Question

A researcher randomly sampled 30 graduates of an MBA program and their starting salaries. It is found the sample mean of the starting salary is $60,960/year with a sample standard deviation of $12,987.

a. Please test the hypothesis that the starting salaries of MBA graduate is more than $50,000/year at 5% level of significance.

b. Further breakdown of the data shows that Male Female
?? = 12 ?? = 18

̅̅
? = 65,000.00 ? = 58,266.70

??

? = 12,751.29 ? = 13,270.34 ??

Please test whether male graduates enjoy higher starting salaries than female graduates at 5% level of significance.

c. An officer from the career office claimed that there is strong gender discrimination in the market, and the starting salary of a male graduate is $10,000 more than that of a female graduate. Do you think that the officer’s claim is true?

Answer 1

a) Let \(\mu\) denotes the average starting salaries of MBA graduates.

Let \(\mu\) denotes the population mean. Hypotheses :

To test null hypothesis \(H_{0}: \mu=50000\) against alternative hypothesis \(H_{1}: \mu>50000\) Let \(\bar{x}\) denotes the sample mean, s denotes the sample standard deviation and \(\mathrm{n}\) denotes the sample size. Here,

$$ \bar{x}=60960.000, s^{2}=168662169.000000, s=12987.00, n=30 $$

The test statistic can be written as:

$$ t=\frac{\sqrt{n}(\bar{x}-50000)}{s} $$

which under \(H_{0}\) follows a t distribution with \(\mathrm{n}-1\) df. Decision rule / Rejection region :

We reject \(H_{0}\) at 0.05 level of significance if \(\mathrm{P}\) -value \(<0.05\) or if \(t_{o b s}>t_{0.05, n-1}=1.699127\)

Now, \(\underline{\text { Value of the test statistic : }}\) The value of the test statistic

$$ t_{o b s}=\frac{\sqrt{30}(60960.0000-50000)}{12987.0000}=4.622345 \approx 4.62 $$

associated degrees of freedom \(=29\) The critical value \(t_{\text {critical}}=t_{0.05, n-1}=t_{0.05,29}=1.699127 \approx 1.699\)

$$ P-\text {value}=P\left(t_{29}>4.622345\right)=1-P\left(t_{29} \leq 4.622345\right)=1-0.999964=0.000036 \approx 0.0000 $$

Conclusion:

since \(\mathrm{p}\) -value \(<0.05\) and \(t_{o b s}>t_{\text {critical}}=1.699127,\) so we reject \(H_{0}\) at 0.05 level of significance Hence, we can conclude that population mean is significantly greater than 50000

b)

Let \(\mu_{1}, \mu_{2}\) denote average starting salaries of male graduates and female graduates respectively.

Hypotheses :

To test null hypothesis \(H_{0}: \mu_{1}-\mu_{2}=0\) against alternative hypothesis \(H_{1}: \mu_{1}-\mu_{2}>0\) This is a one tailed right sided test as the alternative hypothesis contains ' \(^{\prime}\) ' sign. Here,

$$ \overline{x_{1}}=65000.00, s_{1}=12751.29, n_{1}=12, \overline{x_{2}}=58266.70, s_{2}=13270.34, n_{2}=18 $$

The test statistic can be written as:

$$ t=\frac{\left(\overline{x_{1}}-\overline{x_{2}}\right)-0}{\sqrt{\frac{s_{1}^{2}}{n_{1}}+\frac{s_{2}^{2}}{n_{2}}}} $$

which under \(H_{0}\) follows a t distribution with 24 df.

where

$$d  f=\frac{\left(\frac{s_{1}^{2}}{n_{1}}+\frac{s_{2}^{2}}{n_{2}}\right)^{2}}{\frac{\left(\frac{s_{1}^{2}}{n_{1}}\right)^{2}}{n_{1}-1}+\frac{\left(\frac{s_{2}^{2}}{n_{2}}\right)^{2}}{n_{2}-1}}=\frac{\left(\frac{12751.29^{2}}{12}+\frac{13270.34^{2}}{18}\right)^{2}}{\frac{\left(\frac{12751.22^{2}}{12-1}\right)^{2}}{12-1}+\frac{\left(\frac{13270.34^{2}}{18}\right)^{2}}{18-1}} \approx 24  $$

and standard error for the difference between means

$$  SE=\sqrt{\frac{s_{1}^{2}}{n_{1}}+\frac{s_{2}^{2}}{n_{2}}}=\sqrt{\frac{12751.29^{2}}{12}+\frac{13270.34^{2}}{18}}=4830.430270  $$

