Question

UCF is a major Metropolitan University located in Orlando Florida. UCF is advertising their bachelor in Economics with the statistic that the starting salary of a graduate with a bachelor in economics is $ 48,500 according to Payscale (2013-13). The Director of Institutional Research at UCF is interested in testing this information. She decides to conduct a survey of 50 randomly selected recent graduate economic students. The sample mean is $43,350 and the sample standard deviation is 15,000. Alpha = 0.01

Answer 1

Answer:

There is not enough evidence to suggest that mean starting salary of a graduate with a bachelor in economics is different from $48,500.

Step-by-step explanation:

In this case we need to test whether the mean starting salary of a graduate with a bachelor in economics is $48,500.

The information provided are:

$\bar x=\ 43350\\s=\ 15000\\n=50\\\alpha =0.01$

The hypothesis for the test can be defined as follows:

H₀: The mean starting salary of a graduate with a bachelor in economics is $48,500, i.e. μ = 48500.

Hₐ: The mean starting salary of a graduate with a bachelor in economics is different from $48,500, i.e. μ ≠ 48500.

As the population standard deviation is not known we will use a t-test for single mean.

Compute the test statistic value as follows:

 $t=\frac{\bar x-\mu}{s/\sqrt{n}}$

   $=\frac{43350-48500}{15000/\sqrt{50}}\\\\=-2.42773\\\\\approx -2.43$

Thus, the test statistic value is -2.43.

Compute the p-value of the test as follows:

 $p-value=2\times P(t_{49}>-2.43)=0.0188$

*Use a t-table.

Thus, the p-value of the test is 0.0188.

Decision rule:

If the p-value of the test is less than the significance level then the null hypothesis will be rejected and vice-versa.

p-value = 0.0188 > α = 0.01

The null hypothesis will not be rejected at 1% level of significance.

Thus, concluding that there is not enough evidence to suggest that mean starting salary of a graduate with a bachelor in economics is different from $48,500.