Rishi S. answered • 06/26/23
Stanford Undergrad Specializing In Math and Test Prep
This problem is about finding a sample size for a confidence interval estimate of a population mean. A confidence interval provides a range of values that likely contains the true population mean.
In this problem, we want to be 90% confident that the sample mean is within 8 IQ points of the true mean. The standard deviation (σ) is given as 18, and the desired error (E), also called margin of error, is 8 IQ points.
The formula for the sample size (n) needed for estimating a mean is:
n = (Zσ/E)²
where:
- Z is the Z-value from the standard normal distribution corresponding to the desired confidence level (for a 90% confidence level, the Z-value is approximately 1.645).
- σ is the population standard deviation (given as 18).
- E is the desired margin of error (given as 8).
Substituting the given values into the formula:
n = (1.645*18/8)² ≈ 9.41
Since the sample size cannot be a decimal or a fraction, we'll round this number up to the nearest whole number because we want to ensure the confidence level and margin of error requirements are met. Therefore, the sample size needed is 10.
In terms of whether this is a reasonable sample size for a real-world calculation, it depends on the context. If it's feasible to gather a sample of 10 statistics students for the study, then it would be considered reasonable. However, given the variability inherent in IQ scores and the diversity likely present in any population of students, a larger sample size might provide a more accurate and reliable estimate of the true mean IQ score.
Rishi S.
You are absolutely right, my apologies!! I believe I did all the steps correct; however I multipled wrong when I did n = (1.645*18/8)². I initially put that it was 9.41 but it is actually around 13.7, so the sample size needed is actually 14. My apologies again and thank you so much for catching my mistake, please let me know if you spot any others!!06/26/23
Kristel U.
This is wrong unfortunately06/26/23