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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

Expert verified

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

01

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:

P1 =

= 0.1603

PTwo=

=0.05

02

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:

H0: (P1 鈥P2) = 0

Versus

Ha:(笔鈧-笔鈧)&驳迟;0

This is a right-tailed test.

Also, we set a = 0.05 level of significance.

03

Sample distribution of (P1-P2)

As per the requirement, we find the following value.

n1p1= x1

= 209(>15)

n2q2= n1 鈥 x1

= 1304 - 209

= 1095(>15)

n2q2= x2

= 356(>15)

n2q2= n2鈥 x2

= 7120 鈥 356

=6764(>15)

Thus, the given samples are large. Therefore, the sampling distribution of (P1-P2) will be approximately normal.

04

Test statistics

The statistical tests for the null hypothesis are as follows:

z=

where = =

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.

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Test and CI for two Variances: Content vs Site

Method

Null hypothesis 12=1

Alternative hypothesis 121

F method was used. This method is accurate for normal data only.

Statistics

Site N St Dev Variance 95% CI for St Devs

1 25 3.067 9.406 (2.195,4.267)

2 25 3.339 11.147 (2.607,4.645)

Ratio of standard deviation =0.191

Ratio of variances=0.844

95% Confidence Intervals

Method CI for St Dev Ratio CI Variance Ratio

F (0.610, 1.384) (0.372, 1.915)

Tests

Method DF1 DF2 Test statistic p-value

F 24 24 0.84 0.681

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