Session 4: How can I perform bivariate analysis?

A Slower Introduction to R

Yea-Hung Chen, PhD, MS

UCSF Library

Thursday, October 30, 2025

Continuous-versus-continuous analysis

Systolic blood pressure versus diastolic blood pressure

in this section, we’ll examine the association between systolic blood pressure and diastolic blood pressure

class(nhanes$bpxosy1)
#|  [1] "integer"
head(nhanes$bpxosy1)
#|  [1] 135 121 111  NA  NA 110
class(nhanes$bpxodi1)
#|  [1] "integer"
head(nhanes$bpxodi1)
#|  [1] 98 84 79 NA NA 72

Scatter plots

plot(x=nhanes$bpxodi1,y=nhanes$bpxosy1,xlab='Diastolic (mmHg)',ylab='Systolic (mmHg)')

Formula notation

we can equivalently use a y~x formula notation, like this:

plot(nhanes$bpxosy1~nhanes$bpxodi1,xlab='Diastolic (mmHg)',ylab='Systolic (mmHg)')

formula notation is used in several R functions, including:
- lm() for linear regression
- t.test() for t-tests

Formula notation

plot(nhanes$bpxosy1~nhanes$bpxodi1,xlab='Diastolic (mmHg)',ylab='Systolic (mmHg)')

Pearson’s correlation coefficient

cor(nhanes$bpxodi1,nhanes$bpxosy1)
#|  [1] NA
cor(nhanes$bpxodi1,nhanes$bpxosy1,use='complete.obs')
#|  [1] 0.6057034

the use=complete.obs argument is similar to na.rm=TRUE in mean(), sd(), and other functions you saw last week
- if either value in a pair is NA, then the pair is not used
the order of the variables does not matter

Other measures of correlation

cor(nhanes$bpxodi1,nhanes$bpxosy1,use='complete.obs',method='kendall')
#|  [1] 0.4277319
cor(nhanes$bpxodi1,nhanes$bpxosy1,use='complete.obs',method='spearman')
#|  [1] 0.5832935

Simple linear regression

lm(nhanes$bpxosy1~nhanes$bpxodi1)
#|  
#|  Call:
#|  lm(formula = nhanes$bpxosy1 ~ nhanes$bpxodi1)
#|  
#|  Coefficients:
#|     (Intercept)  nhanes$bpxodi1  
#|          49.165           0.979

Simple linear regression: `confint()`

my_model<-lm(nhanes$bpxosy1~nhanes$bpxodi1)
confint(my_model)
#|                      2.5 %    97.5 %
#|  (Intercept)    46.7210856 51.608631
#|  nhanes$bpxodi1  0.9467311  1.011178

Simple linear regression: `summary()`

summary(my_model)
#|  
#|  Call:
#|  lm(formula = nhanes$bpxosy1 ~ nhanes$bpxodi1)
#|  
#|  Residuals:
#|      Min      1Q  Median      3Q     Max 
#|  -32.629 -10.323  -2.271   7.361  89.203 
#|  
#|  Coefficients:
#|                 Estimate Std. Error t value Pr(>|t|)    
#|  (Intercept)    49.16486    1.24660   39.44   <2e-16 ***
#|  nhanes$bpxodi1  0.97895    0.01644   59.56   <2e-16 ***
#|  ---
#|  Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
#|  
#|  Residual standard error: 14.68 on 6121 degrees of freedom
#|    (2030 observations deleted due to missingness)
#|  Multiple R-squared:  0.3669,    Adjusted R-squared:  0.3668 
#|  F-statistic:  3547 on 1 and 6121 DF,  p-value: < 2.2e-16

Simple linear regression: `summary()`

as with the t.test() function, it is possible to extract specific values, using dollar-sign notation

Simple linear regression: `summary()`

names(summary(my_model))
#|   [1] "call"          "terms"         "residuals"     "coefficients" 
#|   [5] "aliased"       "sigma"         "df"            "r.squared"    
#|   [9] "adj.r.squared" "fstatistic"    "cov.unscaled"  "na.action"
summary(my_model)$coefficients
#|                   Estimate Std. Error  t value      Pr(>|t|)
#|  (Intercept)    49.1648582 1.24659908 39.43919 2.462317e-303
#|  nhanes$bpxodi1  0.9789544 0.01643749 59.55620  0.000000e+00
summary(my_model)$r.squared
#|  [1] 0.3668766

