Question

A sample of 1600 computer chips revealed that 50% of the chips do not fail in the first 1000 hours of their use. The company's promotional literature states that 53% 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. State the null and alternative hypotheses.

Answer 1

Solution:

Given:

n = sample size = 1600

Sample Proportion  of the chips do not fail in the first 1000 hours of their use. =

Population Proportion of  the chips do not fail in the first 1000 hours of their use = p = 0.53

Claim: the actual percentage that do not fail is different from the stated percentage.

Part a) State the null and alternative hypotheses.

Since claim is non-directional, this is two tailed test.

Vs

Part b) Test statistic:

Part c) critical value:

Level of significance =

Since this is two tailed test ,

we find : Area =

Look in z table for area = 0.0250 or its closest area and find z value

Area 0.0250 corresponds to -1.9 and 0.06

thus z critical value = -1.96

Since this is two tailed test, we have two z critical values: ( -1.96 , 1.96)

Part d) Decision Rule:
Reject null hypothesis ,if z  test statistic value < z critical value =-1.96 or z  test statistic value > z critical value =1.96 , otherwise we fail to reject H0.

Since z test statistic value = z = - 2.40 < z critical value = -1.96, we reject H0.

Part e) Conclusion:

At 0.05 level of significance , we have sufficient evidence to support  the claim that the actual percentage that do not fail is different from the stated percentage.