Math  /  Data & Statistics

QuestionConstruct a 95%95 \% confidence interval for the population mean, μ\mu. Assume the population has a normal distribution. A sample of 25 randomly selected students has a mean test score of 81.5 with a standard deviation of 10.2. A. (56.12,78.34)(56.12,78.34) B. (77.29,85.71)(77.29,85.71) C. (66.35,69.89)(66.35,69.89) D. (87.12,98.32)(87.12,98.32)

Studdy Solution
Compute the confidence interval:
xˉ±t0.025×SEM=81.5±2.064×2.04 \bar{x} \pm t_{0.025} \times \text{SEM} = 81.5 \pm 2.064 \times 2.04
Calculate the margin of error:
2.064×2.04=4.2096 2.064 \times 2.04 = 4.2096
Calculate the confidence interval:
81.54.2096=77.2904 81.5 - 4.2096 = 77.2904 81.5+4.2096=85.7096 81.5 + 4.2096 = 85.7096
Thus, the 95% 95\% confidence interval is approximately:
(77.29,85.71) (77.29, 85.71)
The correct answer is:
B:(77.29,85.71) \boxed{B: (77.29, 85.71)}

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