determining decision rule for rejecting null hypothesis

Question

A sample of 900 computer chips revealed that 53% of the chips do not fail in the first 1000 hours of their use. The company's promotional literature states that 51% 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 0.02 level.

Answer 1

H0: p is not 51%  vs H1: p1 = 51%

 

 

standard deviation is:

σ = sqrt[ P * ( 1 - P ) / n ] = sqrt[ .51*.49/900 ] = .01666333+

 

 

z -score is (p-P)/σ = (.51-.53)/.01666333+ = -1.2

 

CONCLUSION:

 There is statistical evidence that the proportions are different. H0 accepted