Below are the final exam scores of twenty introductory statistics students.
57, 66, 69, 71, 72, 73, 74, 77, 78, 78, 79, 79, 81, 81, 82, 83, 83, 88, 89, 94
Create a box plot of the distribution of these scores. The five number summary provided below may be useful.
scores <- c(57, 66, 69, 71, 72, 73, 74, 77, 78, 78, 79, 79, 81, 81, 82, 83, 83, 88, 89, 94)
boxplot(scores, xlab = "Final Exam Scores", main = "Introductory Statistics", horizontal = TRUE)
Describe the distribution in the histograms below and match them to the box plots.
The description is as follows:
For each of the following, state whether you expect the distribution to be symmetric, right skewed, or left skewed. Also specify whether the mean or median would best represent a typical observation in the data, and whether the variability of observations would be best represented using the standard deviation or IQR. Explain your reasoning.
This data has right skewed distribution. The outliers do not have much effect on the data so Median would best represent this data. Due to this IQR would better represent the variability of the data.
This data has symmetrical distribution. There are no outliers and due to the symmetry Mean would better represent this data. Due to this Standard Deviation would better represent the variability of the data.
Since majority of the students are below 21 the number of drinks consumed will keep decreasing as the age decreases. This makes it a right skewed distribution. The outliers do not have much effect on the data so Median would best represent this data. Due to this IQR would better represent the variability of the data.
This data has right skewed distribution. The outliers do not have much effect on the data so Median would best represent this data. Due to this IQR would better represent the variability of the data.
The Stanford University Heart Transplant Study was conducted to determine whether an experimental heart transplant program increased lifespan. Each patient entering the program was designated an official heart transplant candidate, meaning that he was gravely ill and would most likely benefit from a new heart. Some patients got a transplant and some did not. The variable transplant indicates which group the patients were in; patients in the treatment group got a transplant and those in the control group did not. Of the 34 patients in the control group, 30 died. Of the 69 people in the treatment group, 45 died. Another variable called survived was used to indicate whether or not the patient was alive at the end of the study.
Survival is not independent of whether or not the patient got a transplant. From the mosaic plot and the box plot it is clear that the paitients who got the treatment have a higher survival rate.
The Q1 of treatment group is almost same as the Q3 of control group so the heart transplant treatment has roughly 50% more survival rate.
From the below calculations we can see that 65.22% of patients in the treatment group and 88.24% of patients in the control group died.
n_treat <- subset(heart_transplant, heart_transplant$transplant == 'treatment')
dead_treat <- subset(heart_transplant, heart_transplant$survived == 'dead' & heart_transplant$transplant == 'treatment')
nrow(dead_treat)/nrow(n_treat)
## [1] 0.6521739
n_control <- subset(heart_transplant, heart_transplant$transplant == 'control')
dead_control <- subset(heart_transplant, heart_transplant$survived == 'dead' & heart_transplant$transplant == 'control')
nrow(dead_control)/nrow(n_control)
## [1] 0.8823529
“Heart transplants decrease the number of deaths” is the claim being tested here.
nrow(dead_control)/nrow(n_control) - nrow(dead_treat)/nrow(n_treat)
## [1] 0.230179
We write alive on 28 cards representing patients who were alive at the end of the study, and dead on 75 cards representing patients who were not. Then, we shuffle these cards and split them into two groups: one group of size 69 representing treatment, and another group of size 34 representing control. We calculate the difference between the proportion of dead cards in the treatment and control groups (treatment - control) and record this value. We repeat this 100 times to build a distribution centered at 0. Lastly, we calculate the fraction of simulations where the simulated differences in proportions are 23.02%. If this fraction is low, we conclude that it is unlikely to have observed such an outcome by chance and that the null hypothesis should be rejected in favor of the alternative.
The simulated differences in proportions is small as it is 23.02%. The null hypothesis should be rejected.