A sample of 1600 computer chips revealed that 38% of the chips do not fail in the first 1000 hours of their use. The company's promotional literature claimed that above 35% do not fail in the first 1000 hours of their use. Is there sufficient evidence at the 0.02 level to support the company's claim? State the null and alternative hypotheses for the above scenario.

Answer with explanation:Let p denotes the population proportion.By considering the given information, we haveNull hypothesis : [tex]H_0 : p\leq0.35[/tex]Alternative hypothesis : [tex]H_1 : p>0.35[/tex], since alternative hypothesis is right tailed so the test is a right tailed test.Given : n= 1600 ; [tex]\hat{p}=0.38[/tex]Test statistics for proportion :-[tex]z=\dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}}[/tex][tex]=\dfrac{0.38-0.35}{\sqrt{\dfrac{0.35(1-0.35)}{1600}}}=2.51588360813\approx2.52[/tex]P-value for right tailed test = [tex]P(z>2.52)=0.0058677\approx0.006[/tex]Conclusion : Since the p-value (0.006) is less than significance level (0.02) that means we reject the null hypothesis.Thus , we conclude that we have sufficient evidence at the 0.02 level to support the company's claim.