Question

A sample of 1400 computer chips revealed that 69% of the chips do not fail in the first 1000 hours of their use. The company's promotional literature states that 72% 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 0.10 level.

Answer 1

Solution:

Given,

p = 72% = 0.72

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

i.e p   0.72

So hypothesis can be written as

Ho: p = 0.72

H1: p   0.72

It is z test for the proportion.

Sample information

n = 1400

= 69% = 0.69

Determine the decision rule for rejecting the null hypothesis, H0, at the 0.10 level.

sign in H1 indicates that the two tailed test.

So, the critical values are and

And the decision rule is "Reject Ho if z <   or z > ".

= 0.10 and /2 = 0.05

   = 1.645 (using z table)

The decision rule is "Reject Ho if z < -1.645 or z > 1.645