Stats scores. (2.33, p. 78) 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.
boxplot(scores)
Mix-and-match. (2.10, p. 57) Describe the distribution in the histograms below and match them to the box plots.
This is a symmetric distribution and is unimodal. The single prominant peak being the bin after 60. This histogram matches boxplot (2).
This histogram appears to be uniform, the peaks are about equally prominant. This histogram matches boxplot (2).
This histogram is bimodal and skewed right. This histogram matches boxplot (1).
Distributions and appropriate statistics, Part II. (2.16, p. 59) 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.
The distribution would be right skewed and the mean would not be a good representation of your typical observation relative to the median because of significant observations above $6,000,000. The IQR would best represent the observations because standard deviation weigh larger deviations more heavily.
The data would appear to be distributed symmetrically. Since the data is evenly distributed the mean would be a better representation and standard deviation would best represent the variability.
This data would be right skewed as the most prominant observation would be zero drinks per week. Since the data would be skewed a better representation of an observation would be the median, and for the variability we would represent this with IQR also because the data is skewed.
This data would be skewed right because much of the CEO’s would be earning significantly higher than the average salary. Since the CEO’s would appear as outliers we would use the median to represent the center of distribution. The IQR would be used to identify variance also because this data is skewed.
Heart transplants. (2.26, p. 76) 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.
cntrl <- 4/34
trtmnt <- 24/69
There is a 23% difference in survival rate between the treatment group and control group so we fail to reject the null hypothesis. This evidence suggests that survival is not independent of transplants.
The box plots show that the median of the treatment group have a higher survival time of about 100 days. Also, the Q1 and Q3 values are higher in the treament group suggesting a dependent relationship, or strong efficacy, in heart transplates and survival rates.
cntrl <- 30/34
trtmnt <- 45/69
cntrl
## [1] 0.8823529
trtmnt
## [1] 0.6521739
We assume the difference in survival rates with or without a heart transplant is due to chance.
We write alive on _____35_____ cards representing patients who were alive at the end of the study, and dead on ____13_____ cards representing patients who were not. Then, we shuffle these cards and split them into two groups: one group of size 24___ representing treatment, and another group of size _____24_____ 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 ____75_____. Lastly, we calculate the fraction of simulations where the simulated differences in proportions are _________. 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.
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