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.

Required:
Give the margin of error and find a 95% confidence interval for the mean TRAP amount in young women represented by this sample.

Answer 1

The margin of error is
`E = 6.9`

The 95% confidence interval is
`26.5  <  \mu < 40.3`

Explanation:

From the question we are told that

The sample size is n = 31

The mean is
`\mu = 33.4 \ ng/ml`

The standard deviation is
`\sigma = 19.6 \ ng/ml`

From the question we are told the confidence level is 95% , hence the level of significance is

`\alpha = (100 - 95 ) \%`

=>
`\alpha = 0.05`

Generally from the normal distribution table the critical value of
`(\alpha )/(2)` is

`Z_{(\alpha )/(2) } =  1.96`

Generally the margin of error is mathematically represented as

`E = Z_{(\alpha )/(2) } *  (\sigma )/(√(n) )`

=>
`E = 1.96  *  (19.6  )/(√(31 ) )`

=>
`E = 6.9`

Generally 95% confidence interval is mathematically represented as

`\= x -E <  \mu <  \=x  +E`

=>
`33.4  - 6.9 <  \mu < 33.4  +  6.9`

=>
`26.5  <  \mu < 40.3`