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:Decision rule for rejecting isreject H0 is test statistic <-1.233Step-by-step 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[tex]H_0: p = 29%\\H_a: p <27%[/tex](left tailed test at 10% significance level)p difference = -0.02STd error of p = [tex]\sqrt{0.29*0.71/1300} \\=0.0126[/tex]Test statistic = p diff/std error = -0.1587For 10% significance level for right tailed test we can reject H0 iftest statistic <-1.645Decision rule for rejecting isreject H0 is test statistic <-1.233