Construct a 99% confidence interval for the population mean, $\mu.$ Assume the population has a normal distribution. A group of 19 randomly selected students has a mean age of 22.4 years with a standard deviation
Of 3.8 years.
A)(19.9, 24.9)
B)(16.3, 26.9)
C)(17.2, 23.6)
D)(18.7, 24.1)

Question

Quizplus · Accepted Answer

The 99% confidence interval for the mean can be 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 99% confidence, $z \approx 2.576$), $\sigma$ is the standard deviation, and $n$ is the sample size. Plugging in the given values: $22.4 \pm 2.576\frac{3.8}{\sqrt{19}}$ results in the interval (19.9, 24.9).

Construct a 99% Confidence Interval for the Population Mean μ.\mu.μ. Assume the Population Has a Normal Distribution

Construct a 99% Confidence Interval for the Population Mean $\mu.$ Assume the Population Has a Normal Distribution