Question

Entrepreneurial careers of MBA alumni. Are African American MBA scholars more likely to begin their careers as entrepreneurs than white MBA scholars? This was a question of interest to the Graduate Management Admission Council (GMAC). GMAC Research Reports (Oct. 3, 2005) published the results of a check of MBA alumni. Of the African Americans who responded to the check, 209 reported their employment status after scaling as tone-employed or a small business proprietor. Of the whites who responded to the check, 356 reported their employment status after scaling as tone-employed or a small business proprietor. Use this information to answer the exploration question.

Short answer

We have enough evidence to conclude that African American MBA students are more likely to begin their careers as an entrepreneur than White MBA students.

Step-by-step solution

Step-by-Step Solution Step 1: Sample the proportion of the two groups

in a survey of 1304 African American MBA alumnus, 209 reported their employment status as self-employed or a small business owner. Of 7,120 whites, 356 reported their employment status as self-employed or a small business owner.

Therefore, we have the sample proportion of the two groups as follows:

$\begin{array}{l}{\overline{P}}_1=\frac{209}{1304}\\ =0.1603\\ \\ {\overline{P}}_2=\frac{356}{7120}\\ =0.05\end{array}$

Set the null and alternative hypotheses

Here we have to test whether there is enough evidence to claim that African American MBA students are more likely to begin their careers as entrepreneurs than White MBA students.

As a result, we establish the null as well as alternative hypotheses as shown in:

H_0: (P_{1 –}P₂) = 0

Versus

H_a:(P₁-P₂)>0

This is a right-tailed test.

Also, we set a = 0.05 level of significance.

Sample distribution of (P1-P2)

As per the requirement, we find the following value.

n₁p₁= x₁ $\left(sinceP1=\frac{x_1}{n_1}\right)$

= 209(>15)

n₂q₂= n₁ – x_{1 $\left(\sin ce{\overline{q}}_1=\frac{n_1-x_1}{n_1}\right)$}

= 1304 - 209

= 1095(>15)

n₂q₂= x_{2 $\left(\sin ce{\overline{P}}_2=\frac{x_2}{n_2}\right)$}

= 356(>15)

n₂q₂= n₂– x₂

= 7120 – 356 $\left(\sin ce{\overline{q}}_2=\frac{n_2-x_2}{n_2}\right)$

=6764(>15)

Thus, the given samples are large. Therefore, the sampling distribution of (P_1-P₂) will be approximately normal.

Test statistics

The statistical tests for the null hypothesis are as follows:

$Z=\frac{\overline{P_1}-{\overline{P}}_2}{{\sigma }_{(\overline{P},-{\overline{p}}_2)}}$

where ${\sigma }_{(p_1-p_2)}=\sqrt{\overline{p}\overline{q}(\frac{1}{n_1}+\frac{1}{n})}=\overline{P}=\frac{x_1+x_2}{n_1+n_2}$

Using MINITAB, we conduct the above test in the following steps.

Step 1: Select 2 Proportions... from the Basic Statistics of Stat ribbon.

Step 2: Enter the Events and Trails as 209 and 1304 for the First sample and 356 and 7120 for the second sample in the Summarized data.

Step 3: Click Options..., enter Confidence level as 95.0, Test difference as 0.0, and Alternative as greater than.

Step 4: Check Use pooled estimate of p for test" option.

Step 5: Click Ok.

Test and CI for two proportions

Thus, the resultant output is generated as follows.

Sample	x	n	Sample p
1	209	1304	0.160276
2	356	7120	0.050000

Difference - p (1) - p (2)

Estimate for difference: 0.110276

95% lower bound for difference: 0.0930339

Test for difference = 0 (VA > 01): Z=14.64 P-Value = 0.000

Fisher's exact test: P-Value = 0.000

Test statistics

From the above output, the test statistics are obtained as = 14.64, and the corresponding

P-value is obtained as 0.000.

As the P-value is less than a = 0.05 level of significance, we reject the null hypothesis. Hence, we have enough evidence to conclude that African American MBA students are more likely to begin their careers as an entrepreneur than White MBA students.

Final answer

We have enough evidence to conclude that African American MBA students are more likely to begin their careers as an entrepreneur than White MBA students.