1 Introduction
In this entry, through the analysis of a data set, we examine how to
make histograms, tables, transitive graphs, and scatter plots and the
way to calculate medians and interquartile ranges (IQR).
In health preference research (HPR), it is naturally difficult to
determine patients’ real necessities. This is because it is often the
case that even patients themselves are not necessarily able to tell
specifically what medical procedure they hope to receive, facing complex
choices and variable expected consequences. A discrete choice experiment
(DCE), which presents multiple health descriptions and infers the
patients’ weighting factors from the responses, is one of the effective
ways to address this problem.
In this entry, before entering the details of DCEs, we are going to
analyze the data set accessorily generated in the process of DCE; our
focus here is the time spent on choosing a choice for each question,
which is called the first response time. Respondents may change their
answer to the other before proceeding to the next question, but the
first response time only measures the time from starting the question to
selecting the first choice. In the DCE held in 2016, 4088 US
participants responded to 20 questions regarding hypothetical health
conditions. The data set that we are going to analyze contains each
individual’s first response time for 20 questions.
2 Load the data set and
packages
library(knitr) #this is for the function "kable," which makes well-organized tables.
library(tidyverse) #this is for the function "read_csv."
library(tinytex)
library(gt)
data1 <- read_csv("C:\\Users\\aaa\\OneDrive\\USF\\Dr. Craig\\resp1wave1_220723.csv")Please replace the inside of quotation marks with the address of the
file location.
3 Make histograms
3.1 Mathematical
explanation of histogram
Let \(x_i\) be the value of each
individual observation (\(1 \le i \le
n\)) \[n = \sum_{k=1}^{K}m_k\]
where
\[\begin{aligned}
n &= \text{the total number of observations } \\
m_k &= \text{the number of}\ x_i\ \text{within the}\ k\ \text{th
bin}\ (1\le k \le K)\\
&= \text{the number of}\ x_i\ s.t.\ r_{k-1}\le x_i < r_k\\
&= \text{frequency}\\
r_j &= \text{the upper boundary of the}\ j\ \text{th bin}\
(min(k)-1\le j \le max(k),\ \text{that is,}\ 0\le j \le K)\\
&= \text{the lower boundary of the}\ j+1\ \text{th bin}\\
r_j-r_{j-1} &= \text{the width of each bin}\\
&=\frac{max(r_j)-min(r_j)}{K}=\frac{r_{K}-r_0}{K}
\end{aligned}\]
It is necessary to decide the range, \([min(r_j), max(r_j)) = [r_0, r_K)\), such
that \(r_0=min(r_j) \le min(x_i)\ \wedge\
max(x_i) \le r_{K+1}=max(r_j)\), to create an understandable
histogram.
3.2 Example I
Now, let \(K=50\) (breaks = 50) and
\(x_i = data1\$time_i\), which is the
first response time (\(n=4088\)). We
get the following histogram.
hist(data1$time,
breaks = 50,
main = "Histogram 3.2: Histogram of the first response time",
xlab = "the first response time (milliseconds)",
col = "lightgreen")
3.3 Example II
Now, let \(K=50\) (breaks = 50) and
\(x_i = log(data1\$time_i)\) (\(n=4088\)) instead of \(x_i = data1\$time_i\) . we get the
following histogram.
hist(log(data1$time),
breaks = 50,
main = "Histogram 3.3: Log of the first response time",
xlab = "Log of the first response time",
col = "lightblue") 
Comparing the two histograms above, we observe that by taking a
logarithmic scale of the variable, the first response time
(milliseconds), the originally right-skewed histogram, Histogram 3.1,
was converted into the more bellshaped histogram, Histogram 3.2.
4 Make a table that
contains Median and IQR for each question
- Definition : the k-th percentile indicates how many
percentages of observations have a smaller value than that number. For
example, if the 25th percentile is 1256, then 25% of observations are
smaller than 1256. - Definition: The median is the 50 percentile (Q2).
- Definition: The interquartile range (IQR) is the difference
between the 25th percentile (Q1) and the 75th percentile (Q3). That is,
\(IQR = Q3-Q1\).
4.1 Mathematical
explanation of Median and IQR
Let \(n = \text{the total number of
observations }\).
Assume we have data set, \(x[1]\le x[2]\le
...\le x[n]\).
4.1.1 Median
When \(n\) is even, that is, \(\exists\ a \in Z \ s.t.\ n = 2a\), \(median =
\frac{1}{2}(x[\frac{n}{2}]+x[\frac{n}{2}+1])\).
When \(n\) is odd, that is, \(\exists\ b \in Z \ s.t.\ n = 2b+1\), \(median = x[\frac{n+1}{2}]\).
