Question

Assume researchers are trying to find out the mean concentration levels of a specific drug in a population of individuals’ blood during a clinical trail. The researchers find that the sample population has a mean concentration of 6.7 mcg/mL with a standard deviation of 1.3 mcg/mL. Calculate a 95% confidence interval for the mean concentration level of the medication for the population’s blood and interpret it

The researchers are 95.0% confident the true mean concentration of medication in the population’s blood is between 4.152 mcg/mL and 9.248 mcg/mL.

The researchers are 95.0% confident the true mean concentration of medication in the population’s blood is between 4.555 mcg/mL and 8.845 mcg/mL.

The researchers are 95.0% confident the true mean concentration of medication in the population’s blood is between 4.568 mcg/mL and 8.832 mcg/mL.

The researchers are 95.0% confident the true mean concentration of medication in the population’s blood is between 4.5615 mcg/mL and 8.8385 mcg/mL.

Answer 1

Here it is given that sample mean = $\overline{x}=6.7$ and sample deviation =$\sigma =\frac{\sigma }{\sqrt{n}}=1.3$

Now z value for 95% CI is z=1.96

Hence Margin of Error = $E=z*\sigma=1.96*1.3=2.548$

So Confidence interval $CI=\overline{x} \pm E=6.7 \pm 2.548=(4.152,9.248)$

The researchers are 95.0% confident the true mean concentration of medication in the population’s blood is between 4.152 mcg/mL and 9.248 mcg/mL.