A random sample of 60 adults is chosen and their mean serum cholesterol level is found to be 196 milligrams per deciliter. Assuming that the population standard deviation is $\sigma = 43$ , compute a 99%
Confidence interval for the mean serum cholesterol level for adults.

Question

Quizplus · Accepted Answer

We will use the formula for a confidence interval:

$$ \text{CI} = \bar{x} \pm z_{\alpha / 2} \frac{\sigma}{\sqrt{n}} $$

where $\bar{x}$ is the sample mean, $\sigma$ is the population standard deviation, $n$ is the sample size, and $z_{\alpha / 2}$ is the $z$-score corresponding to the desired level of confidence.

Plugging in the values we get:

$$ \text{CI} = 196 \pm 2.576 \cdot \frac{43}{\sqrt{60}} \approx (181.8, 210.2) $$

Rounding to the nearest whole number we get the final answer as (182, 210), which corresponds to choice C.

A Random Sample of 60 Adults Is Chosen and Their σ=43\sigma = 43σ=43

A Random Sample of 60 Adults Is Chosen and Their $\sigma = 43$