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.

Answer: Below is the box plot.

library(ggplot2)

scores <- data.frame(scores)

ggplot(scores, aes(y=scores)) + geom_boxplot() + labs(title="Distribution of Final Exam Scores") + ylab("Scores") + theme(legend.position = "none", axis.title.x = element_blank(), axis.text.x = element_blank(), axis.ticks.x = element_blank()) + theme(plot.title = element_text(hjust=0.5))

Mix-and-match. (2.10, p. 57) Describe the distribution in the histograms below and match them to the box plots.

Answer:

- The histogram (a) matches the box plot (2). The distribution of histrogram (a) is symmetric and normally distributed.

- The histogram (b) matches the box plot (3). The distribuion of histogram (b) is multimodal distribution because multiple peaks, which values occur more frequently than any other, in the distribution.

- The histogram (c) matches the box plot (1). The distribution of histogram (c) is right skewed because the tail on the right side of the histrogram is longer than the left side.

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.

  1. Housing prices in a country where 25% of the houses cost below $350,000, 50% of the houses cost below $450,000, 75% of the houses cost below $1,000,000 and there are a meaningful number of houses that cost more than $6,000,000.

Answer:

- The distribution is right skewed because the there are more data in the right of the distribution.

- The median would be best to represent a typical observation in the data because the data is skewed.

- The variability of observations would be best represented by Inter Quartile Range (IQR) because the distribution is skewed.

  1. Housing prices in a country where 25% of the houses cost below $300,000, 50% of the houses cost below $600,000, 75% of the houses cost below $900,000 and very few houses that cost more than $1,200,000.

Answer:

- The distribution is symmetric because the difference between the prices of first and second quartile is same as the differencce between the prices of second and third quartile.

- Both mean and median could be used to represent a typical observation in the data because the data is symmetric.

- The variability of observations could be best represented by either Standard Deviation or Inter Quartile Range (IQR) because the distribution is symmetric.

  1. Number of alcoholic drinks consumed by college students in a given week. Assume that most of these students don’t drink since they are under 21 years old, and only a few drink excessively.

Answer:

- The distribution is right skewed because most of the student don’t drink and only a few students drink excessively.

- The median would be best to represent a typical observation in the data because there are outliners present in the data.

- The variability of observations would be best represented by Inter Quartile Range (IQR) because the distribution is skewed.

  1. Annual salaries of the employees at a Fortune 500 company where only a few high level executives earn much higher salaries than the all other employees.

Answer:

- The distribution is left skewed because only a few high level executives earn much higher salaries.

- The median would be best to represent a typical observation in the data because the data is skewed.

- The variability of observations would be best represented by Inter Quartile Range (IQR) because the distribution 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.

  1. Based on the mosaic plot, is survival independent of whether or not the patient got a transplant? Explain your reasoning.

Answer: Survival is not independent of transplant because the survival rates are not the same for both groups (the survival rates in the treatment group is larger than control group).

  1. What do the box plots below suggest about the efficacy (effectiveness) of the heart transplant treatment.

Answer: The box plots suggest that treatment is effective as the distribution of survival time for the treatment group is larger than the control group. The median survival time is higher and interquartile range is larger for the treatment group.

  1. What proportion of patients in the treatment group and what proportion of patients in the control group died?

Answer:

- P(Death|Treatment) = 45/75 = 0.6

- P(Death|Control) = 30/75 = 0.4

  1. One approach for investigating whether or not the treatment is effective is to use a randomization technique.
  1. What are the claims being tested?

Asnswer:

- The Null hypothesis is that there is no difference in proportion of dead patients between treatment and control group. \(H0(p1-p2)=0\).

- The Alternative hypothesis that there is significant difference in proportion of dead patients between treatment and control group. \(HA(p1-p2) \neq 0\).

  1. The paragraph below describes the set up for such approach, if we were to do it without using statistical software. Fill in the blanks with a number or phrase, whichever is appropriate.

We write alive on __________ cards representing patients who were alive at the end of the study, and dead on _________ cards representing patients who were not. Then, we shuffle these cards and split them into two groups: one group of size _________ representing treatment, and another group of size __________ 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 _________. 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.

Answer:

- 28 cards representing patients who were alive…..

- 75 cards representing patients who were not…..

- 69 representing treatment, and another group…..

- 34 representing control. We calculate…..

- 0.2. Lastly, we calculate…..

- 0.2. If this fraction is low…..

  1. What do the simulation results shown below suggest about the effectiveness of the transplant program?

Answer: The simulation results suggest that the difference in proportion of dead patients between treatment and control group is not due to chance. Therefore, we can reject the Null hypothesis.

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