Question

1. The average expenditure per student for a certain school year was $10337 with a population standard deviation of $1560. A survey for the next school year of 150 randomly selected students resulted in a sample mean of $10798. At α =0.01 level of significance, can it be concluded that the average expenditure has changed?

2. (10 pts) Suppose when you were visiting universities and deciding which to attend, the admission officers at Tennessee Wesleyan(TW) claimed TW students graduate in 4.25 years or less. You are now a student at TW and notice that a lot of students are graduating in more than 4.25 years. Therefore, you doubt that you were given accurate graduation rates by the TW admission officers. You collect data and compute graduation rates of TW graduates from the last 10 years (n=10). You obtain a mean graduation rate of 4.74 years and a standard deviation of 0.76 years. Do you have strong enough evidence to prove that TW admissions officers are not providing accurate rates at α =0.05 level of significance? Use hypothesis testing to determine your conclusion.

Answer 1

Solution:

1)

The null and alternative hypothesis are

H0 : = 10337  

H1 :     10337

The test statistic z is given by

z =

= (10798 - 10337) / (1560/150)

= 3.62

Now , observe that ,there is   sign in H1. So , the test is two tailed.

So ,

p value = P( < -3.62) + P(Z > +3.62) = 0.0001 + 0.0001 = 0.0002

p value is less than the α =0.01

Reject H₀

There is sufficient evidence to conclude that the average expenditure has changed from 10337

2)

Here , the population SD is unknown , so use t distribution.

The null and alternative hypothesis are

H0 :   4.25

H1 :   > 4.25

The test statistics t is given by ..

t =  

= (4.74 - 4.25)/(0.76/10)

= 2.039

Now ,

df = n - 1 = 10 - 1 = 9

> sign in H1 indicates that Right tailed test " (One tailed right sided)

So ,

p value = 0.0359

P value is less than the α =0.05

Reject H₀

There is strong enough evidence to prove that TW admissions officers are not providing accurate rates