A standard IQ test has a mean of 98 and a standard deviation of 16. We want to be 90% certain that we are within 8 IQ points of the true mean. Determine the sample size.
(z * sigma / error )^2
round up
assuming that adults have IQ scores that are normally distributed with a mean of 100 and standard deviation of 15. Find the probability that a randomly selected adult has an IQ less than 121.
To determine the sample size, we can use the formula for confidence interval:
Sample Size (n) = (Z * σ / E)^2
Where:
- Z is the z-score corresponding to the desired confidence level (90% confidence level corresponds to a z-score of 1.645)
- σ is the standard deviation
- E is the desired margin of error (half the width of the confidence interval, which is 8 IQ points in this case)
Let's substitute the given values into the formula:
Z = 1.645
σ = 16
E = 8
n = (1.645 * 16 / 8)^2
n = (1.645 * 2)^2
n = 3.29^2
n ≈ 10.8241
Since the sample size should be a whole number, we need to round up to the nearest whole number:
n = 11
Therefore, the sample size required to be 90% confident and within 8 IQ points of the true mean is 11.