Final answer:
The confidence interval with the provided data (mean = 81.5, SD = 10.2, sample size = 25, z-score = 1.96) is 81.5 ± 4.12 . Option b is correct.
Step-by-step explanation:
To construct a 95% confidence interval for the population mean (μ), we use the formula:
Confidence Interval = μ ± (z* × (SD/√N))
Where:
- μ is the sample mean
- z* is the z-score for the desired confidence level
- SD is the sample standard deviation
- N is the sample size
In the given problem, we have a sample mean (μ) of 81.5, a standard deviation (SD) of 10.2, and a sample size (N) of 25. For a 95% confidence level, the z-score (z*) is approximately 1.96 when using the standard normal (Z) distribution.
Let's calculate:
The standard error:
SE = SD/√N)
= 10.2 / √25
= 10.2 / 5
= 2.04
The margin of error (MOE = z* × SE)
= 1.96 × 2.04
≈ 4.12
Therefore, the confidence interval is μ ± MOE = 81.5 ± 4.00
So the 95% confidence interval for the population mean (μ) is (81.5 - 4.12, 81.5 + 4.12)
Hence, option B is correct.