MGT 205 Project 3
KEMAN XIANG
Run the three code blocks below to get the heart data set.
library(tidyverse)## ── Attaching core tidyverse packages ──────────────────────── tidyverse 2.0.0 ──
## ✔ dplyr 1.1.4 ✔ readr 2.1.4
## ✔ forcats 1.0.0 ✔ stringr 1.5.1
## ✔ ggplot2 3.4.4 ✔ tibble 3.2.1
## ✔ lubridate 1.9.3 ✔ tidyr 1.3.0
## ✔ purrr 1.0.2
## ── Conflicts ────────────────────────────────────────── tidyverse_conflicts() ──
## ✖ dplyr::filter() masks stats::filter()
## ✖ dplyr::lag() masks stats::lag()
## ℹ Use the conflicted package (<http://conflicted.r-lib.org/>) to force all conflicts to become errorsheart <- read_csv('heart.csv')## Rows: 303 Columns: 14
## ── Column specification ────────────────────────────────────────────────────────
## Delimiter: ","
## dbl (14): age, sex, cp, trestbps, chol, fbs, restecg, thalach, exang, oldpea...
##
## ℹ Use `spec()` to retrieve the full column specification for this data.
## ℹ Specify the column types or set `show_col_types = FALSE` to quiet this message.head(heart)## # A tibble: 6 × 14
## age sex cp trestbps chol fbs restecg thalach exang oldpeak slope
## <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
## 1 63 1 3 145 233 1 0 150 0 2.3 0
## 2 37 1 2 130 250 0 1 187 0 3.5 0
## 3 41 0 1 130 204 0 0 172 0 1.4 2
## 4 56 1 1 120 236 0 1 178 0 0.8 2
## 5 57 0 0 120 354 0 1 163 1 0.6 2
## 6 57 1 0 140 192 0 1 148 0 0.4 1
## # ℹ 3 more variables: ca <dbl>, thal <dbl>, target <dbl>Data Description and Usage
- age: The person’s age in years
- sex: The person’s sex (1 = male, 0 = female)
- cp: The chest pain experienced (Value 1: typical angina, Value 2: atypical angina, Value 3: non-anginal pain, Value 4: asymptomatic)
- trestbps: The person’s resting blood pressure (mm Hg on admission to the hospital)
- chol: The person’s cholesterol measurement in mg/dl
- fbs: The person’s fasting blood sugar (> 120 mg/dl, 1 = true; 0 = false)
- restecg: Resting electrocardiographic measurement (0 = normal, 1 = having ST-T wave abnormality, 2 = showing probable or definite left ventricular hypertrophy by Estes’ criteria)
- thalach: The person’s maximum heart rate achieved
- exang: Exercise induced angina (1 = yes; 0 = no)
- oldpeak: ST depression induced by exercise relative to rest (‘ST’ relates to positions on the ECG plot. See more here)
- slope: the slope of the peak exercise ST segment (Value 1: upsloping, Value 2: flat, Value 3: downsloping)
- ca: The number of major vessels (0-3)
- thal: A blood disorder called thalassemia (3 = normal; 6 = fixed defect; 7 = reversable defect)
- target: Heart disease (0 = no, 1 = yes)
Input code for and answer all prompts below.
Initial Investigation
Question 1
1a. Using the chol variable, plot the distribution as a histogram.
heart %>%
ggplot(aes(x = chol)) +
geom_histogram()## `stat_bin()` using `bins = 30`. Pick better value with `binwidth`.
1b. Describe the distribution’s shape (e.g. left or right skew, normal, etc).
#The distribution’s shape is right skewedQuestion 2
2a. Using the age and thalach variables, find their correlation coefficient.
cor <- cor(heart[,1],heart[,8],use='complete',method='pearson')
cor## thalach
## age -0.39852192b. Show this relationship in a plot.
heart %>%
ggplot(aes(x=age,y=thalach)) +
geom_point()
2c. Does this correlation make sense for this data? Why/why not?
