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:

The claim is rejected

Step-by-step explanation:

Claim: 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).

Null hypothesis: $H_0: \mu = 48500$

Alternate hypothesis :$H_a : \mu \neq 48500$

n = 50

Since n is more than 30 .

So we will use Z test

x=43350

Standard deviation = 15000

$Z=\frac{x-\mu}{\frac{s}{\sqrt{n}}}\\Z=\frac{43350-48500}{\frac{15000}{\sqrt{50}}}\\Z=-2.42$

Refer the z table

p value = 0.00776

$\alpha = 0.01$

p value < $\alpha$

So, We are failed to accept null hypothesis

Hence The claim is rejected