Import the "CENTENARI e BIOCHIMICA" dataset, which can be found at the following link: https://urly.it/38h1. How many subjects and variables are there in the dataset? library(readxl) esercitazione <- read_excel("CENTENARI_BIOCHIMICA.xls", col_names=TRUE) dim(esercitazione) Remove samples with missing data and save them in a new dataset called DATI. View(esercitazione) DATI <- na.omit(esercitazione) Calculate the measures of position and dispersion for the age variable and the absolute and relative frequency distributions for the group variable. summary(DATI$age) t = table(DATI$age); t[t == max(t)] sd(DATI$age); var(DATI$age); range(DATI$age) IQR(DATI$age) cv <- function(x) { m <- mean(x) sd <- sd(x) R <- (sd / m) * 100 return(R) } cv(DATI$age) table(DATI$Gruppo); prop.table(table(DATI$Gruppo)) Reclassify the categorical variables that R erroneously interpreted as numeric. str(DATI) DATI$Gruppo <- as.factor(DATI$Gruppo) DATI$uo <- as.factor(DATI$uo)  # operational unit DATI$FUMO <- as.factor(DATI$FUMO) DATI$INFARTO <- as.factor(DATI$INFARTO) DATI$INSUFF_RENE <- as.factor(DATI$INSUFF_RENE) DATI$DIABETE <- as.factor(DATI$DIABETE) DATI$TEST_1_DIABETE <- as.factor(DATI$TEST_1_DIABETE) DATI$TEST_2_DIABETE <- as.factor(DATI$TEST_2_DIABETE) Change the variable names “uo” to “unità_operativa” and “age” to “Età”. names(DATI) names(DATI)[2] <- "unità_operativa" names(DATI)[4] <- "Età" names(DATI) # Alternatively, all at once names(DATI)[c(2, 4)] <- c("unità_operativa", "Età") names(DATI) Create a vector called “età_crescente” containing the values sorted in ascending order of the “Età” variable. età_crescente <- sort(DATI$Età) Create a vector called “COL_TOT_decrescente” containing the values sorted in descending order of the COL_TOT variable. COL_TOT_decrescente <- sort(DATI$COL_TOT, decreasing = TRUE) What are the levels of the “Gruppo” variable? levels(DATI$Gruppo) Change the reference level of the “Gruppo” variable, setting "CTRL" (control) as the reference. # First way DATI$Gruppo <- relevel(DATI$Gruppo, "CTRL") levels(DATI$Gruppo) # Second way DATI$Gruppo <- ordered(DATI$Gruppo, levels = c("CTRL", "CENT", "FIGLI")) levels(DATI$Gruppo) # Third way DATI$Gruppo <- factor(v, levels = c("CTRL", "CENT", "FIGLI")) levels(b) How many centenarians have a BMI < 25? dim(subset(DATI, DATI$Gruppo == "CENT" & DATI$BMI < 25)) # Alternatively table(DATI$Gruppo, DATI$BMI < 25) Create a new variable in the dataset called “fumo_ricodificato” that will equal “Fumatore” for FUMO = 1 and “Non_fumatore” for FUMO = 0. table(DATI$FUMO) DATI$fumo_ricodificato <- ifelse(DATI$FUMO == "1", "Fumatore", "Non_fumatore") table(DATI$fumo_ricodificato) Change the levels of the DIABETE variable, replacing 0 with "NO" and 1 with "SI". # First way levels(DATI$DIABETE)[levels(DATI$DIABETE) == "0"] <- "NO" levels(DATI$DIABETE)[levels(DATI$DIABETE) == "1"] <- "SI" levels(DATI$DIABETE) # Second way levels(DATI$DIABETE) <- c("NO", "SI") Create a new variable (separate from the dataset) called glicemia_Ricod that will be equal to: “Ipoglicemia” when the blood glucose value is below 60, “normale” when it is between 60 and 100 (inclusive), and “Iperglicemia” when above 100. Glicemia_Ricod <- ifelse(DATI$GLICEMIA < 60, "ipoglicemia", ifelse(DATI$GLICEMIA >= 60 & DATI$GLICEMIA <= 100, "normale", "iperglicemia")) table(Glicemia_Ricod) How many hypoglycemic, normal, and hyperglycemic subjects are there? table(Glicemia_Ricod) Extract the albumin values only for the controls and create a data.frame with these values only. dati_ridotti <- as.data.frame(subset(DATI, Gruppo == "CTRL")[, 20]) # Alternatively, in parts dati_ridotti <- subset(DATI, Gruppo == "CTRL") dati_ridotti <- dati_ridotti[, 20] Calculate the mean and median for the Creatinine and Urea variables divided by Group. aggregate(DATI[, 7:8], list(DATI$Gruppo), mean) aggregate(DATI[, 4:5], list(DATI$Gruppo), median) Display with a graph if the distribution of the GLICEMIA variable is different between Diabetics and non-Diabetics. boxplot(DATI$GLICEMIA ~ DATI$DIABETE) # With ggplot p <- ggplot(DATI, aes(x = DIABETE, y = GLICEMIA)) + geom_boxplot() p p1 <- ggplot(DATI, aes(x = DIABETE, y = GLICEMIA)) + geom_boxplot(outlier.colour = "red", outlier.shape = 8, outlier.size = 4) p1 p2 <- ggplot(DATI, aes(x = DIABETE, y = GLICEMIA, color = Gruppo, fill = "red")) + geom_boxplot(outlier.colour = "red", outlier.shape = 8, outlier.size = 4) p2 Display with a graph the percentages of subjects belonging to the various operational units. percentuali <- round(prop.table(table(DATI$unità_operativa)) * 100) labels <- paste(levels(DATI$unità_operativa), percentuali, "%", sep = "_") pie(table(DATI$unità_operativa, label = as.factor(labels))) # With ggplot FREQ <- as.data.frame(table(DATI$unità_operativa)) FREQ <- cbind(FREQ, labels) pie <- ggplot(FREQ, aes(x = "", y = Freq, fill = labels)) + geom_bar(stat = "identity", width = 1) pie <- pie + coord_polar("y", start = 0) + geom_text(aes(label = labels), position = position_stack(vjust = 0.5)) pie <- pie + labs(x = NULL, y = NULL, fill = NULL) pie <- pie + theme_classic() + theme(axis.line = element_blank(),  axis.text = element_blank(),  axis.ticks = element_blank(),  plot.title = element_text(hjust = 0.5, color = "#666666")) Display with a graph the relationship between GLICEMIA and HOMA-IR. scatter.smooth(DATI$GLICEMIA, DATI$`HOMA-IR`) plot(DATI$GLICEMIA, DATI$`HOMA-IR`) # With ggplot p1 <- ggplot(DATI, aes(x = DATI$GLICEMIA, y = `HOMA-IR`)) + geom_point(size = 3) + geom_smooth(method = 'lm') p2 <- ggplot(DATI, aes(x = DATI$GLICEMIA, y = `HOMA-IR`, colour = Gruppo)) + geom_point(size = 3) + geom_smooth(method = 'lm') Save the produced graphs in a PDF. Is the number of subjects who had a heart attack significantly different between smokers and non-smokers? Choose the most appropriate test and indicate the p-value. chisq.test(table(DATI$INFARTO, DATI$FUMO)) Is there a difference in total cholesterol values (COL_TOT) between subjects with and without kidney blockage (INSUFFICIENZA_RENE)? Choose the test you find appropriate and interpret the result. # Checking normality and homoscedasticity using the following tests: shapiro.test(DATI$COL_TOT) bartlett.test(DATI$COL_TOT ~ DATI$INSUFF_RENE) # We have a binary categorical variable and a continuous variable, so we apply the t-test. t.test(DATI$COL_TOT ~ DATI$INSUFF_RENE, paired = FALSE) # No significant difference exists in total cholesterol values Choose the most appropriate plot to display the distribution of the variable GLICEMIA. hist(DATI$GLICEMIA) plot(density(DATI$GLICEMIA)) # with ggplot ggplot(data = DATI, aes(DATI$GLICEMIA)) + geom_histogram() Create a barplot showing the mean blood glucose levels among the subject groups CENT, FIGLI, and CTRL, and add a legend. barplot(tapply(DATI$GLICEMIA, DATI$Gruppo, mean), col = c("purple", "green3", "cyan"), main = "Barplot", ylim = c(0, 120)) legend("topright", c("CTRL", "CENT", "FIGLIC"), cex = 0.8, fill = c("purple", "green3", "cyan")) # with ggplot p <- ggplot(DATI, aes(x = Gruppo, y = GLICEMIA, fill = DATI$Gruppo)) + geom_bar(stat = "identity") p Do diabetic individuals have higher blood glucose values than non-diabetic individuals? # Applying the Shapiro test to check for normality of the glucose variable. shapiro.test(DATI$GLICEMIA) wilcox.test(DATI$GLICEMIA ~ DATI$DIABETE, paired = FALSE) Are cholesterol values different across operating units? shapiro.test(DATI$COL_TOT) bartlett.test(DATI$COL_TOT ~ DATI$unità_operativa) aov(DATI$COL_TOT ~ DATI$unità_operativa) summary(aov(DATI$COL_TOT ~ DATI$unità_operativa)) # To evaluate which groups have significant differences: pairwise.t.test(DATI$COL_TOT, DATI$unità_operativa, p.adj = "bonf") # Units 3 and 5 are significantly different from units 1 and 2. Is there a correlation between COL_TOT and BMI, and between Hematocrit and Red Blood Cells? cor.test(DATI$BMI, DATI$COL_TOT) cor.test(DATI$Ematocrito, DATI$Eritrociti) Investigate the correlations present in the dataset graphically. library(corrplot) M <- cor(DATI[, 4:30]) corrplot(M, method = "circle") How many smokers are there among the groups? table(DATI$FUMO, DATI$Gruppo) Calculate the mean age variable in the CTRL group. tapply(DATI$Età, DATI$Gruppo, mean) # or DATI_CTRL <- subset(DATI, DATI$Gruppo == "CTRL") mean(DATI_CTRL$Età) Is there a relationship between heart