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.

  1. What does each row of the data matrix represent? Ans: Each row represents a case study.

  2. How many participants were included in the survey? Ans: 1691 participants

  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.

Variable Type sex Categorical nominal variable age Numerical continuous variable marital Categorical nominal variable grossIncome Numerical continuous variable smoke Categorical nominal variable amtWeekends Numerical discrete variable amtWeekdays Numerical discrete variable

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.

Ans: Population of interest is children between ages of 5 and 15 Sample size is 160 children

  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.

Ans: The results of this study can’t be used to generalize the population because it is not a big enough sample size and I question how random were the children selected

The findings can’t be used to establish causal relationships because there is not an established correlation

Question 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: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-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.

Ans: We can not conclude that smoking causes dementia based on this study. The reasons are that the sample size is not big enough or random enough to conclude. We didn’t also explore all the other factors like genetics and behavoural patterns

  1. Another article titled The School Bully Is Sleepy states the following:62 “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?

ANS: My friend must be ill informed,because the statement is not justified. There is not enough evidence and clear description of the variables measured. The sample size is not clear so there is no conculsion to be made from this article.

1.36

Exercise and mental health. A researcher is interested in the e???ects 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?

Ans: Prospective Study

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

Treatment Group: Patients that were told to exercise twice a week.

Control Group: Patients for whom advice was given as to not to exercise.

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

Ans; yes the blocking variable is age

  1. Does this study make use of blinding?

ANs; Yes

  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.

Ans: Yes the results of the study can be used to establish a causal relationship between exercise and mental health, because it was performed on random set. It can also be generalized to the population at large

  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?

I would not have reservation funding the proposal because it is compilies with principals of experimental design which are replication, to provide an estimate of experimental error; randomization, to ensure that this estimate is statistically valid; and local control, to reduce experimental error by making the experiment more efficient.

1.48

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

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

boxplot(scores,main="Student scores")

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

histogram C Matches to box plot 1 Right Skew distribution histogram B matches to box plot 3 Multimodal distribution histogram A matches to box plot 2 Symmetrical distribution

1.56

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.

Ans: Left skewed, the median would be best to represent a typical observation.The variability would be best represented with InterQuartile range

  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.

Ans: symmetric distribution, the median would be best to represent a typical observation.The variability would be best represented with interquartile range, because it only take one house that is very much more thatn 1.2million to throw off the mean

  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.

Ans: Left skewed, the median would be best to represent a typical observation.The variability would be best represented with InterQuartile range

  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

Ans: symmetric distribution, the median would be best to represent a typical observation.The variability would be best represented with interquartile range,

1.70

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 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.

Ans: THe mosaic plot, shows that the survival is dependent of the patient getting a transplant. This is because the treatment group had more patients that survived with transplant

  1. What do the box plots below suggest about the eficacy (e???ectiveness) of the heart transplant

Ans: The box plot suggest that the heart transplant increases the survival rate for a longer period of time.

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

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

  1. What are the claims being tested? Ans: Does Heart transplant increase survival reate

  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 2828 cards representing patients who were alive at the end of the study, and dead on 7575 cards representing patients who were not. Then, we shuffle these cards and split them into two groups: one group of size 6969 representing treatment, and another group of size 3434 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 45/69???30/34=???0.2301794569???3034=???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

  1. What do the simulation results shown below suggest about the e???ectiveness of the transplant program?

We conclude that the study results provide sufficiently strong evidence to conclude the heart transplant was a success since the difference in between the 100 simulations is centered near zero