BioMed Research International

Volume 2015, Article ID 426847, 7 pages

http://dx.doi.org/10.1155/2015/426847

## Multiple Comparison of Age Groups in Bone Mineral Density under Heteroscedasticity

^{1}Department of Statistics, Faculty of Science, Anadolu University, 26470 Eskisehir, Turkey^{2}Department of Physical Medicine and Rehabilitation, Medical Faculty, University of Uludag, 16059 Bursa, Turkey

Received 22 February 2015; Accepted 23 April 2015

Academic Editor: Heather F. Smith

Copyright © 2015 Ahmet Sezer et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

#### Abstract

Osteoporosis is a silent disease because individuals may not know that they have osteoporosis until their bones become so fragile. Bone mineral density (BMD) test helps to detect osteoporosis and determine the risk fractures. This study covers bone measurement data from total body dual energy X-ray absorptiometry scans for 28,454 persons who participated in the 1996–2006 National Health and Nutrition Examination Survey in USA Dual energy X-ray absorptiometry (DXA) method is known as the primary method for detecting osteoporosis because of its high precision and accuracy. Testing the equality of the means of normal populations when the variances are unknown and unequal is a fundamental problem in clinical trials and biomedical research. In this study we compare age groups based upon BMD in case of unequal variance being present among the groups. First we test equality of variances among the age groups by the Hartley test. And then Scott-Smith test is used to test equality of BMD means for the age groups. Finally, Tukey-Cramer confidence intervals are constructed to detect which groups start to differ from the reference group in which BMD reaches the peak level.

#### 1. Introduction

Osteoporosis is a systemic skeletal disease characterized by low bone density and micro architectural deterioration of bone tissue with a consequent increase in bone fragility [1]. Measurement of BMD can be used to determine fracture risk and monitor the effects of treatment. Early detection of bone loss is essential to preventing osteoporosis. In fact, osteoporosis affects more than 75 million people in Europe, Japan, and USA and causes more than 2.3 million fractures annually in Europe and USA. The lifetime risk for hip, vertebral, and forearm (wrist) fractures has been estimated to be around 40%, very close to that for coronary heart disease. Osteoporosis does not only cause fractures, but also causes people to become bedridden and causes back pain and loss of height. Prevention of the disease and its associated fractures is important for maintaining health, quality of life, and independence among the elderly.

Early osteoporosis is not usually detected and most of the time does not become clinically evident until fractures occur. Loss of bone density occurs with advancing age and rates of fracture increase with age, giving rise to significant morbidity and mortality. Osteoporosis is three times more common in women than in men, because women have a lower peak bone mass and hormonal changes occur at the menopause. Estrogens have an important function in preserving bone mass during adulthood, and bone loss occurs as levels decline, usually around the age of 50 years. In addition, women live longer than men [2] and therefore have greater reductions in bone mass.

Until recently, osteoporosis was considered an inevitable consequence of ageing. With Improvements in diagnostic technology and assessment facilities, now it is easier to detect the disease before fractures occur. The substantial bone loss is usually around age 65 years in men and 50 years in women [3]. Females tend to keep peak mineral content until menopause; after that it drops about 15% per decade. Fracture rates increase rapidly with age and the lifetime. Many studies show that an inadequate supply of calcium over a lifetime contributes to the development of osteoporosis. The body’s demand for calcium is greater during childhood and adolescence and during pregnancy and breastfeeding. Epidemiological studies indicate that a 10% increase in peak bone mass in the Caucasian female population would decrease the risk of hip fracture by about 30%. Clearly, eliminating the risk factors might significantly reduce the burden of osteoporosis. Obvious interventions include raising levels of exercise, stopping smoking, and increasing dietary intake of calcium [4].

#### 2. Materials and Methods

Testing the equality of the means of normal populations when the variances are unknown and unequal is a fundamental problem in clinical trials and biomedical research. A well known case is the Behrens-Fisher (BF) problem, which focuses on two populations. Behrens-Fisher problem is popular because there is no exact solution satisfying the classical criteria for good tests. The problem is seemingly simple, yet much effort has been made to try to solve this problem [5–7].

It is well known that there exists an analysis of variance (ANOVA) -test for the problem of testing the equality of means from several independent samples under the assumptions of normality. It is well known that the usual -test is not robust to the assumption of equal variances. There is, however, no standard procedure for testing this hypothesis when variances are not equal, and various approximate test procedures have been proposed in the literature. The best known procedure is the test proposed by Welch (1947) [8] and its modifications. Other tests have been proposed by James (1951) [9], Scott and Smith (1971) [10], Brown and Forsythe (1974) [11], S.-Y. Chen and H. J. Chen (1998) [12], Rice and Gaines (1989) [13], Krishnamoorthy et al. (2007) [14], Weerahandi (1995) [15], and Xu and Wang (2008) [16, 17].

##### 2.1. Hartley’s Test for Testing Variance

Hartley’s test () was developed by Hartley in 1950 [18]. This test assumes that data within each group are normally distributed and test involves computing the ratio of the largest group variance to the smallest . This ratio will be compared with the critical value from a table of the sampling distribution of . In this test null hypothesis states that all groups have equal variances alternative to at least one group differing from the others. Considerwhere is the number of the group.

Since we have large sample sizes for the groups, we used critical value 1 from Hartley’s table to make our decision. For example Hartley test is calculated for the first group, Non-Hispanic Mexican-American female group; consider

Since calculated test statistic value is bigger than 1 (critical value from the Hartley table), we reject the null hypothesis that equality of variance assumption is violated among age groups of Mexican-American females. It is evident that equal variance assumption is violated for all the cases we consider (Tables 1–6) and that leads us to check equality of means under heteroscedasticity.