A sample of 900 computer chips revealed that 66% of the chips fail in the first
ID: 3246590 • Letter: A
Question
A sample of 900 computer chips revealed that 66% of the chips fail in the first 1000 hours of their use. The company's promotional literature states that 68% 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. Find the value of the test statistic. Round your answer to two decimal places. A sample of 900 computer chips revealed that 66% of the chips fail in the first 1000 hours of their use. The company's promotional literature states that 68% 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. Find the value of the test statistic. Round your answer to two decimal places.Explanation / Answer
answer:
For:
Ho: p = 0.68
Ha: p < 0.68 (one-tailed)
= 0.01
ZCritical = -2.33
ZCalculated = -1.95
For test of hypothesis using pre-determined :
Do not reject the Null hypothesis -
because Zcalc = -1.95 does not fall in the critical region defined by Z < -2.33.
There is insufficient evidence to conclude that the company's claim is false at the 0.01 level.
The level of significance is another question.
To determine the level of significance for Zcalc = -1.95:
Using Standard Normal table, we find the area under the normal curve...
from - to Zcalc = -1.95 is 0.0256.
(for reference, note area under the normal curve from - to Z = -2.33 is 0.0099)
Just as 0.0099 ~ 0.01 (predefined critical area () = 0.01 for a lower tail test)
0.0256 ~ 0.03 - So the level of significance (/ p-value) for this test is ~ 0.03
Now the question is not whether we should reject the Null hypothesis, we just say the test is significant at the 0.03 level.
Or, we could say that there is sufficient evidence to conclude that the company's claim is false at the 0.03 level.