Answer 1 - (Woodward 1.3).

The first step could be a random selection of a sample of subjects. A sample size of 100 could be large enough. At the beginning, I would measure the indications of all the medical end-points at stake. Then, ask each participant to record and report the amount of alcohol the consume per week. Given the experiment is talikg place over a full year, it would be optimific to re-take the measures of indication of the al the medical end-points every four months whci would result in three repeated measurements. Measurementr can also be collected every three months to generate four in total. With these data, longituginal analyses can show how the medical end-points change as a function of the changes in the amountof alcohol intake.

Answer 1.4 - (Woodward 1.4)

A meta analytic method would be appropriate to investigate the effect of white coat hypertension in the two suggested studies. One way to proceed is to collect results from comparative studies that involed cases with white coat hypertension, along with results from comparative studies that did not involve any case of white coat hypertension. The effect of WCH will then be captured by the discrepancies between the two sets of results.

Preparing the data

vitamina <- read.csv(file.choose(), header = T, sep = ",")
library(dplyr) 
## 
## Attaching package: 'dplyr'
## The following objects are masked from 'package:stats':
## 
##     filter, lag
## The following objects are masked from 'package:base':
## 
##     intersect, setdiff, setequal, union
head(vitamina, 3)
##   Retinol
## 1    1.15
## 2    1.36
## 3    0.38
vitamina %>% summarize(
  Mean.Retinol = mean(Retinol),
  Median.Retinol = median(Retinol),
  Standard.dev.Retinol = sd(Retinol),
  mini.retinol = min(Retinol),
  maxi.retinol = max(Retinol)
)
##   Mean.Retinol Median.Retinol Standard.dev.Retinol mini.retinol
## 1      0.76475           0.76            0.3949423         0.24
##   maxi.retinol
## 1          1.9

Answer 3 – The histogram of Retinol

hist(vitamina$Retinol, main = "Histogram", xlab = "Retinol levels", xlim = c(0.2, 2))

Answer 4 - The sample mean retinol level is 0.765

Answer 5 - The median retinol level is 0.76

Answer 6 - The standard deviation of the retinol levels is 0.395

Answer 7 -

library(sm)
## Package 'sm', version 2.2-5.4: type help(sm) for summary information
sm.density(vitamina$Retinol, xlab = "Retinol levels")

The distribution of the retinol levels is bimodal at around 0.25 and 1. It is skewed to the right, and it ranges overall from 0.24 to 1.9.