Construct a 95% confidence interval for the population mean, μ. Assume the population has a normaldistribution. A sample of 25 randomly English majors has a mean test score of 81.5 with a standard deviation of10.2. Round to the nearest hundredth.
A) (77.29, 85.71)
B) (56.12, 78.34)
C) (66.35, 69.89)
D) (87.12, 98.32)

Question

Quizplus · Accepted Answer

The 95% confidence interval for the mean is calculated using the formula: $\bar{x} \pm z\frac{\sigma}{\sqrt{n}}$, where $\bar{x}$ is the sample mean, $z$ is the z-score corresponding to the confidence level (for 95%, $z = 1.96$), $\sigma$ is the standard deviation, and $n$ is the sample size. Plugging in the given values: $81.5 \pm 1.96\frac{10.2}{\sqrt{25}}$, we get the interval (77.29, 85.71).

Construct a 95% Confidence Interval for the Population Mean, μ