Power of a Test Math Example 2

Follow the full solution, then compare it with the other examples linked below.

Example 2

hard
For testing H0:μ=100H_0: \mu=100 vs Ha:μ=105H_a: \mu=105, with σ=10\sigma=10, n=25n=25, α=0.05\alpha=0.05: calculate the rejection region and power of the test.

Solution

  1. 1
    SE =σ/n=10/5=2= \sigma/\sqrt{n} = 10/5 = 2; critical value: z∗=1.645z^* = 1.645 (one-tailed)
  2. 2
    Rejection region: xˉ>100+1.645(2)=103.29\bar{x} > 100 + 1.645(2) = 103.29
  3. 3
    Power = P(xˉ>103.29∣μ=105)=P(Z>103.29−1052)=P(Z>−0.855)=1−0.196=0.804P(\bar{x} > 103.29 | \mu=105) = P\left(Z > \frac{103.29-105}{2}\right) = P(Z > -0.855) = 1 - 0.196 = 0.804
  4. 4
    Power ≈ 0.80; 80% chance of detecting the effect if μ=105\mu=105

Answer

Rejection region: xˉ>103.29\bar{x} > 103.29. Power ≈0.80\approx 0.80 at μ=105\mu=105.
Power is calculated by finding the probability of being in the rejection region under the alternative hypothesis. The rejection region is determined by α; then we compute the probability of landing there if Hₐ is true. Larger effect sizes (farther from null) give higher power.

About Power of a Test

The probability that a hypothesis test correctly rejects a false null hypothesis. Power =P(reject H0∣H0 is false)=1−β= P(\text{reject } H_0 \mid H_0 \text{ is false}) = 1 - \beta, where β\beta is the probability of a Type II error.

Learn more about Power of a Test →

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Example 4 hard

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