Question

For many important processes that occur in the body, direct measurement of characteristics of the process is not possible. In many cases, however, we can measure a biomarker, a biochemical substance that is relatively easy to measure and is associated with the process of interest. Bone turnover is the net effect of two processes: the breaking down of old bone, called resorption, and the building of new bone, called formation. A biomarker for bone formation measured was osteocalcin (OC), measured in the blood. The units are nanograms per milliliter (ng/ml). For the 31 subjects in the study the mean was 33.4 ng/ml. Assume that the standard deviation is known to be 19.6 ng/ml. Report the 95% confidence interval:

Answer 1

Answer:

The 95% confidence interval is between 26.5 ng/ml and 40.3 ng/ml

Step-by-step explanation:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1-0.95}{2} = 0.025$

Now, we have to find z in the Ztable as such z has a pvalue of $1-\alpha$.

So it is z with a pvalue of $1-0.025 = 0.975$, so $z = 1.96$

Now, find the margin of error M as such

$M = z*\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

$M = 1.96*\frac{19.6}{\sqrt{31}} = 6.9$

The lower end of the interval is the sample mean subtracted by M. So it is 33.4 - 6.9 = 26.5 ng/ml

The upper end of the interval is the sample mean added to M. So it is 6.4 + 33.4 + 6.9 = 40.3 ng/ml

The 95% confidence interval is between 26.5 ng/ml and 40.3 ng/ml