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.
smokinghabits <- read.csv("https://raw.githubusercontent.com/jbryer/DATA606Spring2019/master/data/os3_data/Ch%201%20Exercise%20Data/smoking.csv")
a, What does each row of the data matrix represent?
head(smokinghabits)
## gender age maritalStatus highestQualification nationality ethnicity
## 1 Male 38 Divorced No Qualification British White
## 2 Female 42 Single No Qualification British White
## 3 Male 40 Married Degree English White
## 4 Female 40 Married Degree English White
## 5 Female 39 Married GCSE/O Level British White
## 6 Female 37 Married GCSE/O Level British White
## grossIncome region smoke amtWeekends amtWeekdays type
## 1 2,600 to 5,200 The North No NA NA
## 2 Under 2,600 The North Yes 12 12 Packets
## 3 28,600 to 36,400 The North No NA NA
## 4 10,400 to 15,600 The North No NA NA
## 5 2,600 to 5,200 The North No NA NA
## 6 15,600 to 20,800 The North No NA NA
b, How many participants were included in the survey?
nrow(smokinghabits)
## [1] 1691
c, 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.
str(smokinghabits)
## 'data.frame': 1691 obs. of 12 variables:
## $ gender : Factor w/ 2 levels "Female","Male": 2 1 2 1 1 1 2 2 2 1 ...
## $ age : int 38 42 40 40 39 37 53 44 40 41 ...
## $ maritalStatus : Factor w/ 5 levels "Divorced","Married",..: 1 4 2 2 2 2 2 4 4 2 ...
## $ highestQualification: Factor w/ 8 levels "A Levels","Degree",..: 6 6 2 2 4 4 2 2 3 6 ...
## $ nationality : Factor w/ 8 levels "British","English",..: 1 1 2 2 1 1 1 2 2 2 ...
## $ ethnicity : Factor w/ 7 levels "Asian","Black",..: 7 7 7 7 7 7 7 7 7 7 ...
## $ grossIncome : Factor w/ 10 levels "10,400 to 15,600",..: 3 9 5 1 3 2 7 1 3 6 ...
## $ region : Factor w/ 7 levels "London","Midlands & East Anglia",..: 6 6 6 6 6 6 6 6 6 6 ...
## $ smoke : Factor w/ 2 levels "No","Yes": 1 2 1 1 1 1 2 1 2 2 ...
## $ amtWeekends : int NA 12 NA NA NA NA 6 NA 8 15 ...
## $ amtWeekdays : int NA 12 NA NA NA NA 6 NA 8 12 ...
## $ type : Factor w/ 5 levels "","Both/Mainly Hand-Rolled",..: 1 5 1 1 1 1 5 1 4 5 ...
gender - Categorical; age - numerical;maritalStatus - > Continuous; highestQualification - Categorical; nationality - Categorical; ethnicity - Categorical; grossIncome - Categorical; Ordinal region - Categorical; smoke - Categorical; amtweekends - Categorical; amtweekdays - discrete; and type - 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.
a, Identify the population of interest and the sample in this study.
Answer: The population of interest in this study is the children between the ages of 5 to 15. And the sample used in this study is the 160 children between the ages of 5 to 15.
b, 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: The results of this study cannot be generalized to the population, simply because the region where the study took place is unknown. Also, compared to the population, the sample size is relative small. Therefore, a more robust sample size is needed to generalize the results to whole population. Because this is an experiment, so all findings from this study can be used to establish causal relationships.
1.28, Reading the paper. Below are excerpts from two articles published in the NY Times.
a, Based on this study, can we conclude that smoking causes dementia later in life? Explain your reasoning.
Answer: This is an observational study. There was no treatment and control group. We cannot derive causal relationship on observational study. Although the numbers show that the smoking causes dementia, we are not sure about the external factors(cofounding variable) which is involved. So we cannot conclude that smoking causes dementia. Sample popuation 23123: 25% - Dementia (includes 1136 - Alzheimer, 416 - Vascular dementia) and 5781 persons had Dementia.
b, Another article titled The School Bully Is Sleepy states the following: 62
Answer: This is an observational study. The statement is not a valid statement. Because the study shows that the children who had behavior issues are likely to show sleep disorders. But it is not other way around. We cannot state that the sleep disorders lead to bullying in school children.So we can’t make causal relationship on observational study.
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 strati???ed 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.
a, What type of study is this?
Answer: this is an experimental study.
b, What are the treatment and control groups in this study?
Answer: Treatment group contains random half the subjects from all the ages (18-30, 31-40 and 41- 55 year). Control group contains random half the subjects from all the ages (18-30, 31-40 and 41- 55 year).
c, Does this study make use of blocking? If so, what is the blocking variable?
