{n}}$

      $n=[\frac{z_{\alpha/2}\times \sigma }{MOE}]^{2}$

         $=[\frac{18000\times 1.96}{4200}]^{2}$

         $=70.56\\\approx71$

Thus, the minimum sample size required is 71.