attack and kidney failure? table(DATI$INFARTO, DATI$INSUFF_RENE) fisher.test(table(DATI$INFARTO, DATI$INSUFF_RENE)) Studies have been conducted on immune abnormalities in autistic children; the data on serum concentration levels (units/ml) of an antigen are reported for three groups of children under the age of 10. Are there differences among the three groups: autistic, normal, and developmentally delayed? Autistici <- c(755, 383, 380, 215, 400, 343, 415, 360, 345, 450, 410, 435, 460, 360, 225, 900, 365, 440, 820, 400, 170, 300, 325, 345, 230, 370, 285, 315, 195, 270, 305, 375, 220) normal <- c(165, 390, 290, 435, 235, 320, 330, 205, 375, 345, 305, 220, 270, 355, 360, 335, 305, 360, 335, 305, 325, 245, 285, 370, 345) ritardati <- c(380, 510, 315, 565, 715, 380, 390, 245, 155, 335, 295, 200, 105, 105, 245) # Create a vector with all serum concentration values. concentrazionesierica <- c(autistici, normal, ritardati) length(concentrazionesierica) # Create a vector containing the categories corresponding to serum concentration values. GRUPPI <- c(rep("AUT", length(autistici)), rep("NORM", length(normal)), rep("RIT", length(ritardati))) GRUPPI <- as.factor(GRUPPI) # Check for normality: shapiro.test(concentrazionesierica) # Since we do not have a normal distribution, we apply a non-parametric test, specifically the Kruskal-Wallis test (the non-parametric equivalent of the ANOVA test). # Set the continuous variable as Y and the categorical variable as X. kruskal.test(concentrazionesierica ~ GRUPPI) The dietitian Mario Smilzo claims that his diet leads to rapid weight loss. To prove it, he takes a group of 10 people and weighs them before and after the diet. Their weight before and after the diet is: Investigate whether the diet resulted in a significant weight reduction. PESO1 <- c(80, 90, 96, 85, 102, 105, 99, 89, 98, 78, 75, 79, 98, 82, 95, 102, 95, 80, 97, 76) prepost <- c(rep("pre", 10), rep("post", 10)) shapiro.test(PESO1) bartlett.test(PESO1, prepost) t.test(PESO1 ~ prepost, paired = TRUE, alternative = "less") A study aims to evaluate whether an anti-cancer drug is capable of reducing tumor size. The drug is administered to 3 groups of animals: an untreated group (NO_F), a group treated with the study drug (F_SPE), and a group treated with the traditional drug (F_TRA). The results related to the dry weight of the biopsy in mg are shown in the table. # Comparison among groups with ANOVA TRATTAMENTI <- c(NO_F, F_SPE, F_TRA) Gruppi <- c(rep("NO_F", length(NO_F)), rep("F_SPE", length(F_SPE)), rep("F_TRA", length(F_TRA))) shapiro.test(TRATTAMENTI) bartlett.test(TRATTAMENTI, Gruppi) model <- aov(TRATTAMENTI ~ Gruppi) summary(model) TukeyHSD(model) In a hospital, it is hypothesized that living conditions and stress may influence the response to a particular therapy. If so, many therapy outcomes could be improved by addressing living conditions. Therefore, a correlation between quality of life (QOL), evaluated through questionnaire scores (from 1 to 100), and response to therapy (R2T), evaluated with a clinical improvement score (from 1 to 10), was investigated on a sample of 10 subjects: Verify if there is a correlation. # The data are ordinal categorical QOL <- c(10, 90, 20, 99, 35, 55, 46, 80, 75, 66) R2T <- c(1, 8, 2, 9, 3, 5, 4, 7, 7, 6) cor.test(QOL, R2T, method = "kendall")

EXERCISE 1 In a retrospective study aimed at verifying whether coffee consumption influences the risk of myocardial infarction, data were collected on patients residing in a town. Coffee consumption was classified as low (≤3 cups per day) or high (>3 cups per day). The results are shown in the table. I construct my 2x2 contingency table with R. For this purpose, I need to create a matrix object that contains the data. In the hypothetical table I’ll create, the columns will represent exposed/non-exposed, and the rows will represent case/control. The table will be structured as follows: TAB <- matrix(c(78, 1322, 48, 1352), 2) chisq.test(TAB, correct=FALSE) The relationship is highly significant, and high coffee consumption is associated with myocardial infarction. It’s possible to calculate a risk, or better, an Odds Ratio = 1.661566. EXERCISE 2 We want to assess whether habitual alcohol consumption is a risk factor for liver cirrhosis. Among 130 healthy subjects, 55 were regular alcohol consumers, while among 130 sick subjects, 90 were regular alcohol consumers. 