#This correlation makes sense. Because as age increases, human’s physical functions and health reduce, therefore have lower heart rates.Hypothesis Testing
Question 3
3a. Perform a t test for difference in means between genders (sex) on chol.
t.test(chol ~ sex, data = heart)##
## Welch Two Sample t-test
##
## data: chol by sex
## t = 3.0244, df = 134.39, p-value = 0.002985
## alternative hypothesis: true difference in means between group 0 and group 1 is not equal to 0
## 95 percent confidence interval:
## 7.617474 36.406982
## sample estimates:
## mean in group 0 mean in group 1
## 261.3021 239.28993b. Find and show the mean cholesterol of men and women.
heart %>%
group_by(sex) %>%
summarise(mean_cholesterol = mean(chol))## # A tibble: 2 × 2
## sex mean_cholesterol
## <dbl> <dbl>
## 1 0 261.
## 2 1 239.3c. Interpret the results above using a significance threshold of 0.05. Are the results statistically significant?
#Because the p-value = 0.002985 < significance threshold 0.05, the results are statistically significant.Question 4
4a. Perform a t test (hypothesis test) to determine if the blood pressure among those with heart disease is higher or not.
t.test(trestbps ~ target, data = heart, alternative = 'less')##
## Welch Two Sample t-test
##
## data: trestbps by target
## t = 2.5083, df = 272.56, p-value = 0.9936
## alternative hypothesis: true difference in means between group 0 and group 1 is less than 0
## 95 percent confidence interval:
## -Inf 8.448315
## sample estimates:
## mean in group 0 mean in group 1
## 134.3986 129.30304b. Interpret these results. Statistically, is blood pressure higher among those with heart disease? How can you tell?
#The results are not statistically significant. We fail to reject the null hypothesis. There is insufficient evidence to conclude that resting blood pressure is higher among individuals with heart disease.Modelling
Question 5
5a. Separate heart data into a training and testing data set.
sample_indexes <- sample(1:nrow(heart), size = nrow(heart) * 3/4)
heart_train <- heart[sample_indexes, ]
heart_test <- heart[-sample_indexes, ]5b. Construct a LINEAR regression model to predict for blood pressure.
fit <- lm(trestbps ~ ., data = heart_train)
summary(fit)##
## Call:
## lm(formula = trestbps ~ ., data = heart_train)
##
## Residuals:
## Min 1Q Median 3Q Max
## -34.531 -11.214 -2.583 10.220 52.703
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 82.27851 14.69977 5.597 6.66e-08 ***
## age 0.48965 0.14343 3.414 0.000767 ***
## sex -2.98269 2.68222 -1.112 0.267383
## cp 3.09584 1.26592 2.446 0.015276 *
## chol 0.03750 0.02420 1.549 0.122797
## fbs 5.31727 3.01292 1.765 0.079026 .
## restecg -1.60514 2.22819 -0.720 0.472082
## thalach 0.10656 0.05999 1.776 0.077120 .
## exang 0.18955 2.73809 0.069 0.944875
## oldpeak 1.06635 1.28801 0.828 0.408649
## slope -1.47851 2.34178 -0.631 0.528479
## ca 0.57162 1.22524 0.467 0.641308
## thal 0.57033 1.91458 0.298 0.766080
## target -6.66474 3.16091 -2.108 0.036158 *
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 16.26 on 213 degrees of freedom
## Multiple R-squared: 0.1971, Adjusted R-squared: 0.1481
## F-statistic: 4.023 on 13 and 213 DF, p-value: 6.688e-065c. What are the statistically significant variables in this model? What is the value of the Multiple R-squared statistic, and what does this value mean? Overall, is this a good prediction model?
#1. Statistically significant variables: age (because p_value 0.000134 < 0.05) fbs (because p_value 0.015681 < 0.05) oldpeak (because p_value 0.044594 < 0.05)
#2. Multiple R-squared: 0.1531 R^2=1-SSE/SST (SSE:error sum of squares, SST:total sum of squares)
#It refers to the goodness of fit, which is how well the regression line fits the observations. Its value is between 0 and 1, the closer to 1, the better the regression fitting effect. Because Multiple R-squared is 0.1531, the regression fits not well and is not a good prediction model.Question 6
6a. Construct a LOGISTIC regression model to predict for heart disease indicator flag. Calculate the accuracy using the testing data.
fit_class <- glm(target ~ ., data = heart_train, family = binomial(link = 'logit'))
prediction_class <- ifelse(predict(fit_class, newdata = heart_test, type = 'response') < 0.5, 0, 1)
sum(heart_test$target == prediction_class)/length(prediction_class)## [1] 0.81578956b. What percent of the time is this model accurate?
#81.57895% of the time this model is accurate