Page 70 - Basic Statistics

65

Worked Solution:

30 30
 X i  X i

Means : X = i= 1 = 1976 = 65 . 867 X = i= 1 = 2218 = 73 . 933
1
30 30 2 30 30

n
 X i 2 X n − 2

2
Variance : S = i= 1
n − 1

S 1 2 = ( 130646 ) − 30 (65.867) 2 = 17 . 016 S 2 2 = ( 164456 ) − 30 (73.933) 2 = 16 . 271
30 − 1 30 − 1

(a) The test requires a two-tailed test:

Hypotheses: H0 : μ = μ
2
1
H1 : μ  μ
1
2

Significance level:  = 0.01

σ ˆ 2 σ ˆ 2
Standard deviation: σ ˆ = 1 + 2
X 1 − X 2 n n
1 2
(17.016) (16.271)
= + = 1.053
30 30

X X −
Z = 1 2
σ ˆ
X 1 X − 2
65.867 − 73.933
= = -7.66
1.053

The test results:

a). In the standard normal table;  Z/2 =  2.57

Conclusion:  Z  > Z/2, ie 7.66 > 2.57, then H0 is rejected at the 1 percent

significance level. We conclude that there is a significant difference in the

average test scores by two learning models, STAD and TPS.

~~* CHAPTER 4 TESTING HYPOTHESIS *~~