. A random sample is selected from a normal population with a mean of __40 and a standard deviation of __10. After a treatment is administered to the individuals in the sample, the sample mean is found to be M _ 42.
a. How large a sample is necessary for this sample mean to be statistically significant? Assume a two tailed test with _ = .05
b. If the sample mean were M _ 41, what sample size is needed to be significant for a two-tailed test with _ = .05.
To answer these questions, we need to use the concept of hypothesis testing, specifically the t-test for the mean. The t-test compares the sample mean to a hypothesized population mean and determines the statistical significance of the difference.
a. To determine the sample size necessary for the sample mean to be statistically significant in a two-tailed t-test with α = 0.05, we need to calculate the t-value and consult the t-distribution table.
The t-value is calculated using the formula:
t = (M - μ) / (s / √n)
where M is the sample mean, μ is the population mean, s is the population standard deviation, and n is the sample size.
Given:
μ = 40
s = 10
M = 42
α = 0.05 (two-tailed test)
We can substitute these values into the formula and solve for n:
t = (42 - 40) / (10 / √n)
t = 2 / (10 / √n)
t = 2√n / 10
√n = (t * 10) / 2
n = [(t * 10) / 2]^2
To find the critical t-value for a two-tailed test with α = 0.05, we need to divide α by 2 and find the corresponding t-value with degrees of freedom (df) equal to n-1. For a significance level of 0.05, the critical t-value would be obtained from the t-distribution table. Let's assume it is 2.262 for the df equal to n-1.
Using this critical t-value, we can plug it into the formula for n:
n = [(t * 10) / 2]^2
n = [(2.262 * 10) / 2]^2
n = 11.31^2
n ≈ 128.25
Therefore, to make the sample mean of 42 statistically significant, we would need a sample size of at least 129, rounded up to the nearest whole number.
b. To determine the sample size needed for a sample mean of 41 to be statistically significant in a two-tailed t-test with α = 0.05, we follow the same steps as above, but with a different sample mean:
Given:
μ = 40
s = 10
M = 41
α = 0.05 (two-tailed test)
Using the same formula as before, we can solve for n:
t = (41 - 40) / (10 / √n)
t = 1 / (10 / √n)
t = √n / 10
√n = t * 10
n = (t * 10)^2
Using the critical t-value for a two-tailed test with α = 0.05 (1.96 assuming df = n-1):
n = (1.96 * 10)^2
n ≈ 384.16
Therefore, to make the sample mean of 41 statistically significant, we would need a sample size of at least 385, rounded up to the nearest whole number.
To determine the sample size necessary for the sample mean to be statistically significant, we can use the formula for the t-test. The formula is:
t = (M - μ) / (σ / √n)
Where:
t is the t-value
M is the sample mean
μ is the population mean
σ is the population standard deviation
n is the sample size
a. For M = 42, μ = 40, σ = 10, and α = 0.05 for a two-tailed test, we can rearrange the formula to solve for n. The critical t-value for a two-tailed test with α = 0.05 is approximately 2.0.
2.0 = (42 - 40) / (10 / √n)
Simplifying the equation:
2.0 = 2 / (10 / √n)
Multiply both sides by (10 / √n):
20 / √n = 2
Square both sides:
20^2 / n = 2^2
400 / n = 4
Cross-multiply:
4n = 400
Divide by 4:
n = 100
So, a sample size of 100 is necessary for a sample mean of 42 to be statistically significant.
b. For M = 41, everything else remains the same. The critical t-value is still 2.0.
2.0 = (41 - 40) / (10 / √n)
Simplifying the equation:
2.0 = 1 / (10 / √n)
Multiply both sides by (10 / √n):
20 / √n = 1
Square both sides:
20^2 / n = 1^2
400 / n = 1
Cross-multiply:
n = 400
So, a sample size of 400 is necessary for a sample mean of 41 to be statistically significant.