4.1.2 IQR
\(\forall n\ in\ N, \exists\ k \in N\ s.t\
\)
\[ n =
\begin{cases}
{4k}\\
{4k-1}\\
{4k+1}\\
{4k+2}
\end{cases}
\]
\[\begin{aligned}
\text{(i)}\ n=4k \ \ \ \\
Q1&=\frac{3}{4}x[k]+\frac{1}{4}x[k+1]\\
Q3&=\frac{1}{4}x[3k]+\frac{3}{4}x[3k+1]\\
IQR&=Q3-Q1\\
&=(\frac{1}{4}x[3k]+\frac{3}{4}x[3k+1])-(\frac{3}{4}x[k]+\frac{1}{4}x[k+1])
\\
\text{(ii)}\ n=4k-1\\
Q1&=\frac{1}{2}x[k]+\frac{1}{2}x[k+1] \\
Q3&=\frac{1}{2}x[3k+1]+\frac{1}{2}x[3k+2] \\
IQR&=Q3-Q1\\
&=(\frac{1}{2}x[3k+1]+\frac{1}{2}x[3k+2])-(\frac{1}{2}x[k]+\frac{1}{2}x[k+1])
\\
\text{(iii)}\ n=4k+1 \\
Q1&=x[k] \\
Q3&=x[3k] \\
IQR&=Q3-Q1\\
&=x[3k]-x[k] \\
\text{(iv)}\ n=4k+2 \\
Q1&=\frac{1}{4}x[k]+\frac{3}{4}x[k+1] \\
Q3&=\frac{3}{4}x[3k+2]+\frac{1}{4}x[3k+2] \\
IQR&=Q3-Q1\\
&=(\frac{3}{4}x[3k+2]+\frac{1}{4}x[3k+2])-(\frac{1}{4}x[k]+\frac{3}{4}x[k+1])
\\
\end{aligned}\]
4.2 Make a table for
Median and IQR
Let’s suppose we hope to calculate the median and interquartile range
(IQR) for Question 1. Then, we firstly need to gather the individual
values of the first response time for Question 1 from the data set and
secondly, calculate the median and IQR for that question. When we want
to gain the median and IQR for each question, we are going to simply
repeat that process 20 times since there are 20 questions in the
experiment. Here we use function “quantile” to calculate IQR.
m2 =NULL
ir2 = NULL
iqr2 = NULL
for(i in 1:20) {d1 <- data1[data1$task == i, "time"]
m0 = median(d1$time)
m2 = rbind(m2, m0)
ir0 = quantile(d1$time)
ir2 = rbind(ir2, ir0)
iqr0 = IQR(d1$time)
iqr2 = rbind(iqr2, iqr0)
table1 = cbind(m2,iqr2, ir2)
}
rownames(table1) = paste0("Question ", 1:20)
colnames(table1) <- c("Median", "IQR", "0% percentile", "25% percentile",
"50% percentile", "75% percentile ","100% percentile")
table4.2 <- data.frame(cbind(c(1:20),table1))
colnames(table4.2) <- c("n th Question", "Median", "IQR", "0% percentile", "25% percentile", "50% percentile", "75% percentile ","100% percentile")
kable(table4.2, align = "c", caption = "**Table 4.2: Median and IQR**", row.names = FALSE, escape = FALSE, centering = T) | n th Question | Median | IQR | 0% percentile | 25% percentile | 50% percentile | 75% percentile | 100% percentile |
|---|---|---|---|---|---|---|---|
| 1 | 29645.0 | 23640.25 | 2023 | 20152.00 | 29645.0 | 43792.25 | 3e+05 |
| 2 | 20553.0 | 18588.75 | 2050 | 12991.25 | 20553.0 | 31580.00 | 3e+05 |
| 3 | 19209.5 | 17642.50 | 1475 | 11748.75 | 19209.5 | 29391.25 | 3e+05 |
| 4 | 17394.5 | 16838.00 | 1590 | 10983.00 | 17394.5 | 27821.00 | 3e+05 |
| 5 | 16897.5 | 16081.25 | 1998 | 10359.75 | 16897.5 | 26441.00 | 3e+05 |
| 6 | 16023.5 | 15845.00 | 1732 | 9773.25 | 16023.5 | 25618.25 | 3e+05 |
| 7 | 15416.5 | 14887.75 | 1787 | 9423.50 | 15416.5 | 24311.25 | 3e+05 |
| 8 | 15081.0 | 15779.50 | 1700 | 9047.00 | 15081.0 | 24826.50 | 3e+05 |
| 9 | 14317.0 | 14732.75 | 1786 | 8566.75 | 14317.0 | 23299.50 | 3e+05 |
| 10 | 14210.5 | 14608.50 | 1669 | 8305.75 | 14210.5 | 22914.25 | 3e+05 |
| 11 | 13901.0 | 13757.50 | 1723 | 8545.75 | 13901.0 | 22303.25 | 3e+05 |
| 12 | 9884.0 | 10562.50 | 1938 | 5845.75 | 9884.0 | 16408.25 | 3e+05 |
| 13 | 8982.0 | 9519.50 | 1818 | 5109.25 | 8982.0 | 14628.75 | 3e+05 |
| 14 | 8580.0 | 9704.75 | 1944 | 4813.25 | 8580.0 | 14518.00 | 3e+05 |
| 15 | 8158.5 | 9037.50 | 1829 | 4461.75 | 8158.5 | 13499.25 | 3e+05 |
| 16 | 7905.5 | 8648.75 | 1801 | 4377.00 | 7905.5 | 13025.75 | 3e+05 |
| 17 | 7386.0 | 8255.75 | 1847 | 4082.25 | 7386.0 | 12338.00 | 3e+05 |
| 18 | 7250.5 | 8094.75 | 1538 | 4111.00 | 7250.5 | 12205.75 | 3e+05 |
| 19 | 7133.5 | 7991.25 | 1290 | 4044.25 | 7133.5 | 12035.50 | 3e+05 |
| 20 | 6908.5 | 7638.25 | 1731 | 3905.50 | 6908.5 | 11543.75 | 3e+05 |
5 Plot the results of the
table
Now, we try to plot the results of the table above and connect plots
to reveal the transition of median and IQR as the number of questions
increases. The black points represent the median and red circles denote
the 25% percentile and the 75% percentile for each question.
