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. 28 18 15 20 14 25 21 29 20 23 24 26 It is known that the population standard deviation is 6. 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 0.01 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

D. Your decision for the hypothesis test

Reject H0

Do Not Reject H1

Reject H1

Do Not Reject H0.

Answer 1

The calculations for the mean and the standard deviation of the sample are given after the hypothesis test. From the given data:

= 21.917, s = 4.795, n = 12

The Hypothesis:

H0: = 24

Ha: < 24

This is a Left Tailed Test.

(A) The Test Statistic: Since the population standard deviation is unknown, we use the students t test.

The test statistic is given by the equation:

(B) The Rejection Region: The critical value (Left Tail) at = 0.01, for df = n - 1 = 11,

tcritical = -2.718. Therefore if t observed is < -2.718, we reject H0.

The rejection region for the standardized test statistic is therefore (-infty, -2.718)

(C) The p value: For t = -1.505, df = 11, p value (left tail) = 0.0802

(D) The Decision:   Since tobserved (-1.505) is > -tcritical (-2.718), We Fail to Reject H0.

Also since P value (0.0802) is > (0.01) , We Fail to Reject H0.

The Decision is therefore Do Not Reject H0.

___________________________________________________________________________

Mean = Sum of observations / Total Observations

Standard deviation = Sqrt(Variance). Variance = SS/(n - 1), where SS = SUM(X - Mean)².

#	X	Mean	(x - mean)2
1	28	21.917	37.002889
2	18	21.917	15.342889
3	15	21.917	47.844889
4	20	21.917	3.674889
5	14	21.917	62.678889
6	25	21.917	9.504889
7	21	21.917	0.840889
8	29	21.917	50.168889
9	20	21.917	3.674889
10	23	21.917	1.172889
11	24	21.917	4.338889
12	26	21.917	16.670889
Total	263.00		252.91667

n	12
Sum	263
Average	21.917
SS(Sum of squares)	252.916668
Variance = SS/n-1	22.992
Std Dev=Sqrt(Variance)	4.7950