Question

A random sample of 12 second-year university students enrolled in a business statistics course was drawn. At the course's completion, each student was asked how many hours he or she spent doing homework in statistics. The data are listed below.

281726262620292823271820282626292318172620282720

It is known that the population standard deviation is 9. The instructor has recommended that students devote 2 hours per week for the duration of the 12-week semester, for a total of 24 hours. Test to determine whether there is evidence at the 0.03 significance level that the average student spent less than the recommended amount of time. Fill in the requested information below.

A. The value of the standardized test statistic:

Note: For the next part, your answer should use interval notation. An answer of the form (−∞,?)(−∞,a) is expressed (-infty, a), an answer of the form (?,∞)(b,∞) is expressed (b, infty), and an answer of the form (−∞,?)∪(?,∞)(−∞,a)∪(b,∞) is expressed (-infty, a)U(b, infty).

B. The rejection region for the standardized test statistic:

C. The p-value is

D. Your decision for the hypothesis test:

A. Do Not Reject ?1H1.
B. Do Not Reject ?0H0.
C. Reject ?1H1.
D. Reject ?0H0.

Answer 1

Solution:

Given:

Sample size = n= 12

The population standard deviation =  

We have to test to determine whether there is evidence at the 0.03 significance level that the average student spent less than the recommended amount of time. that is test if mean

Part A. The value of the standardized test statistic:

where

x
28
17
26
26
26
20
29
28
23
27
18
20

thus

Thus

Part B) The rejection region for the standardized test statistic:

Since this is left tailed test , Look in z table for area = 0.0300 or its closest area and find corresponding z value.

Area 0.0301 closest to 0.0300 and it corresponds to -1.8 and 0.08

thus z critical value = -1.88

Thus rejection region is:

(-infty, -1.88)

Part C. The p-value is:

P-value = P( Z < z test statistic)

P-value = P( Z < 0.00)

Look in z table for z = 0.0 and 0.00 and find corresponding area.

P( Z< 0.00) = 0.5000

thus

P-value = P( Z < 0.00)

P-value = 0.5000

Part D. Your decision for the hypothesis test:

Since P-value = 0.5000 > 0.03 significance level, we fail to reject H0

B. Do Not Reject ?0