descriptive statistics
Marisol Wen
2026-06-21
Load data
We are loading the data for X-Ray orders from a major US hospital.
dat <- read.csv("https://raw.githubusercontent.com/tmatis12/datafiles/refs/heads/main/RadDat_IMSE.csv")
str(dat)## 'data.frame': 43632 obs. of 16 variables:
## $ Unique.Identifier : int 1 2 3 4 5 6 7 8 9 10 ...
## $ PatientAge : int 75 87 35 51 67 54 34 65 67 40 ...
## $ Radiology.Technician : int 65 65 16 24 37 7 40 2 2 34 ...
## $ CatalogCode : chr "DX Abdomen 2 vw w/single chest" "DX Abdomen 2 vw w/single chest" "DX Abdomen 2 vw w/single chest" "DX Abdomen 2 vw w/single chest" ...
## $ Ordering.Physician : int 4 4 150 130 173 349 4 4 39 4 ...
## $ PatientTypeMnemonic : chr "IP" "IP" "IP" "IP" ...
## $ Priority : chr "Routine" "Routine" "Routine" "Routine" ...
## $ OrderDateTime : chr "12/27/16 10:32" "1/13/17 11:44" "1/2/17 17:19" "11/13/16 10:13" ...
## $ ExamCompleteDateTime : chr "12/27/16 11:19" "1/13/17 12:32" "1/2/17 18:00" "11/14/16 9:34" ...
## $ FinalDateTime : chr "12/28/16 14:32" "1/14/17 16:00" "1/3/17 7:44" "11/14/16 16:40" ...
## $ Ordered.to.Complete...Mins : int 47 48 41 1401 42 129 42 1068 49 47 ...
## $ Ordered.to.Complete...Hours: num 0.783 0.8 0.683 23.35 0.7 ...
## $ Loc.At.Exam.Complete : chr "GTU" "GTU" "3W" "4W" ...
## $ Exam.Completed.Bucket : chr "8a-8p" "8a-8p" "8a-8p" "8a-8p" ...
## $ Section : chr "DX" "DX" "EC DX" "DX" ...
## $ Exam.Room : chr "DX Rm 1" "DX Rm 1" "DX Rm 5 (EC)" "DX Rm 1" ...head(dat)## Unique.Identifier PatientAge Radiology.Technician
## 1 1 75 65
## 2 2 87 65
## 3 3 35 16
## 4 4 51 24
## 5 5 67 37
## 6 6 54 7
## CatalogCode Ordering.Physician PatientTypeMnemonic
## 1 DX Abdomen 2 vw w/single chest 4 IP
## 2 DX Abdomen 2 vw w/single chest 4 IP
## 3 DX Abdomen 2 vw w/single chest 150 IP
## 4 DX Abdomen 2 vw w/single chest 130 IP
## 5 DX Abdomen 2 vw w/single chest 173 IP
## 6 DX Abdomen 2 vw w/single chest 349 IP
## Priority OrderDateTime ExamCompleteDateTime FinalDateTime
## 1 Routine 12/27/16 10:32 12/27/16 11:19 12/28/16 14:32
## 2 Routine 1/13/17 11:44 1/13/17 12:32 1/14/17 16:00
## 3 Routine 1/2/17 17:19 1/2/17 18:00 1/3/17 7:44
## 4 Routine 11/13/16 10:13 11/14/16 9:34 11/14/16 16:40
## 5 STAT 12/13/16 3:22 12/13/16 4:04 12/13/16 3:19
## 6 Routine 1/17/17 5:38 1/17/17 7:47 1/17/17 10:55
## Ordered.to.Complete...Mins Ordered.to.Complete...Hours Loc.At.Exam.Complete
## 1 47 0.7833333 GTU
## 2 48 0.8000000 GTU
## 3 41 0.6833333 3W
## 4 1401 23.3500000 4W
## 5 42 0.7000000 Emergency Ctr
## 6 129 2.1500000 3E
## Exam.Completed.Bucket Section Exam.Room
## 1 8a-8p DX DX Rm 1
## 2 8a-8p DX DX Rm 1
## 3 8a-8p EC DX DX Rm 5 (EC)
## 4 8a-8p DX DX Rm 1
## 5 12a-8a EC DX DX Rm 5 (EC)
## 6 12a-8a DX DX Portabledat<- dat[dat$Ordered.to.Complete...Mins >= 0,]Part 1
Patients that are age 65 or older qualify for Medicare. We are creating a histogram of the time required to complete X-Ray orders for Medicare patients. The allowable range of times is restricted between the first and third quartile.
