The lifetime of certain type of light bulb is normally distributed with a mean of 1000 hours and a standard deviation of 110 hours. A hardware store manager claims that the new light bulb model has a longer average lifetime. A sample of 10 from the new light bulb model is obtained for a test. Consider a rejection region After testing hypotheses, suppose that a further study establishes that, in fact, the average lifetime of the new lightbulb is 1130 hours. Find the probability of a type II error (round off to second decimal place).

Question

BlackTom AI · Accepted Answer

The problem describes a hypothesis test for the mean lifetime of a bulb, with a known standard deviation. The given rejection region is { xbar >= 1100 }. The true mean under the alternative (the scenario described after testing) is mu = 1130, and we are asked for the probability of a Type II error (failing to reject H0 when mu = 1130).

First, note the sampling distribution of the sample mean when the true mean is mu = 1130: xbar ~ N(1130, (sigma^2 / n)) with sigma = 110 and n = 10, so the standard deviation of xbar is sigma_xbar = 110 / sqrt(10) ≈ 34.785.

Compute the probability of not rejecting H0, i.e., P(xbar < 1100) under mu = 1130:
Z = (1100 - 1130) / (110 / sqrt(10)) = (-30) / 34.785 ≈ -0.862.
P(xbar < 1100) = P(Z < -0.862) = Phi(-0.862) ≈ 0.194.

This is the probability of a Type II error (beta) for mu = 1130. Rounding to two decimals gives beta ≈ 0.19.

Why the result makes sense: the rejection region starts at 1100 hours for the sample mean. Since the true mean is 1130 (well above 1100) we would expect a fairly high chance of observing xbar ≥ 1100 and thus rejecting H0; however, there is still a nonzero probability that the sample mean falls below 1100 due to sampling variability, which is exactly the beta we calculated (about 19%).

If there were multiple answer choices, each option would be assessed similarly: compute the sampling distribution under the specified true mean, translate the rejection region into a z-threshold, and evaluate the tail probability corresponding to failing to reject H0. In this case the computed beta is 0.19 when rounded to two decimals.

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