Page 61 - Basic Statistics

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significance level  = 0.05.

Since the population variance is known then the test statistic to be used is the Z
statistic, namely :

10 . 366 − 10
Z = = 1.002 and from tables : Z  2 / = Z . 0 05 2 / = 1.96.
2 / 15

Testing Results : Because Z < Z  2 / , ie 1.002 < 1.96 then we accept H ,
0

meaning that the data is not enough significance to reject the hypothesis H .
0

Worked Example 4.2:

A factory battery (battery) claims that the average battery life to 55 hours.

On the results of tests conducted on a production batch consisting of 40

batteries, gained an average of 50 hours of life, with a standard deviation of 11
734 hours. Perform hypothesis testing with a significance level of 1 percent that

a. Average battery life is 55 hours

b. Average battery life of less than 55 hours

Worked Solution:

Given :  = 55; x = 50; s = 11.734; n = 40
0

a). Test hypothesis : H :  = 55
0
0
H :   55
0
1

Significance level :  = 0.01

s 11 . 734
Standard deviation for X : s = = = 1.86
x
n ( 40 )

−
X  50 − 55
Statistic Z : z = 0 =
s x . 1 86

= -2.69

~~* CHAPTER 4 HYPOTHESIS TESTING *~~