Healthy bones are continually being renewed by two processes. Through boue forma

Question

Healthy bones are continually being renewed by two processes. Through boue formation new bi hough bone resorption. old bone is removed. If one or both of these processes is disease, aging or space travel, for example. bone loss can be the result. The variables VOMINUs measure bone formation and boue bone is disturbed. by variables VOPLUS ad resorption, respectively. Osteocalcin (OC) is a biochemical higher levels of OC. A is used to measure OC, and it is much less expensive to obtain than direct measures of bone tartrate resistant e (TRAP) is a biochemical marker for bone resorption that is also measured in blood. It is in units per liter (U/L). These variables were measured in a study of 31 healthy women aged 11 marker for bone formation: higher levels of bone formation are associated wt fonmation. The units are milligrams of OC per milliliter of blood (nghm o 32 years. The results were published in C. M. Weaver et al., "Quantification of biocheuical markers of bone turuover by kinetic measures of bone formation and resorption in young healthy females." Journal of Bone and Mineral Research. 12 (1997), pp. 1714-1720 (016-022) We are interested in prediction VOPLUS using the other variables. We took the log transformation for all the variables, and below the matrix plot and the correlation matrix for the different variables Correlations Loglvoplus] Log[oc]Logltrap] Loglvominus] Log[voplus] 1.0000 0.7734 Log[trap] 0.7549 oglvominus 0.8396 0.7734 1.0000 0.7953 0.7549 0.7953 1.0000 0.6642 0.8396 0.5544 0.6642 1.0000 Log[oc] 0,30 0:3541 0 900000 Scatterplot Matrix 7.5 7 6.5 Log[voplus] 6 5.5 4 3.5 3 Log[oc] 2.5. 2 3 2.5 2 1.5 7.5 6.5 6 5.5 6 6.5 7 7.5 22.5 3 3.54 11.5 2 2.5 3 6 6.5

Explanation / Answer

Q.18 As to check the utility of the model is calculated by F test. Which tells us the significance of the test.

Here Value of F is 47.96 and associated p- value is < 0.0001

so we will reject the null hypothesis and can say that the model have some utility.

Q.19

Ho : (Log(trap) ) = 0

Ha : (Log(trap) ) 0

Here coefficient of Log(Trap) = 0.0278587

SE(COeffiicent) = .156972

test static

t = 0.0278587/0.156972 = 0.1775

p - value = Pr(t > 0.1775; dF = 27) = 0.8604

so here p -vallue is greater than 0.05 so we can say that it is insignificance in nature.

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