91影视

Let ${渭}_1$ be the mean percentage of female board directors at firms with nominating committees and ${渭}_2$ 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.

$H_0:{渭}_1鈭�{渭}_2=0$

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

$H_a:{渭}_1鈭�{渭}_2>0$

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 ${渭}_1鈭�{渭}_2>0$.

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

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.

The formula for 95% z the confidence interval is

$\stackrel{炉}{x_1}鈭�\stackrel{炉}{x_2}卤z_{伪/2}\sqrt{\left(\frac{{\displaystyle {蟽}_1}}{{\displaystyle n_1}}+\frac{{\displaystyle {蟽}_2}}{{\displaystyle n_2}}\right)}$

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

So $z_{伪/2}=z_{0.025}=1.96$,

Now,

$$\begin{gathered} z=\frac{(\stackrel{炉}{x_1}鈭�{\stackrel{炉}{x}}_2)}{\sqrt{\left(\frac{{{蟽}_1}^2}{n_1}+\frac{{{蟽}_2}^2}{n_2}\right)}} \\ 5.51=\frac{(7.5鈭�4.3)}{\sqrt{\left(\frac{{{蟽}_1}^2}{n_1}+\frac{{{蟽}_2}^2}{n_2}\right)}} \\ \sqrt{\left(\frac{{{蟽}_1}^2}{n_1}+\frac{{{蟽}_2}^2}{n_2}\right)}=\frac{3.2}{5.51} \\ \sqrt{\left(\frac{{{蟽}_1}^2}{n_1}+\frac{{{蟽}_2}^2}{n_2}\right)}=0.581 \end{gathered}$$

So, the 95% z confidence interval is

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

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