We prepare a plain where x-axis is the number of question (Question
in Table 4.2) and y-axis is the first response time
(milliseconds).
To plot medians, \((x,y)=(n\ th\ Question,
Median)\)
To plot 25% percentile, let \((x,y)=(n\ th\
Question,25\%\ percentile)\) To plot 75% percentile, let \((x,y)=(n\ th\ Question,75\%\
percentile)\)
xmax <- 20
xmin <- 1
ymax <- 60000
ymin <- 0
plot(table4.2$'n th Question', table4.2$Median, bty = "l", pch = 16, type ="o",
xlim = c(xmin, xmax), ylim = c(ymin, ymax),
xlab = NA, ylab =NA, )
par(new=T)
plot(table4.2$'n th Question', table4.2$`25% percentile`, bty = "l", pch = 1, col = "red", type ="o",
xlim = c(xmin, xmax), ylim = c(ymin, ymax), xlab = NA, ylab =NA,)
par(new=T)
plot(table4.2$'n th Question', table4.2$`75% percentile`, bty = "l", pch = 1, col = "red", type ="o",
xlim = c(xmin, xmax), ylim = c(ymin, ymax),
xlab = "n th Question", ylab ="The first response time (milliseconds)",
main = "Graph 5: The median and interquartile range of the first response time")
6 Make a scatter
plot
In Graph 5, we can observe the tendency where the median and IQR
decline as the experiment proceeds and the number of questions
increases. Now, then, we focus on each individual. Since each
participant answered 20 questions, let us classify the questions into
two groups: the first ten questions and the second ten questions.
Focusing on a specific individual, We are able to calculate his or her
medians of the first response time for the first group of questions as
well as the second group of questions. Let us call them median1 and
median2. Preparing the plane with median1 in the x-axis and median2 in
the y-axis, we can plot that individual in the plane. We repeat that
process for all participants, and at the end, we can create the scatter
plot with all individuals (Graph 6).
Again, we set \((x,y)=(Median1,Median2)\), where
Median1: Median of the first response time for the first ten
questions
Median2: Median of the first response time for the second ten
questions
inv2 = NULL
for (i in 1:max(data1$survey_id)) {
ften <- subset(data1, subset = survey_id == i & task <10.5)
sten <- subset(data1, subset = survey_id == i & task >10.5)
inv0 = c(median(ften$time), median(sten$time))
inv2 = rbind(inv2, inv0)
}
table3 <- data.frame(cbind(c(1:max(data1$survey_id)),inv2))
colnames(table3) <- c("ID", "First_ten", "Second_ten" )
rownames(table3) = paste0("ID ", 1:max(data1$survey_id))
plot(table3$First_ten, table3$Second_ten, bty = "l", pch = 1, cex = 0.5,
xlab = "Median1",
ylab ="Median2",
main ="Graph 6: Scatter plots of the first median and second median"
)
par(new=T)
x1 <- seq(0:160000)
y1 <- x1
plot(x=x1, y=y1, type ="l", xlab = NA, ylab =NA, xaxt="n", yaxt="n",
bty ="n", col = "red")
The red line is the line where Median1 = Median2. We can observe that
majority of plots are under the red line, and thus, for the majority of
individuals, Median1 < Median2. This indicates that even when
focusing on individual respondents, we can assume the first response
time declines as the number of questions becomes larger.
7 Reference
Jakubczyk, M., Craig, B. M., Barra, M., Groothuis-Oudshoorn, C. G.
M., Hartman, J. D., Huynh, E., Ramos-Goñi, J. M., Stolk, E. A., &
Rand, K. (2017). Choice defines value: A predictive modeling competition
in Health Preference Research. Value in Health, 21(2), 229–238.
https://doi.org/10.1016/j.jval.2017.09.016
8 How to cite this
entry
Okubo, S. (yyyy, month dd). Analysis of initial response time across
20 pair comparisons. R4HPR. https://r4hpr.org/visor/?e=analysis-of-initial-response-time-across-20-pair-comparisons