Question

A sample of 1300 computer chips revealed that 27% of the chips do not fail in the first 1000 hours of their use. The company's promotional literature states that 29% 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

Decision rule for rejecting is

reject H0 is test statistic <-1.233

Explanation:

Given that a sample of 1300 computer chips revealed that 27% of the chips do not fail in the first 1000 hours of their use, as against company literature of 29%

To test this we create hypotheses as

`H_0: p = 29%\\H_a: p <27%`

(left tailed test at 10% significance level)

p difference = -0.02

STd error of p =
`√(0.29*0.71/1300) \\=0.0126`

Test statistic = p diff/std error = -0.1587

For 10% significance level for right tailed test we can reject H0 if

test statistic <-1.645

Decision rule for rejecting is

reject H0 is test statistic <-1.233