1. Is there an association between habitual alcohol consumption and liver cirrhosis? 2. Provide a measure of the strength of association, commenting on the result. I construct the contingency table. The exposed individuals are 55 and 90, so the non-exposed individuals are obtained by subtracting from the totals. TAB <- matrix(c(90, 55, (130-90), (130-55)), 2) c hisq.test(TAB, correct=FALSE) Significant association with p-value < 0.05. EXERCISE 3 In a study, we want to assess the effectiveness of an innovative antibiotic A compared to the standard antibiotic S in treating an infection I. A sample of 600 subjects with infection I is selected, with 300 treated with A and the other 300 treated with S. After 15 days, 90% of the patients treated with A recovered, and 80% of the subjects treated with S recovered. 1. Is this a cohort study or a case-control study? (Justify the answer) 2. Is there an association between the antibiotic used and recovery? 3. Provide the most informative measure of association for the clinician in relation to the proposed study. 1 - The study is a cohort study. A case-control study is retrospective; in this case, a treatment was administered to a cohort, and the result was then observed. 2 - I construct the contingency table where the exposed A and exposed S groups each contain 300 subjects, with 90% recovered among those exposed to A and 80% recovered among those exposed to S. A <- 300*0.9  B <- 300*0.8 C <- 300 - A D <- 300 - B TAB <- matrix(c(A, C, B, D), 2) chisq.test(TAB, correct=FALSE) Significant association with p-value < 0.05. OR <- 270  60 / 240  30 = 2.25 EXERCISE 4 A study was conducted to assess the association between contact with herbicides/pesticides and bladder cancer in Scottish Terrier dogs. A total of 57 dogs with cancer (C+) and 109 dogs without cancer (C-) were considered. Among the C+ subjects, 42 had come into contact with herbicides/pesticides, and among the C- subjects, 41 had come into contact with herbicides/pesticides. 1. Is there evidence of an association between bladder cancer and contact with herbicides/pesticides? 2. Provide a measure of the strength of association. A <- 42 B <- 57 - 42 C <- 41 D <- 109 - 41 TAB <- matrix(c(A, C, B, D), 2) chisq.test(TAB, correct=FALSE)

1- Import the database “CENTENARI e BIOCHIMICA” - Remove samples with missing data, save a new dataset called DATI, and reclassify the factorial variables mistakenly interpreted by R as numeric. I use the read.table function after saving the data as a tab-delimited text file (.txt). I specify that the file has a header and that the separator between the columns is a tab using sep="\t". Note: If you import the data through the graphical interface, be careful about MISSING DATA. They MUST BE INDICATED in the "import options" drop-down menu under NA. Choose how missing data are indicated—in this example, they are marked as "NA," but in other cases, they may be empty cells. DATA <- read.table("DIRECTORY/DATI/CENTENARI_BIOCHIMICA.txt", header=TRUE, sep="\t", na.string="") Note: In this example, an object called DATA was created that contains the data. If you import the CENTENARI_BIOCHIMICA file using R’s graphical interface, the object will be called CENTENARI_BIOCHIMICA instead of DATA. Of course, the subsequent commands need to be adjusted, or you can create a copy called DATA that contains CENTENARI_DATI. I first check the variable types with: str(DATA) Most of the variables are numeric or integers. The only incorrect ones are Gruppo, uo, pid, FUMO, INFARTO, INSUFF_RENE, DIABETE, TEST_1_DIABETE, TEST_2_DIABETE. I convert these to factors. This is important because R will treat them accordingly during analysis. DATA$Gruppo <- as.factor(DATA$Gruppo) DATA$uo <- as.factor(DATA$uo) DATA$pid <- as.factor(DATA$pid) DATA$FUMO <- as.factor(DATA$FUMO) DATA$INFARTO <- as.factor(DATA$INFARTO) DATA$INSUFF_RENE <- as.factor(DATA$INSUFF_RENE) DATA$DIABETE <- as.factor(DATA$DIABETE) DATA$TEST_1_DIABETE <- as.factor(DATA$TEST_1_DIABETE) DATA$TEST_2_DIABETE <- as.factor(DATA$TEST_2_DIABETE) Check again: str(DATA) Now, remove missing data: DATI <- na.omit(DATA) dim(DATI) 2- Display a graph to show if the distribution of the variable calcium is different between centenarians, their children, and controls. There are at least three types of graphs we can use to visualize the distribution of a variable: density plot, boxplot, and histogram-plot. We will use the density plot. The challenge is that I want to visualize the distributions of the variable “Calcio” for the three groups at the same time. To do this, I can use the sm.density.compare() function from the sm package. I install and load the package: install.packages("sm") library(sm) Generate the graph: sm.density.compare(DATI$CALCIO, DATI$Gruppo) The graph shows how the calcium variable is distributed among the three groups: Centenarians, their children, and Controls. However, the legend is missing, which makes it hard to understand the graph. I also want to refine the graph so that all necessary information is included. What’s missing: 1. Titles for the axes 2. A title for the graph 3. Choice of colors 4. A legend 1 - Add axis labels We can add axis labels with xlab=("Calcium Levels") and ylab=("Density"). This is added to the previous command simply by separating it with a comma. sm.density.compare(DATI$CALCIO, DATI$Gruppo, xlab=("Calcium Levels"), ylab=("Density")) 2 - Add a title To add a title, just append the command title(main="Density-Plot of Calcium and Groups") to the previous code. sm.density.compare(DATI$CALCIO, DATI$Gruppo, xlab=("Calcium Levels"), ylab=("Density")) title(main="Density-Plot of Calcium and Groups") 3 - Choose colors I need to determine how many colors to use by checking how many groups we have, i.e., the levels of the factor variable DATI$Gruppo. The command: levels(DATI$Gruppo) Returns "CENT", "CTRL", "FIGLI", so there are three groups. I can now decide to color Centenarians, their children, and Controls in black, red, and blue, respectively. I add the option col=c("red","black","blue") to the sm.density.compare() command: sm.density.compare(DATI$CALCIO, DATI$Gruppo, xlab=("Calcium Levels"), ylab=("Density"), col=c("red","black","blue")) title(main="Density-Plot of Calcium and Groups") 4 - Add the legend Use the legend() command to add a legend. This command needs three pieces of information: a. Where to place the legend on the graph (“topright”)b. The levels of the grouping variable (use levels(DATI$Gruppo))c. The colors used for the groups: fill=c("red","black","blue") The full command is: legend("topright", levels(DATI$Gruppo), fill=c("red","black","blue")) The final graph code will be: sm.density.compare(DATI$CALCIO, DATI$Gruppo, xlab=("Calcium Levels"), ylab=("Density"), col=c("red","black","blue")) title(main="Density-Plot of Calcium and Groups") legend("topright", levels(DATI$Gruppo), fill=c("red","black","blue")) 3- There are five operational units (regions where the subjects were recruited). Some claim that, due to a technical problem, Unit 2 measures different glucose levels compared to the other units. Check if this hypothesis is valid. This exercise involves determining whether subjects recruited by Unit 2 differ from the others in glucose levels. In other words, we compare Unit 2 with all the other subjects. There are various strategies, but I decide to reduce it to a two-group comparison: Unit 2 versus all others. I see that my data includes the variable DATI$uo, which indicates operational units. However, this is not binary as I want it to be (Unit 2 vs. all other units). So, I create a new variable: DATI$uo2BIN <- as.factor(DATI$uo == "2") levels(DATI$uo2BIN) Now I want to rename the levels from "False" and "True" to 0 and 1: levels(DATI$uo2BIN) <- c(0,1) I have now created a binary factor variable that distinguishes Unit 2 from the others. I can now visualize the glucose levels in the two groups I just created. A boxplot or a density plot can show if the distributions are similar or not. boxplot(DATI$GLICEMIA~DATI$uo2BIN) Replace: • The variable to be visualized • The