Question

A sample of 900 computer chips revealed that 81% of the chips do not fail in the first 1000 hours of their use. The company's promotional literature states that 80% 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 more than the stated percentage. Is there enough evidence at the 0.10 level to support the manager's claim?

Step 1 of 7: State the null and alternative hypotheses.

Step 2 of 7: Find the value of the test statistic. Round your answer to two decimal places.

Step 3 of 7: Specify if the test is one-tailed or two-tailed.

Step 4 of 7: Determine the P-value of the test statistic. Round your answer to four decimal places.

Step 5 of 7: Identify the value of the level of significance.

Step 6 of 7: Make the decision to reject or fail to reject the null hypothesis.

Step 7 of 7: State the conclusion of the hypothesis test.

Answer 1

Given that, n = 900 and sample proportion = 0.81

Step 1) The null and alternative hypotheses are,

H0 : p = 0.80

Ha : p > 0.80

Step 2) Test statistic is,

=> Test statistic = 0.75

Step 3) This test is one-tailed.

Step 4) p-value= P(Z > 0.75)=1 - P(Z < 0.75) = 1 - 0.7734 = 0.2266

=> p-value = 0.2266

Step 5) level of significance = 0.10

Step 6) Since, p-value = 0.2266 > 0.10, we fail to reject the null hypothesis.

Step 7) Conclusion : there is insufficient evidence to support the claim that the actual percentage that do not fail is more than the stated percentage. (That is more than 80%).