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.

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

a=2 Unimodal, symmetric distribution; b=3 uniform, symmetric distribution; c=1 Unimodal, right skewed distribution

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.

The distribution of housing prices would be right skewed given the outlier houses that cost more than $6,000,000. The median would best represent a typical observation given the data’s skewed distribution. The variability of observations would best be represented using inter-quartile range, which is a robust statistic that better represents the data distribution, unlike the standard deviations, which would be more susceptible to the extreme outlier house costs above $6,000,000.

  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.

The distribution of housing prices would be symmetrically skewed due to the uniform distribtion of prices below and above the median housing price. The mean or median are both suitable for representing a typical observation in light of the data’s uniformity.The variability of observations could also best be represented using standard deviation or inter-quartile range.

  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.

The distribution of the drink data would be right skewed since the majority don’t drink while an outlying number of students do. The median would better represent a typical observation due the data’s skewed distribution. The variability of observations would best be represented using an inter-quartile range, which is a more robust statistic less susceptible to the influence of the outlying cases compared to the standard deviation.

  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.

The distribution of the salary data would be right skewed since the majority of the workforce likely earns compartively a similiar salary, unlike the executives who make substantially more. The median would better represent a typical observation due the data’s skewed distribution. The variability of observations would best be represented using an inter-quartile range, which is a more robust statistic less susceptible to the influence of the outlying cases compared to the standard deviation.

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.

We can reject the null hypothesis that survial outcomes were due to chance in the study. The mosaic plot shows a clear disparity between the control and treatment group between the proportion of patients who survived and died.It’s possible there is an association between getting a transplant and longevity. Running additional simulations would help to strengthen the alternative hypothesis that transplants yield longer survival outcomes.

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

The box plot shows that the transplant was an effective treatment for part of the patient sample. A significantly larger inter-quartile range of patients who received treatment surived for longer compared to the 25% to 75% survival times of the control group.

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

About 35% of the patients in the treatment group were still alive as of the end of the study, while only about 12% of the patients in the control group survived to the end.

  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?

Null hypothesis: Any disparity in survival rate outcomes between those who receive heart plants and those who do not in populations who are gravely ill is due to chance.

Alternative hypothesis: A heart transplant leads to longer survival rates in patient populations who are gravely ill.

  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 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 near 0. 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.

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

The spread of the data away from 0 provides convincing evidence for the alterantive hypothesis that the transplant had a real effect on increasing longevity of treated patients compared to patients in the control group.

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