Question

The gender diversity of a large corporation’s board of directors was studied in Accounting & Finance (December 2015). In particular, the researchers wanted to know whether firms with a nominating committee would appoint more female directors than firms without a nominating committee. One of the key variables measured at each corporation was the percentage of female board directors. In a sample of $\mathbf{491}$firms with a nominating committee, the mean percentage was $\mathbf{7}\mathbf{.}\mathbf{5}\mathbf{\%}$; in an independent sample of $\mathbf{501}$firms without a nominating committee, the mean percentage was role="math" localid="1652702402701" $\mathbf{4}\mathbf{.}\mathbf{3}\mathbf{\%}$ .

a. To answer the research question, the researchers compared the mean percentage of female board directors at firms with a nominating committee with the corresponding percentage at firms without a nominating committee using an independent samples test. Set up the null and alternative hypotheses for this test.

b. The test statistic was reported as $\mathit{z}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{1}$ with a corresponding p-value of $\mathbf{0}\mathbf{.}\mathbf{0001}$. Interpret this result if $\mathbf{\text{α=0.05}}$.

c. Do the population percentages for each type of firm need to be normally distributed for the inference, part b, to be valid? Why or why not?

d. To assess the practical significance of the test, part b, construct a $\mathbf{95}\mathbf{\%}$ confidence interval for the difference between the true mean percentages at firms with and without a nominating committee. Interpret the result.

Short answer

Finance is the administration, production, and analysis of cash or assets.

Step-by-step solution

Step-by-Step Solution Step 1: (a) Set up null and alternate hypotheses

Let ${\mu }_1$ be the mean percentage of female board directors at firms with nominating committees and ${\mu }_2$ be the mean percentage of female board directors at firms without nominating committees.

Null Hypothesis: The firms with a nominating committee would appoint female directors, same as firms without nominating committee.

$H_0:{\mu }_1-{\mu }_2=0$

Alternate Hypothesis: The firms with a nominating committee would appoint more female directors than the firms without nominating committee.

$H_a:{\mu }_1-{\mu }_2>0$

(b) Interpret the result

The z value is given to be 5.51 , and the p-value is $0.0001$.

The level of significance is $0.05$.

As the p-value is less than the significance level, the null hypothesis should be rejected.

Therefore, the data provide sufficient evidence to indicate that ${\mu }_1-{\mu }_2>0$.

So, the firms with a nominating committee would appoint more female directors than the firms without nominating committee.

(c) State the reason

The population percentage for each type of firm is not normally distributed for inference part (b) to be valid because the sample sizes for both types of firms are greater than 30. So, the Central Limit Theorem is applied to the test. The assumption that the population percentages for each type of firm are normally distributed is not required for the inference.

(d) Find the confidence interval

The formula for 95% z the confidence interval is

$\overline{x_1}-\overline{x_2}\pm z_{\alpha /2}\sqrt{\left(\frac{{\displaystyle {\sigma }_1}}{{\displaystyle n_1}}+\frac{{\displaystyle {\sigma }_2}}{{\displaystyle n_2}}\right)}$

For confidence level, the level of significance is $0.05$.

So $z_{\alpha /2}=z_{0.025}=1.96$,

Now,

$$\begin{gathered} z=\frac{(\overline{x_1}-{\overline{x}}_2)}{\sqrt{\left(\frac{{{\sigma }_1}^2}{n_1}+\frac{{{\sigma }_2}^2}{n_2}\right)}} \\ 5.51=\frac{(7.5-4.3)}{\sqrt{\left(\frac{{{\sigma }_1}^2}{n_1}+\frac{{{\sigma }_2}^2}{n_2}\right)}} \\ \sqrt{\left(\frac{{{\sigma }_1}^2}{n_1}+\frac{{{\sigma }_2}^2}{n_2}\right)}=\frac{3.2}{5.51} \\ \sqrt{\left(\frac{{{\sigma }_1}^2}{n_1}+\frac{{{\sigma }_2}^2}{n_2}\right)}=0.581 \end{gathered}$$

So, the 95% z confidence interval is

$\begin{array}{l}=3.2\pm 1.96\left(0.581\right)\\ =3.2\pm 1.13876\end{array}$

Therefore, the confidence interval is $(2.06,4.34)$.