RISK FACTORS FOR DIABETES MELLITUS IN INDONESIA: ANALYSIS OF IFLS DATA 2014

Diabetes Mellitus Risk Factor IFLS 2014 Logistic Regression

Authors

  • Alfiana Miranda Nur Afifah
    alfianamirandsanurafifah@gmail.com
    Faculty of Public Health, Universitas Airlangga, 60115 Surabaya, East Java, Indonesia
  • Diah Indriani Faculty of Public Health, Universitas Airlangga, 60115 Surabaya, East Java, Indonesia
  • Susy Kartikana Sebayang Faculty of Public Health, Universitas Airlangga, 60115 Surabaya, East Java, Indonesia
  • Erni Astutik Faculty of Public Health, Universitas Airlangga, 60115 Surabaya, East Java, Indonesia
November 1, 2022
06

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Diabetes Mellitus (DM) remains a public health problem that needs attention in various countries, including in Indonesia that has the 4th largest DM cases globally. The World Health Organization (WHO) estimates that, by 2030, the number of people with DM will be twice higher than the current number which is 180 million people worldwide. Diabetes mellitus can be prevented and avoided by taking control of the risk factors. The purpose of this study was to assess the risk factors of Diabetes Mellitus in Indonesia. This is a secondary data analysis of the Indonesian Family Life Survey (IFLS) 5 which was a cross sectional survey. To analyze risk factors of DM, Chi-square and logistic regression test were used. The risk factors analyzed included sex, age, marital status, history of hypertension, cholesterol levels, obesity, smoking habit, employment status, habit of consuming fast food, consumption of sweet foods, and consumption soft drinks. Significant risk  factors  of  diabetes mellitus included age (OR = 5.28, 95% CI: 4.37 – 6.37; p value = 0.001), marital status (OR = 1.69, 95% CI: 1.36 – 2.09; p value = 0.001), history of hypertension (OR = 2.67, 95% CI: 2.25 – 3.17; p value = 0.001), cholesterol levels (OR = 4.36, 95% CI: 3.58 – 5.31; p value = 0.001), employment status (OR = 1.52, 95% CI: 1.29 – 1.80; p value = 0.001), and habit of consuming sweet foods (OR = 0.63, 95% CI: 0.48 – 0.83; p value = 0.002).

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