Accountants’ salary survey. Each year, ManagementAccountingreports the results of a salary survey of themembers of the Institute of Management Accountants(IMA). One year, the 2,112 members responding had a salarydistribution with a 20th percentile of \(35,100; a medianof \)50,000; and an 80th percentile of \(73,000.

1. Use this information to determine the minimum samplesize that could be used in next year’s survey toestimate the mean salary of IMA members towithin\)2,000 with 98% confidence. [Hint: To estimate s,first applyChebyshev’s Theorem to find ksuch thatat least 60% of the data fall within kstandard deviations of $\mathbf{\mu }$. Then find data-custom-editor="chemistry" $\mathbf{s}\mathbf{\approx }$(80^thpercentile–20^thpercentile)/2k.]
2. Explain how you estimated the standard deviation requiredfor the sample size calculation.
3. List any assumptions you make.

A survey was conducted by the management accounting to the members of the Institute of Management Accountants. Here the number of members n=2112. The median of salary distribution is 50000. The 20^th percentile of the salary is 35100. The 80^th percentile of the salary distribution is 73000.

a.

Let’s consider by the Chebyshev’s theorem, at least$\left(1-\frac{1}{{\mathrm{k}}^2}\right)$ of the observations fall within the k standard deviations of the mean.

So,

$\begin{array}{rcl}\left(1-\frac{1}{{\mathrm{k}}^2}\right)& =& 0.60\\ & \Rightarrow & \frac{1}{{\mathrm{k}}^2}=1-0.60\\ & \Rightarrow & {\mathrm{k}}^2=\frac{1}{0.40}\\ & \Rightarrow & \mathrm{k}=\sqrt{2.5}\\ & \Rightarrow & k=1.58\end{array}$

Therefore,

$\begin{array}{rcl}\mathrm{s}& =& \frac{{80}^{\mathrm{th}}\mathrm{percentile}-{20}^{\mathrm{th}}\mathrm{percentile}}{2\mathrm{k}}\\ & =& \frac{73000-35100}{2\times 1.58}\\ & =& 11985.32\end{array}$

Thus, the sample standard deviation is 11985.32.

Now for a confidence coefficient,

$\begin{array}{rcl}\left(1-\mathrm{\alpha }\right)& =& 0.98\\ & \Rightarrow & \mathrm{\alpha }=1-0.98\\ & \Rightarrow & \mathrm{\alpha }=0.02\\ & \Rightarrow & \frac{\mathrm{\alpha }}{2}=0.01\end{array}$

Therefore, the z-statistics is

$\begin{array}{rcl}{\mathrm{Z}}_{1-\frac{\mathrm{\alpha }}{2}}& =& {\mathrm{Z}}_{1-0.01}\\ & =& {\mathrm{Z}}_{0.99}\\ & =& 2.33\end{array}$

So, the required minimum sample size is,

$\begin{array}{rcl}\mathrm{n}& =& \left[\frac{{\mathrm{Z}}_{0.99}{\left(\mathrm{s}\right)}^2}{\mathrm{ME}}\right]\\ \mathrm{n}& =& \left[\frac{2.33\times {\left(11985.32\right)}^2}{2000}\right]\\ & \approx & 195\end{array}$

Thus, the required sample size is approximately 195.

b.

Referring to the first part of part a.

At first, let’s consider there at least$\left(1-\frac{1}{{\mathrm{k}}^2}\right)$ of the observations fall within k standard deviations of the mean by Chebyshev’s inequality. Then calculate the value of k and then calculate the sample standard deviation by the formula,

$\mathrm{s}=\frac{{80}^{\mathrm{th}}\mathrm{percentile}-{20}^{\mathrm{th}}\mathrm{percentile}}{2\mathrm{k}}$.

c.

There is considered only one assumption that is the estimate of the standard deviation is accurate.