Question:
A standard placement test has a mean of 105 and a standard deviation of ó = 13. Determine the minimum sample size if we want to be 99% certain that we are within 3 points of the true mean.
Answer:
N = ( zc ó/E)^2
= (2.575 x 13/3)^2
= 121
To determine the minimum sample size required, we can use the formula:
N = (zc σ/E)^2
Where:
N = minimum sample size
zc = z-score for a given level of confidence
σ = standard deviation
E = desired margin of error
In this case, the given standard deviation (σ) is 13 and the desired margin of error (E) is 3 points.
To find the z-score for a 99% level of confidence, we can use a standard normal distribution table or a calculator. The z-score for a 99% confidence level is approximately 2.575.
Now we can substitute the values into the formula:
N = (2.575 * 13/3)^2
N = (2.575 * 4.333)^2
N = 11.162^2
N ≈ 121
Therefore, the minimum sample size required to be 99% certain that we are within 3 points of the true mean is 121.