Statistics sample sizes

[tedpark1212]

So I made an attempt to solve question 1, but I am stuck with question 2. I know that I need to use binominal distribution, but am unclear of the proper steps. Can anyone provide some help?

1. Suppose you want to show the significance of an effect size of 0.20 between your sample mean and a hypothesized mean. You intend to conduct a two-sided one-sample t-test. The sample variance is σ2 = 1.0. You assume that the population is large, or effectively infinite. If you want the test to have a power of 0.80, what is the required sample size for the following?

Attempt to solve:

• A level of significance of α = 0.05

n = 2 σ2 (Z (beta) + Z (alpa/2)) 2/ effect size2
n = 2(1) ( 1.28 + 1.96) 2/ 0.202------ for α = 0.05
n = 2 (20.9952) / 0.04
n=524.88
n= 525

The required sample size is 525 with a level of significance of α = 0.05.
• A level of significance of α = 0.01
n = 2 σ2 ( Z (beta) + Z (alpha/2) ) 2/ effect size2
n = 2(1) ( 1.28 + 2.575) 2/ 0.202------ for α = 0.01
n= 2 (14.86) / 0.04
n=743.05
n= 743
The required sample size is 743 with a level of significance of α = 0.01.

2. After some research, you determine that the size of the population is N = 350. If you want the test to have a power of 0.80, what is the required sample size for the following?

• A level of significance of α = 0.05


• A level of significance of α = 0.01

 

[mighty2000]

Hi there,

Thank you for sharing your attempt at solving the first question. It looks like you have the right formula, but there are a few errors in your calculations. First, the critical values for alpha/2 and beta may be different depending on the sample size. Second, the effect size should be squared in the formula.

For the first question, using a significance level of 0.05, the correct formula would be:

n = (2*1*(1.96+0.842)^2)/(0.20^2) = 829

So the required sample size would be 829.

For the second question, we would need to make some assumptions about the population size and the effect size. Assuming a large population and the same effect size of 0.20, the formula would be:

n = (2*1*(1.96+0.842)^2)/(0.20^2*(1-350/829)) = 283

So the required sample size would be 283.

I hope this helps! Let me know if you have any further questions or need clarification on anything.