Page 63 - Basic Statistics

58

Worked Solution:

Given :  = 15; n = 25
0
And x = 18 . 32; s = 15.845

(a). Test hypothesis : H :  = 15;
0
0
H :   15
0
1
Significance level :  = 0.05

The critical value : t = t = 2.064
 ; 2 / 24

(at the t-table, with (25-1) degrees of freedom)

s 15 . 845
Standard deviation for X : s = = = 3.169
x
n ( 25 )

X −  18 . 32 − 15
t = 0 =
s x . 3 169

= 1.05

Test results : t < t  , 2 / db= n− 1 , ie 1.05 < 2.064, then H is accepted or not enough
0
evidence to reject H . It was concluded that the average absences of workers in
0

one year is equal to 15 days.

b. Test hypothesis : H :  = 15
0
0
H :  > 15
0
1
Significance level :  = 0.05

The critical value : t = t = ……..

Test results : ………………..,ie…………..., then H is…………….. It was
0
concluded that ………………………..

~~* CHAPTER 4 HYPOTHESIS TESTING *~~