Simple linear regression: the `data` argument

lm(bpxosy1~bpxodi1,data=nhanes)
#|  
#|  Call:
#|  lm(formula = bpxosy1 ~ bpxodi1, data = nhanes)
#|  
#|  Coefficients:
#|  (Intercept)      bpxodi1  
#|       49.165        0.979

several R functions support the data argument
it is one way to omit dollar-sign notation

Exercise 1

Create a script for today’s exercises. Save the script. Include code at the top for importing the NHANES data.
Create a scatter plot for the relationship between the first two measurements of pulse (bpxopls1 and bpxopls2), placing the second measurement on the y axis. Label both axes.

Exercise 1

Create a scatter plot for the relationship between diastolic blood pressure (bpxodi1) and pulse (bpxopls1), placing diastolic blood pressure on the y axis. Label both axes.
Fit a simple linear model for the relationship between diastolic blood pressure (bpxodi1) and pulse (bpxopls1), with diastolic blood pressure as the y variable. Use summary() to examine the model.

Continuous-versus-dichotomous analysis

Systolic blood pressure versus smoking

in this section, we’ll examine the association between systolic blood pressure and smoking

class(nhanes$bpxosy1)
#|  [1] "integer"
head(nhanes$bpxosy1)
#|  [1] 135 121 111  NA  NA 110
class(nhanes$smoking)
#|  [1] "character"
head(nhanes$smoking)
#|  [1] "History of smoking" "History of smoking" "No"                
#|  [4] "No"                 "No"                 "History of smoking"

Box plots

boxplot(nhanes$bpxosy1~nhanes$smoking,xlab='',ylab='Systolic (mmHg)')

t-tests

t.test(nhanes$bpxosy1~nhanes$smoking)
#|  
#|      Welch Two Sample t-test
#|  
#|  data:  nhanes$bpxosy1 by nhanes$smoking
#|  t = 7.4317, df = 4887.9, p-value = 1.258e-13
#|  alternative hypothesis: true difference in means between group History of smoking and group No is not equal to 0
#|  95 percent confidence interval:
#|   2.672803 4.588225
#|  sample estimates:
#|  mean in group History of smoking                 mean in group No 
#|                          124.7512                         121.1207

t-tests

my_test<-t.test(nhanes$bpxosy1~nhanes$smoking)
names(my_test)
#|   [1] "statistic"   "parameter"   "p.value"     "conf.int"    "estimate"   
#|   [6] "null.value"  "stderr"      "alternative" "method"      "data.name"
my_test$estimate
#|  mean in group History of smoking                 mean in group No 
#|                          124.7512                         121.1207
my_test$conf.int
#|  [1] 2.672803 4.588225
#|  attr(,"conf.level")
#|  [1] 0.95

Exercise 2

Create a box plot for the relationship between pulse (bpxopls1) and asthma (asthma), adding a label for the y axis, and removing the label for the x axis.
Obtain a 95% confidence interval for the difference in mean pulse between the two asthma groups.

Exercise 2

Is the confidence interval for a yes-minus-no difference or for a no-minus-yes difference?
Obtain a 90% confidence interval for the difference in mean pulse between the two asthma groups, by adding the conf.level=0.90 argument to t.test().

Exercise 2

The wilcox.test() function implements the Wilcoxon rank-sum test (also known as the Mann-Whitney U test), which tests whether the “location” of a continuous variable’s distribution differs between two groups. Use the function, and formula notation, to compare the distribution of pulse between the two asthma groups.

