Question

Corporate sustainability of CPA firms. Refer to the Business and Society (March 2011) study on the sustainability behaviours of CPA corporations, Exercise 1.28 (p. 51). Corporate sustainability, recall, refers to business practices designed around social and environmental considerations. The level of support senior managers has for corporate sustainability was measured quantitatively on a scale ranging from 0 to 160 points. The study provided the following information on the distribution of levels of support for sustainability:$\mathbf{\mu }\mathbf{=}\mathbf{68}$ , $\mathbf{\sigma }\mathbf{=}\mathbf{27}$. Now consider a random sample of 45 senior managers and let x represent the sample mean level of support.

a. Give the value of ${\mu }_{\overline{x}}$, the mean of the sampling distribution of$\overline{x}$ , and interpret the result.

b. Give the value of${\sigma }_{\overline{x}}$ , the standard deviation of the sampling distribution of $\overline{x}$, and interpret the result.

c. What does the Central Limit Theorem say about the shape of the sampling distribution of$\overline{x}$ ?

d. Find $P\left(\overline{x}>65\right)$.

Short answer

a..The mean 68 indicates the average value of sample mean level of support.

b. 4.0249 denotes the amount of spread of the distribution from the mean.

c. The sampling distribution of $\overline{x}$is of normal shape since$n=45>30$

d. The probability that $\overline{x}$greater than 65 is 0.7734.

Step-by-step solution

Given information

A random sample of 45 senior managers is selected. So, $\mu =68,\sigma =27$and the sample size is 45

Calculating the mean of sampling distribution of x¯

a. Let $\overline{x}$denote the mean salary for the sample which is distributed with mean ${\mu }_x=\mu$and standard deviation$\sigma =\frac{\sigma }{\sqrt{n}}$

Hence,

$\begin{array}{rcl}{\mu }_x& =& \mu \\ & =& 68\\ & & \end{array}$

The mean indicates the central position of the data. Here, the mean 68 indicates the average value of sample mean level of support.

Calculating the standard deviation of sampling distribution of x¯

b.

Let $\overline{x}$denote the mean salary for the sample, which is distributed with mean ${\mu }_x=\mu$and standard deviation$\sigma =\frac{\sigma }{\sqrt{n}}$

Hence,

$\begin{array}{rcl}\sigma & =& \frac{27}{\sqrt{45}}\\ & =& 4.0249\\ & & \end{array}$

Thus, 4.0249 denotes the amount of spread of the distribution from the mean.

Finding the shape of sampling distribution of x¯ 

c.

According to Central Limit Theorem, it n is larger, then$\overline{x}$ follows normal distribution with mean${\mu }_x=\mu$ and standard deviation${\sigma }_x=\frac{\sigma }{\sqrt{n}}$

Hence, the sampling distribution of$\overline{x}$ is of normal shape since$n=45>30$

Calculating the probability 

d. Let,

$\begin{array}{rcl}P\left(\overline{x}>65\right)& =& P\left(\frac{\overline{x}-\mu }{\frac{\sigma }{\sqrt{n}}}>\frac{65-68}{4.0249}\right)\\ & =& P\left(z>\frac{-3}{4.0249}\right)\\ & =& P\left(z>-0.75\right)\\ & & \end{array}$

$$\begin{gathered} =0.5+0.2734 \\ =0.7734 \end{gathered}$$

Hence, the probability that $\overline{x}$greater than 65 is 0.7734.