Applied Analytics Assignment 2
Heart Disease Patients Analysis
Denis Bharatbhai Vaghasia - s3858391
Last updated: 28 May, 2021
Descriptive Statistics and Visualisation
- As said, all the variables are equally significant. We will labels the values 1 and 0 by converting to factor variables. Also, mutating variable which should be in order.
- Then listed the variable data type and its structure
heart_patient_data <- read.csv('heart_disease_patients.csv')
heart_patient_data <- heart_patient_data %>% mutate(
sex = factor(sex, levels=c(1,0), labels=c('Male','Female')),
fbs = factor(fbs, levels=c(1,0), labels=c('Yes','No')),
exang = factor(exang, levels=c(1,0), labels=c('Yes','No')),
cp = factor(cp, levels = c(1,2,3,4), labels=c(1,2,3,4), ordered=TRUE))
str(heart_patient_data)
## 'data.frame': 303 obs. of 12 variables:
## $ id : int 1 2 3 4 5 6 7 8 9 10 ...
## $ age : int 63 67 67 37 41 56 62 57 63 53 ...
## $ sex : Factor w/ 2 levels "Male","Female": 1 1 1 1 2 1 2 2 1 1 ...
## $ cp : Ord.factor w/ 4 levels "1"<"2"<"3"<"4": 1 4 4 3 2 2 4 4 4 4 ...
## $ trestbps: int 145 160 120 130 130 120 140 120 130 140 ...
## $ chol : int 233 286 229 250 204 236 268 354 254 203 ...
## $ fbs : Factor w/ 2 levels "Yes","No": 1 2 2 2 2 2 2 2 2 1 ...
## $ restecg : int 2 2 2 0 2 0 2 0 2 2 ...
## $ thalach : int 150 108 129 187 172 178 160 163 147 155 ...
## $ exang : Factor w/ 2 levels "Yes","No": 2 1 1 2 2 2 2 1 2 1 ...
## $ oldpeak : num 2.3 1.5 2.6 3.5 1.4 0.8 3.6 0.6 1.4 3.1 ...
## $ slope : int 3 2 2 3 1 1 3 1 2 3 ...
- Now, checking the null values in dataset. 0 indicates we don’t have any Null values
sum(is.na(heart_patient_data))
## [1] 0
Descriptive Statistics and Visualisation Cont.
- Below is descriptive and statistical summary of cholesterol with respect to sex. We will visualize this summary with the help of boxplot.
heart_patient_data %>% group_by(sex) %>% summarise(Min= min(chol, na.rm = TRUE),
Q1= quantile(chol, probs = 0.25, na.rm = TRUE),
Median = median(chol, na.rm = TRUE),
Q2 = quantile(chol, probs = 0.75, na.rm = TRUE),
Max = max(chol, na.rm=TRUE),
Mean = mean(chol, na.rm = TRUE),
SD = sd(chol, na.rm = TRUE),
n=n(),
Missing = sum(is.na(chol))) -> table_chol
knitr::kable(table_chol)
| Male |
126 |
208.75 |
235 |
268.5 |
353 |
239.6019 |
42.64976 |
206 |
0 |
| Female |
141 |
215.00 |
254 |
302.0 |
564 |
261.7526 |
64.90089 |
97 |
0 |
- We can observe that Female tends to have high cholesterol than Male. We can see there is outlier in female Cholesterol level. This could be an extreme case where a patient would be critical with his/her heart disease.
heart_patient_data %>% boxplot(chol ~ sex, data = ., main = 'Box Plot of Patient Cholestrol by Sex', ylab='Cholestrol Level', xlab='Sex', col='#1ABC9C')
Descriptive Statistics and Visualisation Cont.
- Now, We will analyze the relationship between Resting blood pressure and age. It’s obvious that person who is aged will have problem of blood pressure. Hence, we will look at summary and then visualize through scatter plot.
heart_patient_data %>% summarise(Min= min(trestbps, na.rm = TRUE),
Q1= quantile(trestbps, probs = 0.25, na.rm = TRUE),
Median = median(trestbps, na.rm = TRUE),
Q2 = quantile(trestbps, probs = 0.75, na.rm = TRUE),
Max = max(trestbps, na.rm=TRUE),
Mean = mean(trestbps, na.rm = TRUE),
SD = sd(trestbps, na.rm = TRUE),
n=n(),
Missing = sum(is.na(trestbps))) -> table_trestbps
knitr::kable(table_trestbps)
| 94 |
120 |
130 |
140 |
200 |
131.6898 |
17.59975 |
303 |
0 |
- Scatter plot of Blood pressure with respect to age. We can observe, the blood pressure problem start from age of 30 till 70 above. Also, the patient who is 50 to 65 years old have recorded high blood pressure.
