Question

A sample of 1600 computer chips revealed that 47% of the chips fail in the first 1000 hours of their use. The company's promotional literature states that 44% of the chips fail in the first 1000 hours of their use. The quality control manager wants to test the claim that the actual percentage that fail is different from the stated percentage. Is there enough evidence at the 0.01 level to support the manager's claim?

Answer 1

Answer: In the given question we construct our null and alternative hypothesis as H0: P = 0.44 vs Ha: P not equal to 0.44, where "P" is the unknown population proportion of the chips failing in the first 1000 hours of their usage. The test statistic used to test the above is Z = (p - P) /σ; where P is the hypothesized value of population proportion in the null hypothesis, p is the sample proportion, and σ is the standard deviation of the sampling distribution. σ = sqrt[ P * ( 1 - P ) / n ], n is the sample size.

We reject H0 iff |Zobserved| > tau(alpha/2) where tau(alpha/2) is the upper alpha/2 point of a standard normal distribution. Alpha is the level of significance.

Here Z = 2.417469 and tau(alpha/2) =  2.575829, so we see that |Zobserved| < tau(alpha/2). Thus we do not reject H0 and conclude on the basis of the given sample measures at a 1% level of significance that there is not enough evidence to support the manager's claim.

[The answer is obtained using R-software. The code and output are attached below].