CUNY 606 Week 1 Homework Assignment

1.8 Smoking habits of UK residents. A survey was conducted to study the smoking habits of UK residents. Below is a data matrix displaying a portion of the data collected in this survey. Note that “£” stands for British Pounds Sterling, “cig” stands for cigarettes, and “N/A” refers to a missing component of the data.

  1. What does each row of the data matrix represent?

Each row repesents an unique identifier (each UK resident).

  1. How many participants were included in the survey?

A total of 1691 participants were included in the survey.

  1. Indicate whether each variable in the study is numerical or categorical. If numerical, identify as continuous or discrete.

Sex: Categorical. Age: Discrete Numerical. Marital: Categorical. grossIncome: Categorical. Smoke: Categorical. amtWeekends: Discrete Numerical. amtWeekdays: Discrete Numerical.

1.10 Cheaters, scope of inference. Exercise 1.5 introduces a study where researchers studying the relationship between honesty, age, and self-control conducted an experiment on 160 children between the ages of 5 and 15. The researchers asked each child to toss a fair coin in private and to record the outcome (white or black) on a paper sheet, and said they would only reward children who report white. Half the students were explicitly told not to cheat and the others were not given any explicit instructions. Differences were observed in the cheating rates in the instruction and no instruction groups, as well as some differences across children’s characteristics within each group.

  1. Identify the population of interest and the sample in this study.

Children between the ages of 5 and 15. Sample size is 160 children between 5 and 15.

  1. Comment on whether or not the results of the study can be generalized to the population, and if the findings of the study can be used to establish causal relationships.

This is an observation study. This cannot be used to establish causal relationships. The study could be generalized to the public if the sample was truly randomized and the sample was drawn thoughout the nation (though I suspect that n = 160 may be too low.)

1.28 Reading the paper. Below are excerpts from two articles published in the NY Times:

  1. An article titled Risks: Smokers Found More Prone to Dementia states the following:61 “Researchers analyzed data from 23,123 health plan members who participated in a voluntary exam and health behavior survey from 1978 to 1985, when they were 50-60 years old. 23 years later, about 25% of the group had dementia, including 1,136 with Alzheimer’s disease and 416 with vascular dementia. After adjusting for other factors, the researchers concluded that pack-a- day smokers were 37% more likely than nonsmokers to develop dementia, and the risks went up with increased smoking; 44% for one to two packs a day; and twice the risk for more than two packs.”

Based on this study, can we conclude that smoking causes dementia later in life? Explain your reasoning.

This is an observational study, and not an experiment. It would be inappropriate to imply a causation from an observation. In addition, there is bias in this observation, as they had volunteers for the study. It was not randomized.

  1. Another article titled The School Bully Is Sleepy states the following: “The University of Michigan study, collected survey data from parents on each child’s sleep habits and asked both parents and teachers to assess behavioral concerns. About a third of the students studied were identified by parents or teachers as having problems with disruptive behavior or bullying. The researchers found that children who had behavioral issues and those who were identified as bullies were twice as likely to have shown symptoms of sleep disorders.”

A friend of yours who read the article says, “The study shows that sleep disorders lead to bullying in school children.” Is this statement justified? If not, how best can you describe the conclusion that can be drawn from this study?

Again, an observation cannot prove a relationship. The conclusion is that there is a correlation but correlation does NOT imply causation.

1.36 Exercise and mental health. A researcher is interested in the effects of exercise on mental health and he proposes the following study: Use stratified random sampling to ensure representative proportions of 18-30, 31-40 and 41- 55 year olds from the population. Next, randomly assign half the subjects from each age group to exercise twice a week, and instruct the rest not to exercise. Conduct a mental health exam at the beginning and at the end of the study, and compare the results.

  1. What type of study is this?

Prospective experiment.

  1. What are the treatment and control groups in this study?

Treatment: excercising twice a week. Control: Not excercising at all.

  1. Does this study make use of blocking? If so, what is the blocking variable?

Yes. The blocks are ages 18-30, 31-40, and 41-55.

  1. Does this study make use of blinding?

This study does NOT use blinding. (It would be difficult to blind since the subjects are aware if they are excercising or not.)

  1. Comment on whether or not the results of the study can be used to establish a causal relationship between exercise and mental health, and indicate whether or not the conclusions can be generalized to the population at large.

If the study was truly randomized as noted in the question stem and the sample size was large enough, this could be generalized to the population at large. Since this is an experiment, a causal relationship could be established.

  1. Suppose you are given the task of determining if this proposed study should get funding. Would you have any reservations about the study proposal?

Restricting a group from exercising could be ethically wrong.

1.48 Stats scores. 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.

Min Q1 Q2 (Median) Q3 Max 57 72.5 78.5 82.5 94

scores <- c(57, 66, 69, 71, 72, 73, 74, 77, 78, 78, 79, 79, 81, 81, 82, 83, 83, 88, 89, 94)
summary(scores)
##    Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
##   57.00   72.75   78.50   77.70   82.25   94.00
boxplot(scores, main = "Final Exam Scores for Intro to Stats", xlab = "Student Scores", ylab = "Grades")

1.50 Mix-and-match. Describe the distribution in the histograms below and match them to the box plots.

  1. Symmetrical and Unimodal. Matches with 2.

  2. Symmetrical. Multimodal. Matches with 3.

  3. Right Skew and Unimodal. Matches with 1.

1.56 Distributions and appropriate statistics, Part II . 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.

This data would be best presented with median and with IQR. Given that most homes are below $1,000,000, the $6,000,000 is most certainly an outlier. A home like this will definitely change the mean of this data drastically and would alter the summary of this information. The median and IQR are more robust and would not change with this outlier. In addition, because of the $6,000,000 home, this data would be right 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.

This data set could be presented either with mean/SD or median/IQR. Because it is symetrical, the mean/SD would suffice and perhaps best present the data.

  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.

Right skewed. Likely better explained with median/IQR. Most college students drink perhaps 1 to 2 beers. There will be outliers where there are students who drink 8 to 10 beers, but they are few and far inbetween. Given the right skew and the outliers, we need more robust statistics, and median/IQR is the better fit.

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

Again, this will be right skewed. Most employees will make an average salary (likely < 100,000 dollars/year), whereas the CEOs and executives will make greater than a million per year. This is right skewed given these executive outliers. Again, this data will be better explained with median/IQR as they are robust.

1.70 Heart transplants. 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 o cial 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. Another variable called survived was used to indicate whether or not the patient was alive at the end of the study. Of the 34 patients in the control group, 30 died. Of the 69 people in the treatment group, 45 died.

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

Based on the mosaic plot, it appears the survival is dependent on whether or not the patient was transplanted. It appears that transplanted patients were more likely to be alive at the end of the study.

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

The box plots suggest that the treatment group survived longer than the control group.

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

Control group: 30/34 = .8824 Treatment group: 45/69 = .6522

  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?

(Hypothesis Null HO). The transplant does not alter the survivability of these patients.

(Hypothesis Alternative HA). The claim was that transplanted patients were more likely to survive than non-transplanted patient.

  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 79cards 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 .2302 . 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?

Simulation results in the assignment show that a large different such as .2302 is unlikely to occur by chance and that null hypothesis should be rejected. (Occurs 3 out of 100 times!)