Question

A sample of 16001600 computer chips revealed that 50%50% of the chips do not fail in the first 10001000 hours of their use. The company's promotional literature states that 48%48% of the chips do not fail in the first 10001000 hours of their use. The quality control manager wants to test the claim that the actual percentage that do not fail is more than the stated percentage. Is there enough evidence at the 0.020.02 level to support the manager's claim?

Step 1 of 7 :

State the null and alternative hypotheses.

Answer 1

Solution :

This is the right tailed test .

The null and alternative hypothesis is

H0 : p = 0.48

Ha : p > 0.48

= 0.50

n =1600

P0 = 0.48

1 - P0 = 1-0.48 =0.52

Test statistic = z

= - P0 / [P0 * (1 - P0 ) / n]

= 0.50 -0.48 / [0.48*(0.52) / 1600]

= 1.60

P(z > 1.60) = 1 - P(z < 1.60) = 0.0548

P-value = 0.0548

= 0.02

The critical value = 2.054

0.0548 >  0.02

Fail to reject the null hypothesis .

There isnot enough evidence at the 0.02 level to support the manager's claim.