Bone densitometry






Test results

The bone densitometry test determines the bone mineral density (BMD). BMD is compared to two norms: healthy young adults (a T-score) and age-matched (a Z-score).

First, the BMD result is compared with the BMD results from healthy 25- to 35-year-old adults of the same sex and ethnicity as the patient. The standard deviation (SD) is the difference between the patient’s BMD and that of the healthy young adults. This result is the patient’s T-score. Positive T-scores indicate the bone is stronger than normal; negative T-scores indicate the bone is weaker than normal.



In general, the risk for bone fracture doubles with every SD below normal. Thus, a person with a BMD of one SD below normal (T-score of -1) has twice the risk for bone fracture as a person with a normal BMD. A person with a T-score of minus two has four times the risk for bone fracture as a person with a normal BMD. When this information is known, people with a high risk for bone fracture can be treated with the goal of preventing future fractures. Severe (established) osteoporosis is defined as having a bone density that is more than two and a half SD below the young adult mean with one or more past fractures due to osteoporosis.

Secondly, the patient’s BMD is compared to an age-matched norm. This is called the Z-score. Z-scores are calculated in the same way, but the comparisons are made to someone of the patient’s age, sex, race, height, and weight.





Bone densitometry testing may also be used:

  • To confirm a diagnosis of osteoporosis if a patient has already had a fracture
  • To predict chances of fracturing in the future
  • To determine the rate of bone loss and/or monitor the effects of treatment