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

The claim is rejected

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 \\eq 48500`

n = 50

Since n is more than 30 .

So we will use Z test

x=43350

Standard deviation = 15000

`Z=(x-\mu)/((s)/(√(n)))\\Z=(43350-48500)/((15000)/(√(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