Chapter 2 - Summarizing Data

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.

1) goes with C). The histogram shows skewed right, so we would expect
the boxplot to show many outliers on the right, which is what C) shows.
2) goes with A). We can tell based on the range of data where it is
mostly in 50-70, and also since A) looks normal we do expect having
some outliers in the extreme values. 3) goes with B) because of the
data range of 0-100 and there appear to be no outliers. IQR is about
75 - 25 = 50. Since outliers are beyond 1.5 IQR from third or first
quartile, that puts the outlier bounds at about 150 and -50. No
datapoints are beyond these bounds so no outliers in the boxplot

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.

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.

Q1: 350k, median: 450k, Q3: 1000k, a lot of houses about 6000k.
IQR 650k.
Expect distribution to be skewed right since it has outliers to the
right with expensive houses. The median is the robust statistic
for skewed distributions so I expect the median to be more of a typical
observation. Likewise, the IQR is a more robust statistic for skewed
distributions so I expect it to better represent the variability
of the observations

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.

Q1: 300k, median: 600k, Q3: 900k, very few houses above 1200k
IQR: 600k
This distribution is largely symmetric with very few outliers.
The distiance from median to Q1 and median to Q3 is the same which also
suggests symmetric. The mean is about the same as the median for
a symmetric distribution so either should be fine. Because we have
a symmetric distribution, we can use the standard deviation
since we don't have to worry about extreme skewness.

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.

Assuming most students don't drink since under 21 means more than 50% of
students don't drink
Assuming Q1: 0, median: 0 Q3: 3, and a few excessive outliers beyond 7
IQR: 3
Similar to 2a) this looks like a skewed right distribution.
As it is skewed, median is better as it is a more robust statistic in
terms of representing a typical observation. Also for the same
reasoning, IQR would better represent the variability of the observations.

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.

This is also probably similar distribution to 2a) so a skewed right
distribution.
The median is more robust for skewed distributions with outliers
and the IQR better represents the variability than standard deviation
for skewed distributions.

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.

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

It looks like survival is not independent of the transplant (treatment)
because we see different rates of survival between the control and 
treatment groups. Specifically, it looks like those in the treatment
group have higher likelihood of survival. We can see this by the
different proportions of dead and alive under the control and treatment
groups.

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

The boxplots suggest the getting a treatment (new heart) 
extends survival time in days. In the control, we see the upper whisker
is something like 100 days, and the vast majority in the control did
not survive beyond this time. Compare that to the upper whisker of the
treatment group, which is nearly 1500 days. Also, looking at the median
survival time, the control looks like they barely survive up to 10 days
typically compared to 150 or so days in the treatment group.

What proportion of patients in the treatment group and what proportion of patients in the control group died?
```
30/34 died in control group = 88.23529% died
45/69 died in treatment group = 65.21739% died
```
One approach for investigating whether or not the treatment is effective is to use a randomization technique.

What are the claims being tested?

The claim being tested is whether or not the heart transplant (treatment)
had an impact on survival rates compared to no treatment (control).
H-nought = there is no difference between between treatment
H-A = treatment has a difference on survival rate

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, since we assume no difference between H-o and H-A except random noise. Lastly, we calculate the fraction of simulations where the simulated differences in proportions are as negative or more than actual H-A - H-nought = 0.6521739 - 0.8823529 = -0.230179. 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.

since we are told to compare treatment - control, and we think the treatment
has a LOWER proportion of dead cards, if treatment - control is negative,
then treatment had a smaller proportion of dead compared to control.
Actual H-A - H-nought = 0.6521739 - 0.8823529 = -0.230179

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

We see that our actual difference is H-A - H-nought = 0.6521739 - 0.8823529 = -0.230179
The simulations don't have anything with a difference greater than -.19 (4 trials).
So we would reject H-o in favor of H-A (reject the idea that there is no difference
between survival rates between treatment and control group). This suggests
that hearttransplants are effective in increasing lifespan/survival rate.