power and sample size estimation

For a one-sided test: $$H_0: \mu_1 \leq \mu_2 \ H_1: \mu_1 >\mu2$$

$$\alpha=0.025,\beta=0.1$$

How to calculate the sample size needed? Is this information enough to calculate the sample size?

• No, it's not. You need an effect size. Commented May 23, 2022 at 4:32

Suppose that this is a one-sided, one-sample t test. You want to detect whether $$\mu_2 - \mu_1 = \Delta = 2$$ or more and you estimate that the population standard deviation is $$\sigma = 10.$$ You are working at level 2.5% and want power 90% $$(\beta = 0.1).$$

Minitab, R, and many other statistical programs (and some on-line sites) have 'power and sample size' procedures that allow you to find the sample size implied by the constraints mentioned above. (These program implement a formula that uses the non-central t distribution.)

It turns out that the quantity $$\Delta/\sigma$$ is used in the formula, so if all else remains the same, $$\Delta =1, \sigma = 5$$ would lead to the same sample size: $$n=199$$ in the current problem.

Here is a relevant printout from Minitab:

Power and Sample Size

1-Sample t Test

Testing mean = null (versus > null)
Calculating power for mean = null + difference
α = 0.025  Assumed standard deviation = 10

Sample  Target
Difference    Size   Power  Actual Power
2     199     0.8      0.801690


We can use simulation of $$100\,000$$ experiments in R to verify that $$n = 199, \Delta = 2, \sigma = 10, \alpha=0.025,$$ lead to a simulated power of about 80%.

set.seed(2022)
pv = replicate(10^5,
t.test(rnorm(199,100,10), mu=98, alt="g")$p.val) mean(pv <= 0.025) [1] 0.8031 # aprx power 80%  Changing $$\Delta$$ and $$\sigma$$ as mentioned above. set.seed(522) pv = replicate(10^5, t.test(rnorm(199,100,5), mu=99, alt="g")$p.val)
mean(pv <= 0.025)
[1] 0.80179