My Discussion Question

1.23 Haters are gonna hate, study confirms. A study published in the Journal of Personality and Social Psychology asked a group of 200 randomly sampled men and women to evaluate how they felt about various subjects, such as camping, health care, architecture, taxidermy, crossword puzzles, and Japan in order to measure their dispositional attitude towards mostly independent stimuli. Then, they presented the participants with information about a new product: a microwave oven. This microwave oven does not exist, but the participants didn’t know this, and were given three positive and three negative fake reviews. People who reacted positively to the subjects on the dispositional attitude measurement also tended to react positively to the microwave oven, and those who reacted negatively also tended to react negatively to it. Researcher concluded that “some people tend to like things, whereas others tend to dislike things, and a more thorough understanding of this tendency will lead to a more thorough understanding of the psychology of attitudes.” 60 (a) What are the cases?

The data collected for the each of the 200 randomly selected which would represent 1 row on a data frame for these data.

  1. What is (are) the response variable(s) in this study?

The response variable in this study is the subject’s attitude toward the fictional microwave oven.

  1. What is (are) the explanatory variable(s) in this study?

The explanatory variables are the subject’s attitude toward the the various independent stimuli such as camping, health care, and crossword puzzles, etc.

  1. Does the study employ random sampling?

The problem text does state that the sample was obtained by random sampling.

  1. Is this an observational study or an experiment? Explain your reasoning.

This would be an observational study, because it is conducted by survey, and experimental methods such as control groups, treatment groups and blinding are not present in the methodology.

  1. Can we establish a causal link between the explanatory and response variables?

Since this is an observational study a causal relationship cannot be established, there could be a confounding variable not accounted for such as what mood the person was in that day

  1. Can the results of the study be generalized to the population at large?

Yes, if the sample was truly random, then you can generalize it to the population at large. You need to make sure that the sample was randomly selected from everyone, and not just as an example, psychology undergrads at the college where the PI teaches.

The Main 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.

Each row represents a case, data from one individual, in the study.

This is equal to the number of rows which is 1691 participants.

Categorical variables are: sex, marital, and smoke. None of these are ordinal.

Numerical are: age, gross Income, amtWeekends, amtWeekdays. All of these variables are treated as discrete, even though age is really continuous.

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 in the age range 5-15 years old.

  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.

In the text for either 1.5 or 1.10 I did not see any mention of the children being selected at random. This means that the results are not generalization to the population.

This was an experiment and did have features of an experiment such as control and treatment groups, so a casual relationship can be established, but only for the 160 children in the study.

1.28 Reading the paper.Below are excerpts from two articles published in the NY Times: (a) 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-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 was an observational study not an experiment so a causal connection cannot be made. Furthermore their subjects were not random and cannot be generalized to the general population.

  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?

Their statement is not justified since this was an observational study and not an experiment. I would tell the person that they may have the cart before the horse, that remorse from bullying may lead to sleep disorders, or there could be something else, parental attention or nutrition, that the the study missed that is linked to both behaviors.

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?

This is an experiment.

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

The treatment group are those told to exercise, and the control group are those told not to exercise.

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

Age is put into three blocks, 18-30, 31-40, 41-55. Prehaps because instances of mental illness may be different in these groups.

  1. Does this study make use of blinding?

For the participants it would be nearly impossible to blind them, they would be well aware if they exercised or not. Furthermore, there is no mention of the researchers being blinded as to who has exercised or not. No, this is not a blinded experiment.

  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.

Given that this is an experiment with treatment and control groups and these groups were selected at random, you can both state a causal connection and generalize it to the population.

  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?

My sole reservation is the length of the study, which is not mentioned in the text, and the fact that they are only examined twice. I would recommend to the P.I. that both groups be examined at set intervals over the course of a year or more.

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, horizontal = TRUE)

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

  1. Normally distributed, symmetric and unimodal, boxplot 2

  2. Uniformly distributed, boxplot 3

  3. Right Skewed and unimodal, boxplot 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 would be a right skewed distribution, since the median is $450,000 but a significant number of houses are greater than $6M. Since it would have a large number of outliers

outlier = 1000000 + 1.5*(1000000-650000)
outlier
## [1] 1525000

In cases with a lot of outliers the median and IQR give more meaningful center and spread since they are less sensitive to outliers.

  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.

As seen below $ 1.2M is not an outlier to these housing prices. Also the quartiles happen at very even intervals of $ 300,000. This would suggest that the distributio is fairly symmteric and as such mean and standard deviation are good indicators of center and spread.

outlier = 900000 + 1.5*(900000-300000)
outlier
## [1] 1800000
  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.

“Assume that most students don’t drink sicne they are under 21.” The Author must have gone to a way different college as an undergrad than I did.

This distribution would have a strong mode at zero drinks per week since most students are under 21 assuming they start at age 17 or 18. Since there are students who do drink, and a few who drink heavily, this data set will be right skewed. As such median and IQR will be the best indicators of center and spread since the data is asymmetric and has some outliers.

  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.

These data sets tend to be right skewed with several high outliers representing executive salaries. Once again median and IQR are better indicators of center and spread due to their robustness against outliers.

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.

The area of the alive box in the treatment group is much bigger than the area of the alive box in the control. Furthermore the area of the alive box in the treatment group is larger than the area of the dead box of the treatment group. In the control group the area of the the alive box is very small compared to the dead box. This indicates that the heart transplant was very effective in extending the life of the recipient. So, no survival is not independent of recieving a heart transplant.

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

Consider that the upper whisker of the control box plot is less than the median of the treatment box plot, and that all but one outlier of the control group exists within the IQR of the treatment group. These data suggest that there is great efficacy in heart transplants.

  1. What proportion of patients in the treatment group and what proportion of patients in the control group died? > It looks like 85% of the control group died and about 60% of the treatment group died.

  2. One approach for investigating whether or not the treatment is effective is to use a randomization technique.

  1. What are the claims being tested?

That for a gravely ill heart patient, receiving a heart transplant increases survivablity.

  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 95 cards representing patients who were alive at the end of the study, and dead on 205 cards representing patients who were not. Then, we shuffle these cards and split them into two groups: one group of size 200 representing treatment, and another group of size 100 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 0.25. 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.

It is very unlikely that the difference in survival rate (0.4 - 0.15 = 0.25) is from random chance.