Page 77 - Basic Statistics

72

Significance level:  = 0.05

If there is an indication that the variance is not the same, then standard

deviation is found by computing:

S 2 S 2 16.75 39.46
S = 1 + 2 = + = . 1 4994
X 1 − X 2 n 1 n 2 25 25

X − X 64 − 76.72
t = 1 2 = = − . 8 48
S 1.4994
X 1 − X 2

2
[ (S / n ) + (S / n )] 2
2
Effective degrees of freedom: Dbef = 1 1 2 2
2
2
(S / n ) 2 (S / n ) 2
1 1 + 2 2
n 1 − 1 n 1 − 1
(
( 16 . 75 / 25 ) + ( 39 . 46 / 25 ) ) 2
=
( 16 . 75 / 25 ) 2 /( 25 − ) ) 1 + ( 39 . 46 / 25 ) 2 /( 25 − ) ) 1
(
(

Dbef = 41.264

In Table t-student: t/2;Df = t0.025;41
= 2.021

Therefore  t > t/2;41, which is 8.48 > 2.021 then H0 is rejected at the 5 percent

significance level. It was concluded that there is a significant difference between

the mean test scores of both models.

b) This the test requires one-tailed test:

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

Significance level:  = 0.05

In Table t-student: t/2;Df = t0.05;41

= -1.684

~~* CHAPTER 4 TESTING HYPOTHESIS *~~