Answer: Yes. This study makes use of blocking. The blocking variable is age.
d, Does this study make use of blinding?
Answer: No. It does not explicitly mention that the study is using blinding.
e, 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.
Answer: This is an experimental study. It can use used to find out the causal relationship. It depends on the sample size. The study can be generalized to the population if the sample size is large enough.
f, 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: suppose I am given the task of determining, I would recommend to fund the study. I would recommend a good number of sample size for this study.
1.48, Stats scores. Below are the final exam scores of twenty introductory statistics students.
Create a box plot of the distribution of these scores. The five number summary provided below may be useful.
statscores <- c(57, 66, 69, 71, 72, 73, 74, 77, 78, 78, 79, 79, 81, 81, 82, 83, 83, 88, 89, 94)
boxplot(statscores)
1.50, Mix-and-match. Describe the distribution in the histograms below and match them to the box plots.
Answer: (a) The match box plot number is 2. The distribution is unimodel which has one peak. And the histogram is symmetric.
(b) The match box plot number is 3. The distribution is multimodel which has many peak. And the histogram is symmetric.
(c) The match box plot number is 1. The distribution is unimodel which has one peak. And the histogram is right skewed.
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.
a, 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.
Answer: 1. Right skewed; 2. Median - Because the study is right skewed; 3. Standard Deviation - Because we need to best represent the variablity
b, 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.
Answer: 1. Symmentrical distribution; 2. Mean - Because the study is symmentrical; 3. IQR - Because all the data can be showed in a single chart with variability
c, 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.
Answer: 1. Symmentrical distribution; 2. Median - Because the study is multimodel; 3. Standard Deviation
d, 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. Right skewed; 2. Median - Because the study is right skewed; 3. Standard Deviation - Because we need to best represent the variablity
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.
heattrans <- read.csv("https://raw.githubusercontent.com/jbryer/DATA606Fall2016/master/Data/Data%20from%20openintro.org/Ch%201%20Exercise%20Data/heartTr.csv")
head(heattrans)
## id acceptyear age survived survtime prior transplant wait
## 1 15 68 53 dead 1 no control NA
## 2 43 70 43 dead 2 no control NA
## 3 61 71 52 dead 2 no control NA
## 4 75 72 52 dead 2 no control NA
## 5 6 68 54 dead 3 no control NA
## 6 42 70 36 dead 3 no control NA
a, Based on the mosaic plot, is survival independent of whether or not the patient got a transplant? Explain your reasoning.
summary(heattrans)
## id acceptyear age survived
## Min. : 1.0 Min. :67.00 Min. : 8.00 alive:28
## 1st Qu.: 26.5 1st Qu.:69.00 1st Qu.:41.00 dead :75
## Median : 49.0 Median :71.00 Median :47.00
## Mean : 51.4 Mean :70.62 Mean :44.64
## 3rd Qu.: 77.5 3rd Qu.:72.00 3rd Qu.:52.00
## Max. :103.0 Max. :74.00 Max. :64.00
##
## survtime prior transplant wait
## Min. : 1.0 no :91 control :34 Min. : 1.00
## 1st Qu.: 33.5 yes:12 treatment:69 1st Qu.: 10.00
## Median : 90.0 Median : 26.00
## Mean : 310.2 Mean : 38.42
## 3rd Qu.: 412.0 3rd Qu.: 46.00
## Max. :1799.0 Max. :310.00
## NA's :34
mosaicplot(table(heattrans$transplant,heattrans$survived))
From the graph of mosaic plot, the survival of the patient is dependet on transplant.
b, What do the box plots below suggest about the efficacy (effectiveness) of the heart transplant treatment.
Answer: From the graph of box plots, the efficacy of heart transplant treatment is very bad. Most(30) of the patients were dead. There were some outlies in this study. So some of them were alive might be due to the chance or error in experiment.
c, What proportion of patients in the treatment group and what proportion of patients in the control group died?
control_died <- (30/75)
control_died
## [1] 0.4
treatment_died <- (45/75)
treatment_died
## [1] 0.6
d, One approach for investigating whether or not the treatment is effiective is to use a randomization technique.
What are the claims being tested?
Answer: Whether the trasplant is successful or not.
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.
Answer: 1, Random; 2, Random; 3, half; 4, half, 5, Mean or Zero; 6, Low
What do the simulation results shown below suggest about the effectiveness of the transplant program?
Answer: From the simulation results, it shows that the effectiveness of happening is 3 times.