Sample of 1700 computer chips revealed that 51% of the chips do not fail in the first 1000 hours of their use. The company's promotional literature states that 48% 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 different from the stated percentage. Determine the decision rule for rejecting the null hypothesis, H0, at the 0.01 level.

Decision Rule:

If P-value < significance level ⇒ Reject null hypothesis

If P-value > significance level ⇒ Do not Reject null hypothesis

Step-by-step explanation:

We are given that the company's promotional literature states that 48% of the chips fail in the first 1000 hours of their use. Also, a sample of 1700 computer chips revealed that 51% of the chips fail in the first 1000 hours of their use.

And the quality control manager wants to test the claim that the actual percentage that fail is different from the stated percentage, i.e;

Null Hypothesis, : p = 0.48 {means that the actual percentage that fail is same as the stated percentage}

Alternate Hypothesis, : p  0.48 {means that the actual percentage that fail is different from the stated percentage}

Decision Rule ;

If P-value of the test is less than the significance level of 0.01, then we will reject null hypothesis, .

If P-value of the test is more than the significance level of 0.01, then we will not reject the null hypothesis, .

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