在残疾人和无残疾人之间使用的宫癌查的比较
Chia-Yu Chen1, Pei-Tseng Kung2,3, Li-Ting Chiu4
1Graduate Institute of Public Health, China Medical University, Taichung 406040, Taiwan.
Healthcare (Basel, Switzerland)
|May 27, 2023
概括
患有残疾的女性比没有残疾的女性有较低的帕普查率 (PST). 这项研究强调了残疾人宫癌查的差异,特别是那些有认知障碍的人.
科学领域:
- 公共卫生 公共卫生
- 妇科瘤学 妇科瘤学
- 健康差异 在健康上的差异
背景情况:
- 宫癌是全球女性的重大健康问题.
- 高水平的宫癌查率对于早期发现和预防至关重要.
- 了解弱势群体之间的查差异至关重要.
研究的目的:
- 为了比较台湾残疾人和没有残疾人的帕普查 (PST) 使用情况.
- 为了确定与较低的PST率相关的特定残疾类型.
主要方法:
- 一个回顾性队列研究,利用台湾的残疾登记档案和NHIRD.
- 倾向分数匹配 (PSM) 创建了一个1:1队列,包括186,717名残疾人和186,717名对照.
- 条件后勤回归分析了接受PST的几率,控制了相关变量.
主要成果:
- 与没有残疾的人 (21.82%) 相比,残疾人的PST率较低 (16.93%).
- 对于残疾人来说,接受PST的几率明显较低 (OR = 0.74).
- 对于那些有智力/发育残疾 (OR = 0.38),痴呆症 (OR = 0.40) 和多重残疾 (OR = 0.52) 的人来说,观察到的几率较低.
结论:
- 医疗保健提供者必须承认残疾人对有效查的独特需求.
- 认知障碍对宫癌查构成重大障碍.
- 需要有针对性的干预措施来改善残疾人群体中PST的发生率.
更多相关视频
07:13Comparison of Predictive Performance of Three Lymph Node Staging Systems in Colorectal Signet Ring Cell Carcinoma Based on Machine Learning Model
Published on: April 18, 2025
150
07:31A Computerized Functional Skills Assessment and Training Program Targeting Technology Based Everyday Functional Skills
Published on: February 13, 2020
7.0K
相关概念视频
Comparing the Survival Analysis of Two or More Groups
229
Survival analysis is a cornerstone of medical research, used to evaluate the time until an event of interest occurs, such as death, disease recurrence, or recovery. Unlike standard statistical methods, survival analysis is particularly adept at handling censored data—instances where the event has not occurred for some participants by the end of the study or remains unobserved. To address these unique challenges, specialized techniques like the Kaplan-Meier estimator, log-rank test, and...
229
Test for Homogeneity
2.0K
The goodness–of–fit test can be used to decide whether a population fits a given distribution, but it will not suffice to decide whether two populations follow the same unknown distribution. A different test, called the test for homogeneity, can be used to conclude whether two populations have the same distribution. To calculate the test statistic for a test for homogeneity, follow the same procedure as with the test of independence. The hypotheses for the test for homogeneity can...
2.0K
Preventive Healthcare Services
1.1K
Preventive healthcare services keep people healthy via frequent check-ups, screening, and counseling. They primarily aid in disease prevention rather than treating an acute or chronic illness. Preventive treatment also keeps individuals productive and energetic, allowing them to work well into their retirement years. Examples of preventive care services include:
1.1K
