AI trong phân tích dữ liệu với R - Viện Phương Nam (6-11/1/2026)
Thach Tran
2025-12-28
Ngày 2: So sánh 2 nhóm
Việc 1. Đọc dữ liệu vào R
df = read.csv("C:\\Thach\\VN trips\\2026_1Jan\\PN Institute\\Datasets\\Bone data.csv")
dim(df)## [1] 2162 9head(df)## id sex age weight height prior.fx fnbmd smoking fx
## 1 1 Male 73 98 175 0 1.08 1 0
## 2 2 Female 68 72 166 0 0.97 0 0
## 3 3 Male 68 87 184 0 1.01 0 0
## 4 4 Female 62 72 173 0 0.84 1 0
## 5 5 Male 61 72 173 0 0.81 1 0
## 6 6 Female 76 57 156 0 0.74 0 0Việc 2. So sánh mật độ xương giữa nam và nữ
2.1 Vẽ biểu đồ histogram đánh giá phân bố mật độ xương
library(lessR)## Warning: package 'lessR' was built under R version 4.3.3##
## lessR 4.3.9 feedback: gerbing@pdx.edu
## --------------------------------------------------------------
## > d <- Read("") Read text, Excel, SPSS, SAS, or R data file
## d is default data frame, data= in analysis routines optional
##
## Many examples of reading, writing, and manipulating data,
## graphics, testing means and proportions, regression, factor analysis,
## customization, and descriptive statistics from pivot tables
## Enter: browseVignettes("lessR")
##
## View lessR updates, now including time series forecasting
## Enter: news(package="lessR")
##
## Interactive data analysis
## Enter: interact()Histogram(fnbmd,
data = df,
fill = "blue",
xlab = "Mật độ xương ở cổ xương đùi (g/cm2)",
ylab = "Số người",
main = "Phân bố mật độ xương")
## >>> Suggestions
## bin_width: set the width of each bin
## bin_start: set the start of the first bin
## bin_end: set the end of the last bin
## Histogram(fnbmd, density=TRUE) # smoothed curve + histogram
## Plot(fnbmd) # Violin/Box/Scatterplot (VBS) plot
##
## --- fnbmd ---
##
## n miss mean sd min mdn max
## 2122 40 0.829 0.155 0.280 0.820 1.510
##
##
## --- Outliers --- from the box plot: 33
##
## Small Large
## ----- -----
## 0.3 1.5
## 0.3 1.5
## 0.4 1.4
## 0.4 1.4
## 0.4 1.4
## 0.4 1.4
## 0.4 1.4
## 0.4 1.4
## 0.4 1.3
## 0.4 1.3
## 0.4 1.3
## 1.3
## 1.3
## 1.3
## 1.3
## 1.2
## 1.2
## 1.2
##
## + 15 more outliers
##
##
## Bin Width: 0.1
## Number of Bins: 14
##
## Bin Midpnt Count Prop Cumul.c Cumul.p
## ---------------------------------------------------
## 0.2 > 0.3 0.25 1 0.00 1 0.00
## 0.3 > 0.4 0.35 9 0.00 10 0.00
## 0.4 > 0.5 0.45 15 0.01 25 0.01
## 0.5 > 0.6 0.55 103 0.05 128 0.06
## 0.6 > 0.7 0.65 306 0.14 434 0.20
## 0.7 > 0.8 0.75 522 0.24 956 0.44
## 0.8 > 0.9 0.85 534 0.25 1490 0.69
## 0.9 > 1.0 0.95 371 0.17 1861 0.86
## 1.0 > 1.1 1.05 183 0.08 2044 0.95
## 1.1 > 1.2 1.15 48 0.02 2092 0.97
## 1.2 > 1.3 1.25 21 0.01 2113 0.98
## 1.3 > 1.4 1.35 6 0.00 2119 0.98
## 1.4 > 1.5 1.45 2 0.00 2121 0.98
## 1.5 > 1.6 1.55 1 0.00 2122 0.982.2 So sánh mật độ xương giữa nam và nữ
library(table1)##
## Attaching package: 'table1'## The following object is masked from 'package:lessR':
##
## label## The following objects are masked from 'package:base':
##
## units, units<-table1(~ fnbmd | sex, data = df)| Female (N=1317) |
Male (N=845) |
Overall (N=2162) |
|
|---|---|---|---|
| fnbmd | |||
| Mean (SD) | 0.778 (0.132) | 0.910 (0.153) | 0.829 (0.155) |
| Median [Min, Max] | 0.770 [0.280, 1.31] | 0.900 [0.340, 1.51] | 0.820 [0.280, 1.51] |
