Chapter 1

Question 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.

ID sex age marital grossIncome smoke amtWeekends amtWeekdays
1 Female 42 Single Under£2,600 Yes 12 cig/day 12 cig/day
2 Male 44 Single £10,400 to £15,600 No N/A N/A
3 Male 53 Married Above£36,400 Yes 6 cig/day 6 cig/day
… … … … … … … …
1691 Male 40 Single £2,600 to £5,200 Yes 8 cig/day 8 cig/day
  1. What does each row of the data matrix represent?
  2. How many participants were included in the survey?
  3. Indicate whether each variable in the study is numerical or categorical. If numerical, identify as continuous or discrete. If categorical, indicate if the variable is ordinal.
Answer
  1. Each row of the data matrix represents a UK resident’s response to their smoking habits.
  2. There were 1691 participants in the survey.
  3. The variables of the study are:
sex age marital grossIncome smoke amtWeekends amtWeekdays
categorical continuous categorical ordinal categorical discrete discrete
Question 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.
  2. 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.
Answer
  1. The population of interest is all children between the age of 5 and 15. The sample in this study is of 160 of such children.
  2. The sample size is quite small to provide a a very good representative sample of the generalized population bewteen the of 5 and 15. However, since the study is experimental, the findings can be used to establish causal relationships.
Question 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:

“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-aday 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.

  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?

Answer
  1. No, smoking does not have a causal relationship with developing dementia. This study was observational rather than experimental, therefore, there could be more factors that can influence developing this disease.
  2. No, this statement is implying a causal relationship from an observational study. A better conclusion from this study would be that there is a association that children with symptoms of sleeping disorder tend to be identified as bullies.
Question 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 old 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?
  2. What are the treatment and control groups in this study?
  3. Does this study make use of blocking? If so, what is the blocking variable?
  4. Does this study make use of blinding?
  5. 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.
  6. Suppose you are given the task of determining if this proposed study should get funding. Would you have any reservations about the study proposal?
Answer
  1. This is an example of an blocked randomized experiment.
  2. The treatment group is the participants who are to exercise twice a week, and the control group is the participants who are not to exercise.
  3. This study did use blocking. The blocking variable is age, ranging from 18-30 years old, 31-40 years old and 41-55 years old.
  4. This study did not indicate the use of blinding because the participants knows if they are in the group that will be exercising or not.
  5. This study can be used to establish a causal relationship between exercising and mental health. In addition, it can be used to make conclusions on the general public because it was done by random sampling. However, the sample size may be an issue if it is not large enough, and whether or not fairly active participants where sent to the non-exercising group and vice-versa.
  6. I find that this study may not lead to accuracy results because it is unclear whether the effect will arise from stationary participant in the treatment group, or from fit participants in the control group. There is also not defined how much exercising that two weeks the treatment group will do, so fairly inactive participants might not match those who are active, and vice versa. Therefore, the reservation I would make is to have the control group continue their regular exercising regime and the treatment group increases their than their normal regime. This way in the control group, inactive will remain inactive and active will remain the same, while in the treatment group inactive participants will become active and active will be more active.
Question 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
Answer
scores <- c(57, 66, 69, 71, 72, 73, 74, 77, 78, 78, 79, 79, 81, 81, 82, 83, 83, 88, 89, 94)
boxplot(scores, ylab="Final Scores", xlab="Scores")
title("Boxplot of Final Exam Scores")

Question 1.50

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

Answer

Histogram A has a distribution that appears to be normal, it is unimodal and symmetric, therefore it matches with boxplot 2 because the mean, median, and mode all see to occur at the same point.

Histogram B looks more like multimodel than uniform since it starts high, dips, high again before it dips once more and forms another smaller mode and it is symmetric, therefore it matches with boxplot 3 because the interquartile range is very wide.

Histogram C is unimodal and skewed to the right, therefore it matches with boxplot 1 because upper whisker is further than where most of the data is, with many outliers.

Question 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.
  2. 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.
  3. 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.
  4. 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

  1. This distribution is expected to be extremely left skewed since many houses are priced over $6,000,000. The median would be a better represention of a typical obsevation in this data. The IQR would depict the variability of observations because it is not affected by the outliers in the 6 million price range.

  2. This distribution is expected to be symmetric since there is not many houses that are overpriced, and the quartiles fall evenly. The mean would be a better represention of a typical obsevation in this data. The standard deviation would depict the variability of observations because there is not many extreme prices, either too high or too low.

  3. This distribution is expected to be extremely right skewed since many students don’t drink. The median would be a better represention of a typical obsevation in this data. The IQR would depict the variability of observations because it is not affected by those who do drink.

  4. This distribution is expected to be right skewed since there is a few executives with a high salary. The median would be a better represention of a typical obsevation in this data. The IQR would depict the variability of observations because it is not affected by the high salary values.

Question 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 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. 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.
  2. What do the box plots below suggest about the efficacy (effectiveness) of the heart transplant treatment.
  3. What proportion of patients in the treatment group and what proportion of patients in the control group died?
  4. One approach for investigating whether or not the treatment is effective is to use a randomization technique.
  1. What are the claims being tested?
  2. 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 [xxx] cards representing patients who were alive at the end of the study, and dead on [xxx] cards representing patients who were not. Then, we shuffe these cards and split them into two groups: one group of size [xxx] representing treatment, and another group of size [xxx] 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 [xxx]. Lastly, we calculate the fraction of simulations where the simulated differences in proportions are [xxx]. 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.
iii. What do the simulation results shown below suggest about the effectiveness of the transplant program?

Answer
  1. Survival is dependent of whether or not the patient got a transplant because the width of the treatment group of who survived is nearly 3 times the width of those who are in the control group meaning that those who received the transplant lived longer.
  2. The boxplot show that the median of the treatment group is greater than the control group meaning that patients live longer suggesting that heart transplant treatment was effective.
library(openintro)
## Please visit openintro.org for free statistics materials
## 
## Attaching package: 'openintro'
## The following objects are masked from 'package:datasets':
## 
##     cars, trees
data(heartTr)
table1<-table(heartTr$survived,heartTr$transplant)
table1[2,]/colSums(table1)
##   control treatment 
## 0.8823529 0.6521739

The proportion of patients in the control group that dies is 88.2%. The proportion of patients in the treatment group that dies is 65.2%

  1. The claims being test is: Null: The success rate (survival) does not depend on the heart transplant treatment and observed difference in proportions is due to chances. Alternative: The success rate (survival) depends on the heart transplant treatment and observed difference in proportions is not due to chances.
  2. Sentence 1: 28 and 75. Sentence 2: 69 and 34. Sentence 4: 0 Sentence 5: greater than or equal to (45/69)-(30/34) = -0.23017.
  3. The stacked dot plot of differences from the 100 simulations shows that the proportions of dead tends to be greater in the control group suggesting that the data provide strong evidence that the heart transplant treatment is effective.