heart_patient_data %>% plot(trestbps ~ age, data = ., xlab = 'Patient Age', ylab = 'Resting Blood Pressure (mm)')
# Descriptive Statistics and Visualisation Cont.
- Increase in heart rate can also be a symptoms of heart rate disease. Hence, we will look at statistics with the help of summary and then plot histogram.
heart_patient_data %>% summarise(Min= min(thalach, na.rm = TRUE),
Q1= quantile(thalach, probs = 0.25, na.rm = TRUE),
Median = median(thalach, na.rm = TRUE),
Q2 = quantile(thalach, probs = 0.75, na.rm = TRUE),
Max = max(thalach, na.rm=TRUE),
Mean = mean(thalach, na.rm = TRUE),
SD = sd(thalach, na.rm = TRUE),
n=n(),
Missing = sum(is.na(thalach))) -> table_thalach
knitr::kable(table_thalach)
| 71 |
133.5 |
153 |
166 |
202 |
149.6073 |
22.875 |
303 |
0 |
- With histogram, we will also plot the mean line to know what is an average heart beat rate of heart disease patient. We can see the the average heart rate of patient is 149.6. We can also observe that values are normally distributed.
heart_patient_data$thalach %>% hist(col = '#F39C12', xlim=c(50,250), xlab="Maximum heart rate achieved", main= "Histogram of Heart Rate achieved")
heart_patient_data$thalach %>% mean() %>% abline(v=., col="#1C2833", lwd=2, lty=5)
- We have gone through three attributes that are comparison of cholesterol level with sex, Resting blood pressure relationship with age and frequency of maximum heart beat rate of patients. - Females has more cholesterol problems and patient with age above 40 years comes with more blood pressure. This tells us that person with age above 40 mostly have high chances of getting heart disease and also cholesterol level is moreover high to Female. If the person also comes with maximum heart rate above 120 till 200, this cause an trouble for person and can be the patient of disease.
Hypothesis Testing
- This section is hypothesis testing with the help of one-sample t-test.
- Initially we will check, whether the data is normally distributed and based on that we will move further and state the statistical hypothesis.
- Many researchers and doctors has concept that heart disease patient has cholesterol level of 240 mg/dl.
- So, we will check whether the average cholesterol level belief was correct or not.
- We will start with summary.
heart_patient_data$chol %>% summary()
## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 126.0 211.0 241.0 246.7 275.0 564.0
- Now, we will plot the qqPlot of cholesterol and analyze it.
heart_patient_data$chol %>% qqPlot(dist = 'norm')
## [1] 153 49
- It is said, when all the values are between curve, it is normally distributed. However, as we have large sample size that is more than 30 sample count hence normality assumption is generally not required.
Hypothesis Testing Cont.
- So, We can summarize as :
\[H_0: \mu_1 < \ 240 \] - We will use Confidence Interval Approach. The specialty of this approach is when we test Ho for one sample t-test, it will also test for two-tailed hypothesis test. - So, we will calculate 95% CI for sample mean 246.7. Since, we don’t know standard deviation we will use s/sqrt(n). Hence, below mentioned formula will calculate 95% CI.
t.test(heart_patient_data$chol, conf.level = .95)$conf.int
## [1] 240.8397 252.5465
## attr(,"conf.level")
## [1] 0.95
t.test(heart_patient_data$chol, mu = 240, alternative = 'two.sided')
##
## One Sample t-test
##
## data: heart_patient_data$chol
## t = 2.2501, df = 302, p-value = 0.02516
## alternative hypothesis: true mean is not equal to 240
## 95 percent confidence interval:
## 240.8397 252.5465
## sample estimates:
## mean of x
## 246.6931
So, as per the confidence Interval rule if the 95% CI does not capture Ho, we reject Ho.
To conclude, we remember Ho = 240 and we can see that Ho value is not captured by the 95% CI [240.8397, 252.5465]. Hence, the test is statistically significant.