grouping variable • The legend with new terms • Modify colors (there are now only two groups) sm.density.compare(DATI$GLICEMIA, DATI$uo2BIN, xlab=("Glucose Levels"), ylab=("Density"), col=c("red","black")) title(main="Density-Plot of Glucose and Groups") legend("topright", levels(DATI$uo2BIN), fill=c("red","black")) The graphs seem to indicate a small difference in glucose values between Unit 2 and the others. The two distributions are slightly different. However, to be sure this difference is statistically significant, we need to apply a statistical test. First: What test should I use—parametric or non-parametric? I run a Shapiro test to check for normality: shapiro.test(DATI$GLICEMIA) Since the p-value is < 0.05, I conclude that glucose is not normally distributed. Therefore, I apply a non-parametric test for independent samples: wilcox.test(DATI$GLICEMIA~DATI$uo2BIN, paired=FALSE) The result shows that the difference is significant. 4- Study the relationship between glucose and diabetes. Use a regression model where Y=GLICEMIA, X=DIABETE. Interpret the result. The regression model is: model <- glm(DATI$GLICEMIA~DATI$DIABETE, family="gaussian") summary(model) The result shows an association between glucose levels and diabetic status. Specifically, the analysis tells us that the intercept (status of diabetic=0) has a mean glucose value of 86.35, and there is an increase of 90.68 units in glucose when moving from diabetic status 0 to 1. In other words, becoming diabetic is associated with an increase in glucose of 90.68 units. 5- Assuming that the glucose variable is normally distributed, calculate if the average glucose values differ between diabetics and non-diabetics. Compare the results with the previous exercise and comment. If we assume that the glucose variable is normally distributed and we want to assess if there are significant differences between diabetics and non-diabetics, the appropriate statistical test is an independent samples t-test. t.test(DATI$GLICEMIA~DATI$DIABETE) The result tells us that the difference in average glucose values between the two groups is significant. The average glucose value in the healthy group is 86.35, while in the diabetic group it is 177.04. The difference between the average values of the two groups is 90.68 units. This result, when compared with the previous one, provides a better understanding and interpretation of the regression result. 6- Which variables are normally distributed? To test if the variables in my dataset are normally distributed, I can use the Shapiro-Wilk test. I should apply it to all numerical variables that I assume have a normal distribution and want to test. This test is not applicable to categorical or qualitative variables. The formula to use is shapiro.test(VARIABLE) for all variables I want to test, but it could be a long process if there are many variables. Alternatively, I can use the lapply function. The formula is lapply(DATAFRAME, test-to-apply). The dataframe must be composed of variables where the function we want to run can be applied. In other words, the dataframe must contain only columns compatible with the function we want to use. In this case, the shapiro.test function does not make sense for categorical variables, so we will create a dataframe that contains only the numerical variables to test for normality with shapiro.test. I can do this by creating a new dataframe by selecting the numerical variables I am interested in. In this case, the numerical variables are between columns 4 and 30 inclusive, so the command will be: lapply(DATI[,4:30], shapiro.test) I will receive an output list of results that I will need to observe. The result suggests that the only normally distributed variable is COL_TOT, as the p-value is < 0.05. 