The incubation period of Ebola virus is known to have a normal distribution with a mean of 10 days and a standard deviation of 7 days Suppose a group of researchers claimed that the true mean is shorter than 10 days. A test is conducted using a random sample of 23 patients. Consider a rejection region Suppose the test failed to reject . If the true mean incubation period is 6 days, what is the type II error probability of the test using (round off to third decimal place)?

Question

The incubation period of Ebola virus is known to have a normal distribution with a mean of 10 days and a standard deviation of 7 days Suppose a group of researchers claimed that the true mean is shorter than 10 days. A test is conducted using a random sample of 23 patients. Consider a rejection region Suppose the test failed to reject . If the true mean incubation period is 6 days, what is the type II error probability of the test using  (round off to third decimal place)?

BlackTom AI · Accepted Answer

First, I restate the setup in my own words to ensure we’re aligned: we have a test for the mean incubation period with known sigma = 7, sample size n = 23, and the rejection region is {X̄ ≤ 8}. The null hypothesis is that the true mean is 10 or more days (the group claimed the mean is shorter than 10, so we test H0: μ ≥ 10 vs Ha: μ < 10). We are asked for the Type II error when the true mean is μ = 6.

Step 1: Identify the sampling distribution under the alternative value μ = 6.
- The sample mean X̄ is normally distributed: X̄ ~ N(μ, σ²/n) with σ = 7 and n = 23. So X̄ ~ N(6, (7²/23)).
- The standard deviation of X̄ is σ_X̄ = 7 / √23 ≈ 1.459.

Step 2: Convert the rejection region into a probability under μ = 6.
- The rejection region is X̄ ≤ 8. Under μ = 6, we want the probability of NOT rejecting H0, which corresponds to X̄ > 8.
- The Type II error β(μ = 6) = P(X̄ > 8 | μ = 6).

Step 3: Standardize and compute the probability.
- Compute the z-value: z = (8 − 6) / (7/√23) = 2 / (7/√23) ≈ 2 / 1.459 ≈ 1.371.
- Therefore, P(X̄ > 8 | μ = 6) = 1 − Φ(1.371).
- Look up or approximate Φ(1.371). Φ(1.371) ≈ 0.915.

Step 4: Finish the calculation.
- β(μ = 6) ≈ 1 − 0.915 = 0.085.

Result interpretation: with true mean 6 days, the probability of failing to reject H0 given the rejection region X̄ ≤ 8 is about 0.085 (8.5%). This matches the requested rounding to three decimal places: 0.085.

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