Find the margin of error.
-A laboratory tested twelve chicken eggs and found that the mean amount of cholesterol was 193 milligrams with s = 15.4 milligrams. Construct a 95 percent confidence interval for the true mean cholesterol content of all such eggs.
A) $183.2

Question

Find the margin of error.
-A laboratory tested twelve chicken eggs and found that the mean amount of cholesterol was 193 milligrams with s = 15.4 milligrams. Construct a 95 percent confidence interval for the true mean cholesterol content of all such eggs. 
A) $183.2 < \mu < 202.8$
B) $185.0 < \mu < 201.0$
C) $183.1 < \mu < 202.9$
D) $183.3 < \mu < 202.7$

Quizplus · Accepted Answer

The margin of error (E) for a 95% confidence interval when the population standard deviation is unknown and the sample size is small (n < 30) is calculated using the t-distribution. The formula for the margin of error in this context is $E = t_{\alpha/2} \times \frac{s}{\sqrt{n}}$, where $t_{\alpha/2}$ is the t-score that corresponds to the desired confidence level (95% in this case), $s$ is the sample standard deviation, and $n$ is the sample size.Given: $s = 15.4$ milligrams, $n = 12$, and for a 95% confidence interval with 11 degrees of freedom (n-1), the t-score ($t_{\alpha/2}$) can be found in a t-distribution table or calculated using statistical software. Typically, for 11 degrees of freedom, the t-score is approximately 2.201.Calculating the margin of error:$$E = 2.201 \times \frac{15.4}{\sqrt{12}} \approx 9.8$$The mean amount of cholesterol is given as 193 milligrams. Therefore, the 95% confidence interval for the true mean is:$$193 \pm 9.8 = (183.2, 202.8)$$This matches choice A: $183.2 < \mu < 202.8$.

Find the Margin of Error 183.2<μ<202.8183.2 < \mu < 202.8183.2<μ<202.8 B) 185.0<μ<201.0185.0 < \mu < 201.0185.0<μ<201.0

Find the Margin of Error $183.2 < \mu < 202.8$ B) $185.0 < \mu < 201.0$