7- HOMA-IR indicates insulin resistance. Do you expect that insulin resistance values will differ among diabetics? How could you test this hypothesis? In theory, I expect there to be differences in insulin resistance between diabetics and non-diabetics. There are essentially two ways to test this hypothesis: 1- I evaluate whether there are differences in the mean or median values of HOMA-IR between diabetics and non-diabetics using an independent samples t-test or a non-parametric test if the variable does not meet the assumptions for the t-test. 2- I evaluate whether there is a relationship between the continuous variable HOMA-IR and the binary variable DIABETE. In this case, I apply a regression model, and depending on whether the binary variable is considered dependent (i.e., Y) or independent (i.e., X), we speak of logistic regression (Y=DIABETE) or simple regression (DIABETE=X). The interpretation of the result will, of course, change depending on what is being evaluated. In particular, if DIABETE=X, I am evaluating how insulin resistance changes when moving from non-diabetics to diabetics. If DIABETE=Y, I am evaluating the risk of being diabetic versus not being diabetic for each unit change in HOMA-IR. I apply the three tests: t.test(DATI$'HOMA-IR'~DATI$DIABETE) Note the use of quotes in DATI$'HOMA-IR'. These quotes are essential because the word contains a hyphen, a symbol that could be interpreted by R as a minus. For this reason, when words contain special symbols, quotes are used to tell R "consider it as a single word." In the regression model, the command is: model<-glm(DATI$'HOMA-IR'~DATI$DIABETE, family="gaussian") summary(model) The result shows that the relationship between HOMA-IR and DIABETE is significant, and there is an increase of 3.82 units of HOMA-IR when moving from non-diabetic to diabetic. Note the result of the regression and that of the t-test. model<-glm(DATI$DIABETE~DATI$'HOMA-IR', family="binomial") summary(model) This result actually gives us information regarding the risk of being diabetic with respect to a unit increase in HOMA-IR. 8- A rapid test has been proposed to identify diabetic subjects. This test is cheap and fast and is called test_2, while test_1 consists of a laboratory measurement of glucose levels. Test_1 represents the gold standard. Calculate the sensitivity, specificity of test_2, and the predictive value of the test. The exercise involves comparing a field test with a validated laboratory test. I determine from the comparison of the two tests the false positives and false negatives, and then calculate sensitivity and specificity. To do this, I need to use the epiR package. install.packages("epiR")  library(epiR) The function is applied to a 2x2 table where I have indicated false positives and false negatives. This table can be obtained using the table() function. epi.tests(table(DATI$TEST_2_DIABETE,DATI$TEST_1_DIABETE)) 9- Save the dataset DATI in a tab-delimited file. I use the write.table() function. write.table(DATI,"filename.txt",sep="\t", col.names=TRUE,row.names=FALSE,quote=FALSE) Let's explain some elements of the command: write.table() is the function. In the first position, I put the name of the dataframe I want to write to a file. In quotes, I put the name I want to save the file with, and I can also add the destination path similarly to the command used to load files. Then, I use the tab symbol as the separator. Next, I indicate that the column names should be included, while I specify that the row names should not be included. In this case, I have not assigned any names to the rows, so R would simply number the rows (which would be an extra column in my file that I don't need). Finally, with the term quote=FALSE, I tell R that each individual cell value should not be enclosed in quotes. 10- Merge the two datasets: 1) dataset CENTENARI and BIOCHIMICA, and 2) dataset DATI, using pid as the row identifier. I use the merge() function. DATI_MERGE<-merge(DATI,CENTENARI_BIOCHIMICA,by.x="pid",by.y="pid",all=T) I create a new object DATI_MERGE, which is the merge of the two files. The two files have been integrated using pid as the identifier. 11- Create a graph to describe whether the median values of smoking (FUMO) are different from the median values of heart attack (INFARTO). Obviously, this is a trick question, and the answer is that it is not possible, and the question does not make sense. 12- Would you be able to estimate a probability measure of being diabetic for each unit increase in HOMA-IR values? Described in exercise 7.