medicare = dat[dat$PatientAge >= 65,]
summary(medicare$Ordered.to.Complete...Mins)## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 0 21 38 1786 90 112168q1 = quantile(medicare$Ordered.to.Complete...Mins, 0.25)
q3 = quantile(medicare$Ordered.to.Complete...Mins, 0.75)
medicare.iqr = medicare[medicare$Ordered.to.Complete...Mins >= q1 & medicare$Ordered.to.Complete...Mins <= q3,]
hist(medicare.iqr$Ordered.to.Complete...Mins, breaks = 25, main = "X-ray Completion Times for Medicare Patients", xlab = "Completion Time in Minutes", ylab = "Number of Observations", col = "red",border = "black")
This applies the filter to the data to get patients aged 65 or older. It then restricts the completion times to the IQR range, which is from the first quartile to the third quartile. The histogram is right-skewed because most X-ray orders were completed in the lower time range, while fewer orders took longer.
Part 2
We are comparing Radiology Technician 62 and Radiology Technician 65 based on the median time to complete an X-ray order.
t62 = dat[dat$Radiology.Technician == 62,]
t65 = dat[dat$Radiology.Technician == 65,]
summary(t62$Ordered.to.Complete...Mins)## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 13.00 48.00 80.00 76.88 117.00 118.00summary(t65$Ordered.to.Complete...Mins)## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 1.0 16.0 27.0 94.9 45.0 35306.0median62 = median(t62$Ordered.to.Complete...Mins)
median65 = median(t65$Ordered.to.Complete...Mins)
cat("Radiology Technician 62's median completion time:", median62,"\nRadiology Technician 65's median completion time:", median65)## Radiology Technician 62's median completion time: 80
## Radiology Technician 65's median completion time: 27It means Technician 62 had a median completion time of 80 minutes, while that of Technician 65 was 27 minutes. It, therefore, means that Technician 65 is faster based on the median completion time.
Part 3
Create a side by side boxplot comparing the ages of patients who receive STAT and Routine orders for an X-ray.
STAT <- dat[dat$Priority == "STAT",]
Routine <- dat[dat$Priority == "Routine",]
boxplot(Routine$PatientAge, STAT$PatientAge, names = c("Routine", "STAT"), main = "Patient Age by X-Ray Order Priority", xlab = "Order Priority",ylab = "Age in Years",col = c("red", "black"))
The boxplot should show that Routine orders have a higher median patient age than STAT orders. The spread is wider for STAT orders, which means patient ages vary more for STAT orders. There are also some younger patients shown as low outliers.
Part 4
We are calculating the mean and standard deviation of the time required to complete an X-Ray order on floor 3W compared to floor 4W.
f3W = dat[dat$Loc.At.Exam.Complete == "3W",]
f4W = dat[dat$Loc.At.Exam.Complete == "4W",]
summary(f3W$Ordered.to.Complete...Mins)## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 2.0 38.0 70.0 1463.1 402.5 29637.0summary(f4W$Ordered.to.Complete...Mins)## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 1.00 31.75 58.00 1675.45 610.75 37572.00mean3W = mean(f3W$Ordered.to.Complete...Mins)
mean3W## [1] 1463.051sd3W = sd(f3W$Ordered.to.Complete...Mins)
sd3W## [1] 3894.639mean4W = mean(f4W$Ordered.to.Complete...Mins)
mean4W## [1] 1675.451sd4W = sd(f4W$Ordered.to.Complete...Mins)
sd4W## [1] 4387.644The mean completion time of floor 3W is about 1463.05 minutes while that of floor 4W is about 1675.45 minutes. Hence, X-Ray order is normally completed later on floor 4W than on floor 3W.
The standard deviation indicates the spread of completion times. A floor with a larger standard deviation has more variation in completion times.