Sample of 17001700 computer chips revealed that 51Q% of the chips do not fail in the first 10001000 hours of their use. The company's promotional literature states that 48H% of the chips do not fail in the first 10001000 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, H0H0, at the 0.010.01 level.

The decision rule for rejecting H0 is if the test statistic falls outside the region bounded by the critical values.

Explanation:

Null hypothesis: The actual percentage that do not fail is the same as the stated percentage.

Alternate hypothesis: The actual percentage that do not fail is different from the stated percentage.

Test statistic (z) = (p' - p) ÷ sqrt[p(1-p) ÷ n]

p' is sample proportion = 0.51

p is population proportion = 0.48

n = 1700

z = (0.51 - 0.48) ÷ sqrt[0.48(1-0.48) ÷ 1700] = 0.03 ÷ 0.012 = 2.5

The test is a two-tailed test. At 0.01 significance level, the critical values are -2.576 and 2.576

Decision rule:

Reject H0 (null hypothesis) if the test statistic falls outside the region bounded by the critical values -2.576 and 2.576.

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