A random sample of 80 adults is chosen and their mean serum cholesterol level is found to be 202 milligrams per deciliter. Assume that the population standard deviation is $\sigma = 40$ Based on a 95% confidence interval for the mean serum cholesterol, is it likely that the mean serum
Cholesterol is greater than 219? (Hint: you should first construct the 95% confidence interval.)
A) No
B) The likelihood cannot be determined.
C) Yes

Question

Quizplus · Accepted Answer

To construct a 95% confidence interval for the mean serum cholesterol, we use the formula:
$$\bar{x} \pm Z_{\alpha/2} 	imes \frac{\sigma}{\sqrt{n}}$$
where $\bar{x} = 202$, $\sigma=40$, $n=80$, and $Z_{\alpha/2}$ is the z-score that corresponds to the desired confidence level, which is 95%. From a standard normal distribution, we find that $Z_{\alpha/2} = 1.96$.
Plugging in the values, we get:
$$202 \pm 1.96 	imes \frac{40}{\sqrt{80}} \approx 192.6 	ext{ to } 211.4$$
This means we are 95% confident that the true mean serum cholesterol level is between 192.6 and 211.4 milligrams per deciliter.

To determine if the mean serum cholesterol is likely to be greater than 219, we compare 219 to the lower bound of the confidence interval, which is 192.6. Since 219 is greater than 192.6, it is outside the confidence interval, and thus it is not likely that the mean serum cholesterol is greater than 219.

A Random Sample of 80 Adults Is Chosen and Their σ=40\sigma = 40σ=40

A Random Sample of 80 Adults Is Chosen and Their $\sigma = 40$