Two business schools, A and B, located in the same metropolitan area and they are competing for bragging rights. One of the points of competition is average salary of graduating seniors. 30 graduating seniors from A and 25 from B were surveyed. A's students had an average salary of \$62,000, and B's students had an average salary of \$67,000. Based on historical data, the population standard deviation is assumed to be \$10,000 for A and \$15,000 for B. Construct the hypotheses and conduct the appropriate tests that school B could use to claim that its students have a higher average graduating salary that A. Using a 5% level of significance, and the sample data provided, determine if school B can claim that its average graduating salary is greater than that of school A.

#### Solution

Given that the sample size $n_1 = 30$, $n_2 = 25$, sample mean $\overline{x}_1= 62000$,

$\overline{x}_2= 67000$, standard deviation $\sigma_1 = 10000$ and $\sigma_2 = 15000$.

##### Step 1 State the hypothesis testing problem

The hypothesis testing problem is

$H_0 : \mu_1 = \mu_2$ against $H_1 : \mu_1 < \mu_2$ ($\textit{left-tailed}$)

##### Step 2 Define test statistic

The test statistic is

` $$ \begin{aligned} Z=\frac{(\overline{x}_1 -\overline{x}_1)-(\mu_1 -\mu_2)}{\sqrt{\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}}}. \end{aligned} $$ `

The test statistic $Z$ follows standard normal distribution $N(0,1)$.

##### Step 3 Specify the level of significance

The significance level is $\alpha = 0.05$.

##### Step 4 Determine the critical value

As the alternative hypothesis is $\textit{left-tailed}$, the critical value of $Z$ $\text{is}$ $\text{-1.64}$.

The rejection region (i.e. critical region) is $\text{Z < -1.64}$.

##### Step 5 Computation

The test statistic for testing above hypothesis under the null hypothesis is

` $$ \begin{aligned} Z_{obs}&=\frac{(\overline{x}_1 -\overline{x}_1)-0}{\sqrt{\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}}}\\ &= \frac{62000-67000}{\sqrt{\frac{10000^2}{30}+\frac{15000^2}{25}}}\\ &= -1.424 \end{aligned} $$ `

##### Step 6 Decision

Traditional approach:

The rejection region (i.e. critical region) is $\text{Z < -1.64}$. The test statistic is $Z_{obs} =-1.424$ which falls $outside$ the critical region, we $\textit{fail to reject}$ the null hypothesis.

OR

$p$-value approach:

The test is $\text{left-tailed}$ test, so the p-value is the area to the $\text{left}$ of the test statistic ($Z_{obs}=-1.424$) is p-value = $0.0773$.

The p-value is $0.0773$ which is $\textit{greater than}$ the significance level of $\alpha = 0.05$, we $\textit{fail to reject}$ the null hypothesis.

There is no sufficient evidence to conclude that school B has average graduating salary greater than that of school A.

#### Further Reading

- Statistics
- Descriptive Statistics
- Probability Theory
- Probability Distribution
- Hypothesis Testing
- Confidence interval
- Sample size determination
- Non-parametric Tests
- Correlation Regression
- Statistics Calculators