Answer to Question #92086 in Statistics and Probability for Paula

Question

Let X denote the IQ of a randomly selected adult American. Assume, a bit unrealistically, that X is normally distributed with unknown mean μ and standard deviation 15. Take a random sample of n = 25 students, so that, after setting the probability of committing a Type I error at α = 0.01, we can test the null hypothesis H0: μ = 100 against the alternative hypothesis that HA: μ > 100. What is the power of the hypothesis test if the true population mean were μ = 110?

Answer 1

Setting α, the probability of committing a Type I error, to 0.01, implies that we should reject the null hypothesis when the test statistic Z ≥ 2.3263.

We transform the test statistic Z to the sample mean by way of:

$$Z={\overline{x}-\mu \over \sigma/\sqrt{ n}}=>\overline{x}=\mu+Z\sigma/\sqrt{n}$$ $$\overline{x}=100+2.3263\cdot 15\sqrt{25}=106.9789$$

So the observed sample mean is 106.99 or greater.

$$Power=P(\overline{x}\geq106.9789\ when\ \mu=110)=$$

$$=P(Z\geq{106.9789-110\over 15/\sqrt{ 25}})=P(Z\geq-1.00703)=0.8430$$

In summary, we have determined that we have a 84.30% chance of rejecting the null hypothesis H₀: μ = 100 in favor of the alternative hypothesis H_A: μ > 100 if the true unknown population mean is in reality μ = 110.