| Missing | 17 (1.3%) | 23 (2.7%) | 40 (1.9%) |
ttest(fnbmd ~ sex, data = df)##
## Compare fnbmd across sex with levels Male and Female
## Grouping Variable: sex
## Response Variable: fnbmd
##
##
## ------ Describe ------
##
## fnbmd for sex Male: n.miss = 23, n = 822, mean = 0.910, sd = 0.153
## fnbmd for sex Female: n.miss = 17, n = 1300, mean = 0.778, sd = 0.132
##
## Mean Difference of fnbmd: 0.132
##
## Weighted Average Standard Deviation: 0.141
##
##
## ------ Assumptions ------
##
## Note: These hypothesis tests can perform poorly, and the
## t-test is typically robust to violations of assumptions.
## Use as heuristic guides instead of interpreting literally.
##
## Null hypothesis, for each group, is a normal distribution of fnbmd.
## Group Male: Sample mean assumed normal because n > 30, so no test needed.
## Group Female: Sample mean assumed normal because n > 30, so no test needed.
##
## Null hypothesis is equal variances of fnbmd, homogeneous.
## Variance Ratio test: F = 0.023/0.018 = 1.336, df = 821;1299, p-value = 0.000
## Levene's test, Brown-Forsythe: t = 3.449, df = 2120, p-value = 0.001
##
##
## ------ Infer ------
##
## --- Assume equal population variances of fnbmd for each sex
##
## t-cutoff for 95% range of variation: tcut = 1.961
## Standard Error of Mean Difference: SE = 0.006
##
## Hypothesis Test of 0 Mean Diff: t-value = 21.080, df = 2120, p-value = 0.000
##
## Margin of Error for 95% Confidence Level: 0.012
## 95% Confidence Interval for Mean Difference: 0.120 to 0.144
##
##
## --- Do not assume equal population variances of fnbmd for each sex
##
## t-cutoff: tcut = 1.961
## Standard Error of Mean Difference: SE = 0.006
##
## Hypothesis Test of 0 Mean Diff: t = 20.407, df = 1560.981, p-value = 0.000
##
## Margin of Error for 95% Confidence Level: 0.013
## 95% Confidence Interval for Mean Difference: 0.119 to 0.145
##
##
## ------ Effect Size ------
##
## --- Assume equal population variances of fnbmd for each sex
##
## Standardized Mean Difference of fnbmd, Cohen's d: 0.939
##
##
## ------ Practical Importance ------
##
## Minimum Mean Difference of practical importance: mmd
## Minimum Standardized Mean Difference of practical importance: msmd
## Neither value specified, so no analysis
##
##
## ------ Graphics Smoothing Parameter ------
##
## Density bandwidth for sex Male: 0.044
## Density bandwidth for sex Female: 0.034
2.3 Sử dụng chatGPT
2.3.1 Biểu đồ histogram
PROMPT: Dùng gói lệnh ‘lessR’ vẽ biểu đồ histogram mô tả phân bố của mật độ xương (fnbmd), tô màu xanh và đặt tựa trục x là ‘Mật độ xương ở cổ xương đùi (g/cm2)’, trục y là ‘Số người’, tựa là ‘Phân bố mật độ xương’
2.3.2 So sánh mật độ xương
PROMPT: Dùng gói lệnh ‘lessR’ so sánh mật độ xương giữa nam và nữ (sex)
Việc 3. So sánh khác biệt về RER giữa 2 nhóm
3.1 Nhập liệu
placebo <- c(105, 119, 100, 97, 96, 101, 94, 95, 98)
coffee <- c(96, 99, 94, 89, 96, 93, 88, 105, 88)3.2 Đánh giá khác biệt về RER giữa 2 nhóm bằng t-test
