Question

How does an MBA major affect the number of job offers received? An MBA student randomly sampled four recent graduates, one each in finance, marketing, and management, and asked them to report the number of job offers. Can we conclude at the \(5 \%\) significance level that there are differences in the number of job offers between the three MBA majors?

Answer 1

The problem objective is to compare three populations (Finance, Marketing, and Management categories). The parameters are the three population means \(\mu_{1}, \mu_{2}\), and \(\mu_{3} .\) The null hypothesis states that there is no differences between the population means. Hence,

Let \(x_{i j}\) be the \(i^{\text {thobservation of the }} j^{\text {th }}\) sample.

Let \(n_{j}\) ne the numbe rof obseravtions in the sample drawn from the \(j^{\text {th }}\) population.

Let the sample meam of the \(j^{\text {th }}\) sample be \(\bar{x}_{j}=\frac{\sum_{i=1}^{n_{j}} x_{i j}}{n_{j}}\).

Let

\(n_{j}=4\), for \(j=1,2,3 \quad\) (or) \(n_{1}=n_{2}=n_{3}=4\)

The sample mean of the Finance is,

\(\begin{aligned} \bar{x}_{1} &=\frac{3+1+4+1}{4} \\ &=2.25 \end{aligned}\)

The sample mean of the Marketing is,

\(\bar{x}_{2}=\frac{\sum_{i=1}^{n_{j}} x_{i j}}{n_{j}}\)

\(=\frac{1+5+3+4}{4}\)

\(=3.25\)

The sample mean of the Management is,

\(\bar{x}_{3}=\frac{\sum_{i=1}^{n_{j}} x_{i j}}{n_{j}}\)

\(=\frac{8+5+4+6}{4}\)

\(=5.75\)

\(H_{0}: \mu_{1}=\mu_{2}=\mu_{3}\)

The analysis of variance determines whether there is enough statistical evidence to show that the numm hypothesis is false. Con sequently, the alternative hypothesis will always specify the following:

\(H_{1}\) : At least two means differ

Let

\(\begin{aligned} n &=n_{1}+n_{2}+n_{3} \\ &=4+4+4 \\ &=12 \end{aligned}\)

The grand mean is,

\(\overline{\bar{x}}=\frac{\sum_{j=1}^{k} \sum_{i=1}^{n_{i}} x_{i j}}{n}\)

\(=\frac{3+1+4+1+1+5+3+4+8+5+4+6}{12}\)

\(=3.75\)

Let the number of populations be \(k=3\).

The sum of squares for treatments is,

\(\begin{aligned} S S T &=\sum_{j=1}^{k} n_{j}\left(\bar{x}_{j}-\overline{\bar{x}}\right)^{2} \\ &=\sum_{j=1}^{3} n_{j}\left(\bar{x}_{j}-3.75\right)^{2} \\ &=4(2.25-3.75)^{2}+4(3.25-3.75)^{2}+4(5.75-3.75)^{2} \\ &=26 \end{aligned}\)

The sum of squares for error is,

\(S S E=\sum_{j=1}^{k} \sum_{i=1}^{n_{j}}\left(x_{i j}-\bar{x}_{j}\right)^{2}\)

\(\begin{aligned} &=\sum_{i=1}^{n_{1}}\left(x_{i 1}-\bar{x}_{1}\right)^{2}+\sum_{i=1}^{n_{2}}\left(x_{i 2}-\bar{x}_{2}\right)^{2}+\sum_{i=1}^{n_{1}}\left(x_{i 3}-\bar{x}_{3}\right)^{2} \\ &=\left[\begin{array}{l}\left\{(3-2.25)^{2}+(1-2.25)^{2}+(4-2.25)^{2}+(1-2.25)^{2}\right\}+7 \\ \left\{(1-3.25)^{2}+(5-3.25)^{2}+(3-3.25)^{2}+(4-3.25)^{2}\right\}+ \\ \left\{(8-5.75)^{2}+(5-5.75)^{2}+(4-5.75)^{2}+(6-5.75)^{2}\right\}\end{array}\right] \\ &=24.25 \end{aligned}\)

Mean square for treatments is,

\(\begin{aligned} M S T &=\frac{S S T}{k-1} \\ &=\frac{26}{3-1} \\ &=13 \end{aligned}\)

Mean square for error is,

\(M S E=\frac{S S E}{n-k}\)

\(=\frac{24.25}{12-3}\)

\(=\frac{24.25}{9}\)

\(=2.69444\)

The test statistic value will be

\(F=\frac{M S T}{M S E}\)

\(=\frac{13}{2.69444}\)

\(=4.82474\)

\(\approx 4.82\)

The numerator degrees of freedom is,

\(\begin{aligned} v_{1} &=k-1 \\ &=3-1 \\ &=2 \end{aligned}\)

The denominator degrees of freedom is,

\(v_{2}=n-k\)

\(-12-2\)

\(=9\)

Let the level of significance be \(\alpha=0.05\).

Critical value: Using the Excel function formula, the F-critical value is,

\(\begin{aligned} F_{\alpha, k-1, n-k} &=F_{0.05,2.9} \\ &=[=\operatorname{FINV}(0.05,2,9)] \\ &=4.256494729 \\ & \approx 4.26 \end{aligned}\)

P-value: Using the Excel function formula, the \(\mathrm{p}\) - value is,

\(F_{\alpha, k-1, n-k}=F_{0.05,29}\)

\(=[=\mathrm{FDIST}(4.824742,2,9)]\)

\(=0.037678235\)

\(\approx 0.0377\)

Rejection region:

Reject the null hypothesis, if \(F>F_{\alpha, k-1, n-k}(\) or \(F>4.26)\) otherwise do not reject \(H_{0}\).

Conclusion:

The test statistic value \(F=4.82\) is greater than the critical value. Reject the null hypothesis. Similarly, the p-value is less than the significance level of \(0.05 .\) Reject the null hypothesis. Therefore, it can be concluded that there are differences in the number of job offers between the three MBA majors.