Question

Unethical corporate conduct. How complicit are entrylevel accountants in carrying out an unethical request from their superiors? This was the question of interest in a study published in the journal Behavioral Research in Accounting (July 2015). A sample of 86 accounting graduate students participated in the study. After asking the subjects to perform what is clearly an unethical task (e.g., to bribe a customer), the researchers measured each subject’s intention to comply with the unethical request score. Scores ranged from -1.5 (intention to resist the unethical request) to 2.5 (intention to comply with the unethical request). Summary statistics on the 86 scores follow: $\overline{\mathbf{x}}\mathbf{=}\mathbf{2}\mathbf{.42}\mathbf{,}\mathit{s}\mathbf{=}\mathbf{2}\mathbf{.84}$.

a. Estimate $\mathit{\mu }$, the mean intention to comply score for the population of all entry-level accountants, using a 90% confidence interval.

b. Give a practical interpretation of the interval, part a.

c. Refer to part a. What proportion of all similarly constructed confidence intervals (in repeated sampling) will contain the true value of $\mathit{\mu }$?

d. Compute the interval, $\overline{\mathbf{x}}\mathbf{\pm }\mathbf{2}\mathit{s}$. How does the interpretation of this interval differ from that of the confidence interval, part a?

Short answer

1. The 90% confidence interval for the mean intension to comply score for the mean intension to comply score for the population of all entry-level accounts$\left(\mu \right)$ lies between 1.916 and 2.924.

1. There is 90% confident that the mean intention to comply score for the population of all entry-level accounts$\left(\mu \right)$ lies between 1.916 and 2.924.

1. About 90% of all similarly constructed confidence intervals will contain the true value of population mean$\left(\mu \right)$ in repeated sampling.

1. The interval$\overline{x}\pm 2s$ is $\left(-3.26,8.1\right)$.

Step-by-step solution

Given information

Sample size n=86, the mean$\overline{x}=2.42$ and the standard deviation s=2.84.

Estimating the mean μ

Here, the confidence coefficient is 0.90. Therefore,

$\begin{array}{c}\left(1-\alpha \right)=0.90\\ \alpha =0.10\\ \frac{\alpha }{2}=0.05\end{array}$

From table, the required$z_{0.05}$value for 90% confidence level is 1.645.

The 90% confidence interval is obtained is obtained below:

$\begin{array}{c}\overline{\mathrm{x}}\pm {\mathrm{z}}_{\frac{\mathrm{\alpha }}{2}}\left({\mathrm{\sigma }}_{\overline{\mathrm{x}}}\right)=2.42\pm 1.645\left(\frac{2.84}{\sqrt{86}}\right)\\ =2.42\pm 0.504\\ =\left(2.42+0.504,2.42-00.504\right)\end{array}$

That is $\left(1.916,2.924\right)$

Thus, the 90% confidence interval for the mean intension to comply score for the population of all entry-level accounts$\left(\mu \right)$ lies between 1.916 and 2.924.

Interpretation

There is 90% confident that the mean intention to comply score for the population of all entry-level accounts$\left(\mu \right)$ lies between 1.916 and 2.924.

Calculating the proportion

About 90% of all similarly constructed confidence intervals will contain the true value of population mean$\left(\mu \right)$ in repeated sampling.

Computing the confidence interval

$\begin{array}{c}\overline{\mathrm{x}}\pm 2\mathrm{s}=2.42\pm 2\left(2.84\right)\\ =2.42\pm 5.68\\ =\left(2.42+5.68,2.42-5.68\right)\\ =\left(-3.26,8.1\right)\end{array}$

Thus, the interval $\overline{\mathrm{x}}\pm 2\mathrm{s}$is $\left(-3.26,8.1\right)$.

From, part a. the 95% confidence interval provides the range of values for the population mean $\left(\mu \right)$. But, the interval$\overline{\mathrm{x}}\pm 2\mathrm{s}$ provides the range of actual values of X.