Z test assistance

Question

A sample of 1200 computer chips revealed the 30% of the chips do not fail in the first 1000 hours of their use. The company's promotional literature claimed that 28% do not fail in the first 1000 hours of their use.

Is there sufficient evidence at the 0.02 level to refute the company's claim?

 

Enter the value of the z test statistic.

Answer 1

your z statistic is (.3 - .28)/sqr(.28*.72/1200) = .02/.01296 = 1.54.  The .02 level is going to be a lot bigger than 1.54 so there is not sufficient evidence to refute the promotional literature. If the sample size was ten times bigger, say 12,000 cases, then the z value would become 4.88 which is almost 5 standard deviations out and would be significant.