Low power to catch a drop to 980 hours

Question

A lab tests H0: \mu = 1000 against H1: \mu 
eq 1000 for bulb life, with a known \sigma = 50, a sample of n = 25, at the two-sided 5% level. Suppose the true mean life is actually \mu = 980. Find the power of the test (the probability it correctly rejects H0).

Accepted Answer

Standard error: 	ext{SE} = 50/\sqrt{25} = 50/5 = 10. The test rejects when \bar x < 1000 - 1.96(10) = 980.4 or \bar x 1000 + 1.96(10) = 1019.6. Under the true mean \mu = 980, the power comes almost entirely from the lower region: P(\bar x < 980.4 \mid \mu = 980) = P\!\left(Z < \frac{980.4 - 980}{10}ight) = P(Z < 0.04) = 0.516. The upper region contributes P(\bar x 1019.6 \mid \mu = 980) = P(Z 3.96) \approx 0, negligible. So 	ext{Power} \approx 0.52, \qquad \beta \approx 0.48. Conclusion: despite a real 20-hour drop, the test catches it only about half the time. Takeaway. Power is poor whe…

Low power to catch a drop to 980 hours

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