ttest(placebo, coffee)##
## Compare Y across X with levels Group1 and Group2
## Grouping Variable: X
## Response Variable: Y
##
##
## ------ Describe ------
##
## Y for X Group1: n.miss = 0, n = 9, mean = 100.556, sd = 7.699
## Y for X Group2: n.miss = 0, n = 9, mean = 94.222, sd = 5.608
##
## Mean Difference of Y: 6.333
##
## Weighted Average Standard Deviation: 6.735
##
##
## ------ Assumptions ------
##
## Note: These hypothesis tests can perform poorly, and the
## t-test is typically robust to violations of assumptions.
## Use as heuristic guides instead of interpreting literally.
##
## Null hypothesis, for each group, is a normal distribution of Y.
## Group Group1 Shapiro-Wilk normality test: W = 0.780, p-value = 0.012
## Group Group2 Shapiro-Wilk normality test: W = 0.924, p-value = 0.427
##
## Null hypothesis is equal variances of Y, homogeneous.
## Variance Ratio test: F = 59.278/31.444 = 1.885, df = 8;8, p-value = 0.389
## Levene's test, Brown-Forsythe: t = 0.230, df = 16, p-value = 0.821
##
##
## ------ Infer ------
##
## --- Assume equal population variances of Y for each X
##
## t-cutoff for 95% range of variation: tcut = 2.120
## Standard Error of Mean Difference: SE = 3.175
##
## Hypothesis Test of 0 Mean Diff: t-value = 1.995, df = 16, p-value = 0.063
##
## Margin of Error for 95% Confidence Level: 6.731
## 95% Confidence Interval for Mean Difference: -0.397 to 13.064
##
##
## --- Do not assume equal population variances of Y for each X
##
## t-cutoff: tcut = 2.136
## Standard Error of Mean Difference: SE = 3.175
##
## Hypothesis Test of 0 Mean Diff: t = 1.995, df = 14.624, p-value = 0.065
##
## Margin of Error for 95% Confidence Level: 6.782
## 95% Confidence Interval for Mean Difference: -0.449 to 13.116
##
##
## ------ Effect Size ------
##
## --- Assume equal population variances of Y for each X
##
## Standardized Mean Difference of Y, Cohen's d: 0.940
##
##
## ------ Practical Importance ------
##
## Minimum Mean Difference of practical importance: mmd
## Minimum Standardized Mean Difference of practical importance: msmd
## Neither value specified, so no analysis
##
##
## ------ Graphics Smoothing Parameter ------
##
## Density bandwidth for X Group1: 5.642
## Density bandwidth for X Group2: 4.112
3.3 Đánh giá khác biệt về RER giữa 2 nhóm bằng bootstrap
library(simpleboot)## Simple Bootstrap Routines (1.1-7)library(boot)
set.seed(123) # Để kết quả tái lập được
boot_result <- two.boot(placebo, coffee, FUN = mean, R = 10000)
boot.ci(boot_result, type = c("perc", "bca", "norm", "basic"))## BOOTSTRAP CONFIDENCE INTERVAL CALCULATIONS
## Based on 10000 bootstrap replicates
##
## CALL :
## boot.ci(boot.out = boot_result, type = c("perc", "bca", "norm",
## "basic"))
##
## Intervals :
## Level Normal Basic
## 95% ( 0.383, 12.188 ) ( 0.111, 11.778 )
##
## Level Percentile BCa
## 95% ( 0.889, 12.556 ) ( 0.778, 12.444 )
## Calculations and Intervals on Original Scale3.4 Sử dụng chatGPT
PROMPT 1: Tôi có dữ liệu của 18 người trong 2 nhóm như sau: placebo = 105, 119, 100, 97, 96, 101, 94, 95, 98 và coffee = 96, 99, 94, 89, 96, 93, 88, 105, 88. Bạn giúp viết lệnh R từ gói lệnh ‘simpleboot’ và ‘boot’ để phân tích sự khác biệt giữa 2 nhóm bằng phương pháp bootstrap và trình bày nhiều giá trị 95% CI (percentile, BCA, normal, basic) không?
Việc 4 So sánh tỷ lệ gãy xương giữa 2 nhóm
4.1 So sánh tỷ lệ gãy xương giữa nam và nữ
table1(~ as.factor(fx) | sex, data = df)| Female (N=1317) |
Male (N=845) |
Overall (N=2162) |
|
|---|---|---|---|
| as.factor(fx) | |||
| 0 | 916 (69.6%) | 701 (83.0%) | 1617 (74.8%) |
| 1 | 401 (30.4%) | 144 (17.0%) | 545 (25.2%) |
chisq.test(df$fx, df$sex)##
## Pearson's Chi-squared test with Yates' continuity correction
##
## data: df$fx and df$sex
## X-squared = 48.363, df = 1, p-value = 0.0000000000035424.2 Sử dụng chatGPT
PROMPT: So sánh tỉ lệ gãy xương (fx) giữa nam và nữ (sex)