Final answer:
To estimate the population mean with a known population variance of 529, a confidence level of 0.98, and a margin of error of 3, the minimum required sample size is 320.
Step-by-step explanation:
The question is asking for the sample size needed to estimate the population mean with a known population variance and a specific confidence level and margin of error. To calculate the required sample size, we use the formula for the margin of error (E) in terms of the confidence interval for a normally distributed population:
E = Z * (σ/√n)
Where:
- E is the margin of error
- Z is the Z-score corresponding to the confidence level
- σ is the population standard deviation (the square root of the variance)
- n is the sample size
Given a population variance of 529, we first calculate the population standard deviation by taking the square root, which is σ = √529 = 23. The Z-score for a 0.98 confidence level is approximately 2.33 (from Z-tables or a calculator). The desired margin of error E is given as 3. Substituting these values into the formula, we can solve for n:
3 = 2.33 * (23/√n)
Now, solve for n:
n = (2.33 * 23 / 3)2 = (53.59 / 3)2 = 17.8632 = 319.011
Since we cannot have a fraction of a sample, we round up to the nearest whole number. Therefore, the minimum sample size needed is 320.