A sample of size n = 16 is drawn from an approximately normal population whose standard deviation is $\sigma = 10.5$ The sample mean is $\bar { x } = 51.0$ Construct a 99% confidence interval for $\boldsymbol { \mu }$
A) (48.91, 53.09)
B) (38.28, 63.72)
C) 44.24, 57.76
D) (51.00, 57.76)

Question

Quizplus · Accepted Answer

The 99% confidence interval for the mean is calculated using the formula: $\bar{x} \pm z \frac{\sigma}{\sqrt{n}}$. For 99% confidence, $z \approx 2.576$. Thus, $51 \pm 2.576 \times \frac{10.5}{\sqrt{16}}$ gives the interval (44.24, 57.76).

A Sample of Size N = 16 Is Drawn from an Approximately