Question: Magazine subscriber salaries. Each year, the trade magazine Quality Progress publishes a study of subscribers’ salaries. One year, the 223 vice presidents sampled had a mean salary of \(116,754 and a standard deviation of \)39,185. Suppose the goal of the study is to estimate the true mean salary of all vice presidents who subscribe to Quality Progress.

a. If 2,193 vice presidents subscribe to Quality Progress, estimate the mean with an approximate 95% confidence interval.

b. Interpret the result

Let n denotes the number of vice presidents that is n=2193

The mean salary is $\overline{x}=\$116754$

The sample standard deviation is s = $ 39185

The population size is

The 95% confidence intervals for the population mean is as follows,

Consider,

$$\begin{gathered} \frac{n}{N}=\frac{223}{2193} \\ =0.1017 \end{gathered}$$

Here,$\frac{n}{N}=0.1017>0.05$the finite population correction factor should be included in the standard error calculation.

The 95% confidence intervals for the population mean by using the following formula,

$\overline{x}\pm 2\frac{s}{\sqrt{n}}\sqrt{\frac{N-n}{N}}$

Therefore

.$$\begin{gathered} \overline{x}\pm 2\frac{s}{\sqrt{n}}\sqrt{\frac{N-n}{N}}=116754\pm 2\left(\frac{39185}{\sqrt{223}}\times \sqrt{\frac{2193-223}{2193}}\right) \\ =116754\pm 4974.06 \\ =\left(111779.9,121728.1\right) \end{gathered}$$

Hence, the required confidence interval is (111779.9, 121728.1).

The confidence interval for the mean salary of the vice presidents is (111779.9, 121728.1). Thus, it is 95% confident that the true mean salary of the vice presidents falls between $111779.9 and $121728.1.