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.58 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?

Each row in the dataset represents the smoking habits of residents in the United Kingdom.

  1. How many participants were included in the survey?

The survery contains 1691 participants.

  1. 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

Sex: Categorical Age: continuous Matrital: Categorical Gross Income: discrete smoke: Categorical amt Weekends:discrete amt Weekdays: discrete

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. Di???erences were observed in the cheating rates in the instruction and no instruction groups, as well as some di???erences across children’s characteristics within each group.

  1. In this study there are 160 Children between ages 5 and 15.

  2. The study doesn’t state that the children were assigned randomly, so the study couldn’t be used for generalized to the population.

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.

Since this study was voluntary and the sample were not random, we cannot conclude that the study can prove that smoking can cause dementia later in life.

  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 identied 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 identied 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 justied? If not, how best can you describe the conclusion that can be drawn from this study

Explain your reasoning.

The study doesn’t state where or how the students were selected - local/random, etc. This conclusion can be drawn that there is a correlation but cant say there is causation.

1.36 Exercise and mental health. A researcher is interested in the eects of exercise on mental health and he proposes the following study: Use stratied 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?

Stratied random sampling

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

The treament group will exercise twice a week and the control group will not exercise.

  1. Does this study make use of blocking? If so, what is the blocking variable? No

  2. Does this study make use of blinding? No

  3. 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. Yes - for this experiment we used random sampling, took a mental health exam at the beginning and end and compared with results from people who did exercise to those who didn’t.

  4. Suppose you are given the task of determining if this proposed study should get funding. Would you have any reservations about the study proposal? I feel that the only improvment would be to maybe to blind the researches - that way they wouldn’t know which group they were in as to not accidently add any bias.

1.48 Stats scores. Below are the ???nal 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 ???ve number summary provided below may be useful.

scores <- c(57, 66, 69, 71, 72, 73, 74, 77, 78, 78, 79, 79, 81, 81, 82, 83, 83, 88, 89, 94)

boxplot(scores)

summary(scores)
##    Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
##   57.00   72.75   78.50   77.70   82.25   94.00

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

    • 2 symetric
    • 3 uniform
    • 1 right skew

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.

Right skew, median, and IQR. The extreme values over 6mil would skew the data. The Median and IQR are better observations of the data when there are extreme values.

  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.

Symetric distribution. The median and IQR would be better observations because of the few houses above 1.2mil.

  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 skew, median and IQR. The few students who drink excessively would skew the data.

  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.

Right skew, median, and IQR. The few very high salaries would skew the data to the right.

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 ocial heart transplant candidate, meaning that he was gravely ill and would most likely bene???t 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. (See the next page for additional parts to this question.)

Based off the mosaic plot, the survial is dependent on whether the patient got a heart transplant or not. From reading the mosaic plot it shows that patients that received a heart transplant lived longer and more people were alive at the end of this study.

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

On the box plot it shows that the heart transplant is effective for increasing life expectancy.

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

From reading the article it showed:

Control Group

alive = 4

dead = 30

control = 34

Treatment Group

alive = 24

dead = 45

total = 69

  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?

H0 -Heart transplant and survival rate are independent and without no relationship. The survival rate of transplant patients was due to chance.

HA- transplant and survival rates are not independent. The survival rates were not due to chance

  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 28 representing patients who were alive at the end of the study, and dead on 75 cards representing patients who were not. Then, we shu???e 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 . 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 transplant program via the simulation results was effective since the results were centered around 0/ below .23. We then reject H0.