The authors of the paper "Weight-Bearing Activity during Youth Is a More Important Factor for Peak Bone Mass than Calcium Intake" studied a number of variables they thought might be related to bone mineral density (BMD). The accompanying data on
x = weight
at age 13 and
y = bone
mineral density at age 27 are consistent with summary quantities for women given in the paper.
Weight (kg) | BMD (g/cm2) |
---|---|
54.4 | 1.15 |
59.3 | 1.26 |
74.6 | 1.42 |
62.0 | 1.06 |
73.7 | 1.44 |
70.8 | 1.02 |
66.8 | 1.26 |
66.7 | 1.35 |
64.7 | 1.02 |
71.8 | 0.91 |
69.7 | 1.28 |
64.7 | 1.17 |
62.1 | 1.12 |
68.5 | 1.24 |
58.3 | 1.00 |
A simple linear regression model was used to describe the relationship between weight at age 13 and BMD at age 27. The following values are given for this data.
a =
0.558
b = 0.009
n = 15
SSTo = 0.356 SSResid = 0.313
(a)
What percentage of observed variation in BMD at age 27 can be explained by the simple linear regression model? (Round your answer to one decimal place.)
%
(b)
Give an estimate of σ. (Round your answer to three decimal places.)
(c)
Give an estimate of the average change in BMD associated with a 1 kg increase in weight at age 13. (Round your answer to three decimal places.)
g/cm2
(d)
Compute an estimate of the mean BMD at age 27 for women whose weight at age 13 was 60 kg. (Round your answer to three decimal places.)
g/cm2
Here as given is
a = 0.558
b = 0.009
n = 15
SST0 = 0.356
SSresid = 0.313
(a) Here we have to find the coefficient of determination
r2 = (SST0 - SSresid )/SST0 = (0.356 - 0.313)/0.356 = 0.1211 or 12.1%
(b) Here estimate of = sqrt [SSresid/dFresid]
dFresid = n -2 = 15 -2 = 13
estimate of = sqrt (0.313/13) = 0.155
This is a typical deviation of a bone mineral density value in the sample from the value predicted by the least squares line.
(c) Here b = 0.009
so here 1 kg weight increase in weight in age 13 will increase on average 0.009 g/cm2
(d) Here x = 60
y = 0.558 + 0.009 * 60= 1.098 g/cm2
------------------------------------
DEAR STUDENT,
IF YOU HAVE ANY QUERY ASK ME IN THE COMMENT BOX,I AM HERE TO HELPS YOU.PLEASE GIVE ME POSITIVE RATINGS
*****************THANK YOU***************
Get Answers For Free
Most questions answered within 1 hours.