Question

A sample of 1300 computer chips revealed that 42% of the chips do not fail in the first 10001000 hours of their use. The company's promotional literature claimed that above 39% do not fail in the first 1000hours of their use. Is there sufficient evidence at the 0.05 level to support the company's claim?

State the null and alternative hypotheses for the above scenario.

Step 1 of 3:

Find the point estimate for the true difference between the population means.

Step 2 of 3:

Calculate the margin of error of a confidence interval for the difference between the two population means. Round your answer to six decimal places.

Step 3 of 3:

Construct the 90% confidence interval. Round your answers to one decimal place.

Answer 1

Solution:-

From the above test we have sufficient evidence in the favor of the claim that above 39% do not fail in the first 1000 hours of their use.

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: P < 0.39
Alternative hypothesis: P > 0.39

Note that these hypotheses constitute a one-tailed test.

Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method, shown in the next section, is a one-sample z-test.

Analyze sample data. Using sample data, we calculate the standard deviation (S.D) and compute the z-score test statistic (z).

S.D = sqrt[ P * ( 1 - P ) / n ]

S.D = 0.01353
z = (p - P) / S.D

z = 2.22

where P is the hypothesized value of population proportion in the null hypothesis, p is the sample proportion, and n is the sample size.

Since we have a one-tailed test, the P-value is the probability that the z-score is greater than 2.22.

Thus, the P-value = 0.0139

Interpret results. Since the P-value (0.0139) is less than the significance level (0.05), we cannot accept the null hypothesis.

From the above test we have sufficient evidence in the favor of the claim that above 39% do not fail in the first 1000 hours of their use.