Final answer:
To achieve a margin of error of 3 or less with a 95% confidence level and a population standard deviation of 11, the formula for the sample size (n) requires using a Z-score of 1.96 corresponding to the confidence level. After plugging in the values, the calculation suggests a sample size of approximately 51.61, which rounds up to 52, making the correct answer D. 52.
Step-by-step explanation:
The question is asking about determining the sample size required to achieve a margin of error of 3 or less for an estimate with a 95% confidence level, when the population standard deviation is known to be 11. To calculate the sample size, we use the formula for the sample size of a population mean when the population standard deviation is known, given by: n = (Z*σ/E)^2
Where:
- n is the sample size
- Z is the Z-score corresponding to the desired confidence level (which is approximately 1.96 for a 95% confidence level)
- σ is the population standard deviation
- E is the margin of error
Substituting the given values into the formula: n = (1.96 * 11 / 3)^2
Calculating this yields n = (21.56 / 3)^2 = (7.18667)^2 ≈ 51.61.
Since we cannot have a fraction of a sample, we round up to the nearest whole number, giving us n = 52. Therefore, the minimum sample size needed is 52, making the answer D. 52.