A sample of 1300 computer chips revealed that 60% of the chips do not fail in the first 1000 hours of their use. The company's promotional literature states that 58% of the chips do not fail in the first 1000 hours of their use. The quality control manager wants to test the claim that the actual percentage that do not fail is different from the stated percentage. Determine the decision rule for rejecting the null hypothesis, H0, at the 0.01 level.
Solution:
Given:
Claim: the actual percentage that do not fail is different from the stated percentage 58%.
Since claim is non-directional, this is two tailed test.
Thus hypothesis of the study are:
Vs
We have to determine the decision rule for rejecting the null hypothesis, H0, at the 0.01 level.
Since this is two tailed test, find:
Area =
Look in z table for Area = 0.0050 and find corresponding z value.
Area 0.0050 is in between 0.0049 and 0.0051, and both the area are at same distance from 0.005
thus we look for both area and find both z values.
Area 0.0049 corresponds to -2.5 and 0.08 , thus z= -2.58
Area 0.0051 corresponds to -2.5 and 0.07 , thus z= -2.57
Thus average of both z values is = ( -2.57 + -2.58 ) / 2 = -2.575
Thus critical z value is = -2.575
Since this is two tailed test , there are two z critical values = ( -2.575 , 2.575 )
Thus decision rule is:
Reject null hypothesis if z test statistic value < - 2.575 or z test statistic value > +2.575 , otherwise we fail to reject H0.
In other words:
Reject null hypothesis if absolute value of z test statistic value > 2.575 , otherwise we fail to reject H0.
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