Decision rule / Rejection region :

We reject \(H_{0}\) at 0.05 level of significance if \(\mathrm{P}\) -value \(<0.05\) or if \(t_{\text {stat}}>t_{0.05, d f}\)

Now, Value of the test statistic :

The value of the test statistic is

$$  t_{\text {stat }}=\frac{(65000.00-58266.70)-0}{\sqrt{\frac{12751.29^{2}}{12}+\frac{13270.34^{2}}{18}}}=1.393934 \approx 1.394  $$

associated degrees of freedom \(=24\) The critical value \(=t_{c r i t i c a l}=t_{0.05,24}=1.710882 \approx 1.711\)

$$  P-\text {value}=P\left(t_{24}>t_{\text {stat}}\right)=P\left(t_{24}>1.393934\right)=1-P\left(t_{24} \leq 1.393934\right)=1-0.911945=0.088055 \approx 0.0881  $$

Conclusion:

since p-value \(>0.05\) and \(t_{\text {stat}} \not \supset t_{\text {critical}}=1.711,\) so We fail to reject \(H_{0}\) at 0.05 level of significance Hence, we can conclude that mean for first population is not significantly greater than the mean for second population.

There is not sufficient evidence to support the claim that male graduates enjoy higher starting salaries than female graduates.

C)

Hypotheses :

To test null hypothesis \(H_{0}: \mu_{1}-\mu_{2} \leq 10000\) against alternative hypothesis \(H_{1}: \mu_{1}-\mu_{2}>10000\) This is a one tailed right sided test as the alternative hypothesis contains ' \(>\) ' sign. Here,

$$ \overline{x_{1}}=65000.00, s_{1}=12751.29, n_{1}=12, \overline{x_{2}}=58266.70, s_{2}=13270.34, n_{2}=18 $$

The test statistic can be written as:

$$ t=\frac{\left(\overline{x_{1}}-\overline{x_{2}}\right)-10000}{\sqrt{\frac{s_{1}^{2}}{n_{1}}+\frac{s_{2}^{2}}{n_{2}}}} $$

which under \(H_{0}\) follows a t distribution with 24 df. where

$$ d f=\frac{\left(\frac{s_{1}^{2}}{n_{1}}+\frac{s_{2}^{2}}{n_{2}}\right)^{2}}{\frac{\left(\frac{s_{1}^{2}}{n_{1}}\right)^{2}}{n_{1}-1}+\frac{\left(\frac{s_{2}^{2}}{n_{2}}\right)^{2}}{n_{2}-1}}=\frac{\left(\frac{12751.29^{2}}{12}+\frac{13270.34^{2}}{18}\right)^{2}}{\frac{\left(\frac{12751.29^{2}}{12-1}\right)^{2}}{12-1}+\frac{\left(\frac{13270.34^{2}}{18}\right)^{2}}{18-1}} \approx 24 $$

and standard error for the difference between means

$$ S E=\sqrt{\frac{s_{1}^{2}}{n_{1}}+\frac{s_{2}^{2}}{n_{2}}}=\sqrt{\frac{12751.29^{2}}{12}+\frac{13270.34^{2}}{18}}=4830.430270 $$

Decision rule / Rejection region :

We reject \(H_{0}\) at 0.05 level of significance if \(\mathrm{P}\) -value \(<0.05\) or if \(t_{\text {stat}}>t_{0.05, d f}\)

Now, Value of the test statistic :

The value of the test statistic is

$$ t_{\text {stat }}=\frac{(65000.00-58266.70)-10000}{\sqrt{\frac{12751.29^{2}}{12}+\frac{13270.34^{2}}{18}}}=-0.676275 \approx-0.676 $$

associated degrees of freedom \(=24\) The critical value \(=t_{c r i t i c a l}=t_{0.05,24}=1.710882 \approx 1.711\)

$$ P-\text { value }=P\left(t_{24}>t_{\text {stat}}\right)=P\left(t_{24}>-0.676275\right)=1-P\left(t_{24} \leq-0.676275\right)=1-0.252666=0.747334 \approx 0.7473 $$

Conclusion:

since \(p\) -value \(>0.05\) and \(t_{\text {stat }} \not \supset t_{\text {critical}}=1.711\), so We fail to reject \(H_{0}\) at 0.05 level of significance Hence, we can conclude that mean for first population is not significantly greater than the mean for second population.

There is not sufficient evidence to support the officer's claim that the starting salary of a male graduate is \(\$ 10,000\) more than that of a female graduate.