Categorical-versus-categorical analysis

Smoking versus asthma

in this section, we’ll examine the association between smoking and asthma

class(nhanes$smoking)
#|  [1] "character"
head(nhanes$smoking)
#|  [1] "History of smoking" "History of smoking" "No"                
#|  [4] "No"                 "No"                 "History of smoking"
class(nhanes$asthma)
#|  [1] "character"
head(nhanes$asthma)
#|  [1] "No"                "No"                "No"               
#|  [4] "No"                "History of asthma" "No"

Contingency tables

table(nhanes$smoking,nhanes$asthma)
#|                      
#|                       History of asthma   No
#|    History of smoking               636 2603
#|    No                               844 4029

the first variable will be placed along the rows

Contingency tables

table(nhanes$smoking,nhanes$asthma,useNA='always')
#|                      
#|                       History of asthma   No <NA>
#|    History of smoking               636 2603    4
#|    No                               844 4029    5
#|    <NA>                               6   25    1

we can add the useNA argument to show missing values

Row-wise percents

prop.table(
  table(nhanes$smoking,nhanes$asthma),
  margin=1
)*100

notice the two arguments inside prop.table():
- the table()
- margin=1, which indicates we want row-wise percents

Row-wise percents

equivalently, written all in one line:

prop.table(table(nhanes$smoking,nhanes$asthma),margin=1)*100
#|                      
#|                       History of asthma       No
#|    History of smoking          19.63569 80.36431
#|    No                          17.31993 82.68007

Column-wise percents

prop.table(table(nhanes$smoking,nhanes$asthma),margin=2)*100
#|                      
#|                       History of asthma       No
#|    History of smoking          42.97297 39.24910
#|    No                          57.02703 60.75090

the margin=2 argument indicates that we want column-wise percents

Fisher’s exact tests

fisher.test(table(nhanes$smoking,nhanes$asthma))
#|  
#|      Fisher's Exact Test for Count Data
#|  
#|  data:  table(nhanes$smoking, nhanes$asthma)
#|  p-value = 0.008981
#|  alternative hypothesis: true odds ratio is not equal to 1
#|  95 percent confidence interval:
#|   1.038673 1.309340
#|  sample estimates:
#|  odds ratio 
#|    1.166346

notice that this also returns the odds ratio

Chi-squared tests

chisq.test(table(nhanes$smoking,nhanes$asthma))
#|  
#|      Pearson's Chi-squared test with Yates' continuity correction
#|  
#|  data:  table(nhanes$smoking, nhanes$asthma)
#|  X-squared = 6.8411, df = 1, p-value = 0.008908

X-squared is the \(\chi^2\) statistic
df is the degrees of freedom

Exercise 3

Produce a contingency table of asthma (asthma) and gender (gender), with gender along the rows.
Find the row-wise percents.
Conduct a Fisher’s exact test for the association between asthma and gender.

Exercise 3

Store the test, using object assignment. Use names() to examine the values that can be extracted. Extract the odds ratio (estimate).
Extract the 95% confidence interval for the odds ratio (conf.int).

Exercise 3

Next week, we’ll discuss the ggplot2 package, which is a powerful and popular data-visualization package. Install the package, as follows:
- Select Install in the Packages tab.
- Type ggplot2.
- Select the Install button at the bottom.
Equivalently, type the following in the console and press the enter or return key on your keyboard:
```
install.packages('ggplot2')
```

Exercise 3

Verify that the package was successfully installed by checking the Packages tab.
- In the search field at the top right, begin typing ggplot2.
- Verify that the package shows up.

Session 4: How can I perform bivariate analysis?

Continuous-versus-continuous analysis

Systolic blood pressure versus diastolic blood pressure

Scatter plots

Formula notation

Formula notation

Pearson’s correlation coefficient

Other measures of correlation

Simple linear regression

Simple linear regression: confint()

Simple linear regression: summary()

Simple linear regression: summary()

Simple linear regression: summary()

Simple linear regression: the data argument

Exercise 1

Exercise 1

Exercise 1

Continuous-versus-dichotomous analysis

Systolic blood pressure versus smoking

Box plots

t-tests

t-tests

Exercise 2

Exercise 2

Exercise 2

Exercise 2

Categorical-versus-categorical analysis

Smoking versus asthma

Contingency tables

Contingency tables

Row-wise percents

Row-wise percents

Column-wise percents

Fisher’s exact tests

Chi-squared tests

Exercise 3

Exercise 3

Exercise 3

Exercise 3

Exercise 3

Simple linear regression: `confint()`

Simple linear regression: `summary()`

Simple linear regression: `summary()`

Simple linear regression: `summary()`

Simple linear regression: the `data` argument