Use the given degree of confidence and sample data to construct a confidence interval for the population mean µ. Assume that the population has a normal distribution.
-A laboratory tested twelve chicken eggs and found that the mean amount of cholesterol was 198 milligrams with $\mathrm { s } = 10.5$ milligrams. Construct a $95 \%$ confidence interval for the true mean cholesterol content of all such eggs.
A) $191.3 \mathrm { mg }

Question

Use the given degree of confidence and sample data to construct a confidence interval for the population mean µ. Assume that the population has a normal distribution.
-A laboratory tested twelve chicken eggs and found that the mean amount of cholesterol was 198 milligrams with $\mathrm { s } = 10.5$ milligrams. Construct a $95 \%$ confidence interval for the true mean cholesterol content of all such eggs.
A) $191.3 \mathrm { mg } < \mu < 204.7 \mathrm { mg }$
B) $191.4 \mathrm { mg } < \mu < 204.6 \mathrm { mg }$
C) $191.2 \mathrm { mg } < \mu < 204.8 \mathrm { mg }$
D) $192.6 \mathrm { mg } < \mu < 203.4 \mathrm { mg }$

Quizplus · Accepted Answer

The correct answer is A because the confidence interval is calculated as:$$\bar{x} \pm z_{\alpha/2} \frac{\sigma}{\sqrt{n}}$$where $\bar{x}$ is the sample mean, $z_{\alpha/2}$ is the critical value from the standard normal distribution corresponding to the desired confidence level, $\sigma$ is the population standard deviation, and $n$ is the sample size.In this case, we have $\bar{x} = 198$, $z_{0.025} = 1.96$, $\sigma = 10.5$, and $n = 12$. Plugging these values into the formula, we get:$$198 \pm 1.96 \frac{10.5}{\sqrt{12}} = 198 \pm 6.7$$Therefore, the 95% confidence interval is $191.3 \mathrm { mg } < \mu < 204.7 \mathrm { mg }$.

Use the Given Degree of Confidence and Sample Data to Construct