Before begin..

Let’s load the SBP dataset.

dataset_sbp <- read.csv(file = "Inha/5_Lectures/2024/Advanced biostatistics/scripts/BTE3207_Advanced_Biostatistics/dataset/sbp_dataset_korea_2013-2014.csv")

head(dataset_sbp)

##   SEX BTH_G SBP DBP FBS DIS  BMI
## 1   1     1 116  78  94   4 16.6
## 2   1     1 100  60  79   4 22.3
## 3   1     1 100  60  87   4 21.9
## 4   1     1 111  70  72   4 20.2
## 5   1     1 120  80  98   4 20.0
## 6   1     1 115  79  95   4 23.1

The normal distribution

set.seed(1)
sim_100_5 <- rnorm(1000, mean = 100, sd = 5) #makes a data with 1000 elements having mean of 100 and sd of 5

hist(sim_100_5, 100, probability = T) #how does it look like?
curve(dnorm(x, mean=100, sd=5), 
      col="darkblue", lwd=2, add=TRUE, yaxt="n")

mean(sim_100_5)

## [1] 99.94176

sd(sim_100_5)

## [1] 5.174579

normal distribution

Let’s compare some data collected from real people (random 100 Koreans) with the normal distribution

set.seed(1)
sbp_10000 <- sample(dataset_sbp$SBP, 10000) %>%
        data.frame()

hist_sbp_10000 <- ggplot(data = sbp_10000) +
        geom_histogram(aes(x = .,
                           y =..density..),
                       bins = 15,
                       fill = "white",
                       col = "black") +
        ggtitle("Histogram of SBP of 10000 Koreans") +
        ylab("Probability") +
        xlab("SBP (mmHg)")

hist_sbp_10000

hist_sbp_10000 + 
        stat_function(fun = dnorm, args = list(mean = mean(sbp_10000$.), sd = sd(sbp_10000$.)))

sbp_10000$. %>% mean

## [1] 121.7429

sbp_10000$. %>% sd

## [1] 14.68246

sbp_10000$. %>% median

## [1] 120

z-score

How can we say the location of a person with 130 mmHg of SBP, in terms of distribution? We usually calculate z-score, to understand the location of one sample more easilty.

z_sbp_10000 <- (131-mean(sbp_10000$.))/sd(sbp_10000$.)

z_sbp_10000

## [1] 0.6304871

Here, the z-score of a person with SBP of 131 is 0.63.

Using this information, we cancalculate the percentile of him (assuming the SBP dataset is following the normal distribution)

pnorm(z_sbp_10000)

## [1] 0.735812

It’s percentile is 73.6%.

In other words,

29% of samples are having greater value than 130
71% of samples are having smaller SBP value than 130

Meanwhile, the actural 73.6th percentile of our dataset (sbp_10000) was 130. Why?

quantile(sbp_10000$., 0.736)

## 73.6% 
##   130

#Normalization

min-max normalization makes minimum vale to the 0 and maximum value to 1.

dataset_sbp$norm_SBP <- (dataset_sbp$SBP - min(dataset_sbp$SBP))/
        (max(dataset_sbp$SBP) - min(dataset_sbp$SBP))


ggplot(data = dataset_sbp) +
        geom_histogram(aes(x = SBP,
                           y = ..count..),
                       binwidth = 0.01,
                       fill = "white",
                       col = "black") +
        ggtitle("Histogram of SBP of 10000 Koreans") +
        ylab("Count") +
        theme_classic(base_size = 20, base_family = "serif") +
        xlab("SBP (mmHg)")

ggplot(data = dataset_sbp) +
        geom_histogram(aes(x = norm_SBP,
                           y = ..count..),
                       binwidth = 0.01,
                       fill = "white",
                       col = "black") +
        ggtitle("Histogram of Min-Max Normalized\nSBP of 10000 Koreans") +
        ylab("Count") +
        theme_classic(base_size = 20, base_family = "serif") +
        xlab("Normalized SBP (mmHg)")

#        stat_function(fun = dnorm, args = list(mean = mean(dataset_sbp$norm_SBP),
#                                               sd = sd(dataset_sbp$norm_SBP)))

Standardization

Standardization makes mean to 0 and SD into 1.

dataset_sbp$std_SBP <- (dataset_sbp$SBP - mean(dataset_sbp$SBP))/
        sd(dataset_sbp$SBP)

ggplot(data = dataset_sbp) +
        geom_histogram(aes(x = std_SBP,
                           y = ..count..),
                       binwidth = 0.3,
                       fill = "white",
                       col = "black") +
        ggtitle("Histogram of standardized\nSBP of 10000 Koreans") +
        ylab("Probability") +
        xlab("Standardized SBP (mmHg)") +
        theme_classic(base_family = "serif", base_size = 20)

        #stat_function(fun = dnorm, args = list(mean = mean(dataset_sbp$std_SBP), sd = sd(dataset_sbp$std_SBP)))

Skewed data and z-score

We can calculate z-score of skewed data, but their calculation does not match with our assumption “the data is approximately normal”, therefore it results in wrong predictions.

dataset_sbp$FBS %>%
        hist(probability = T, main = "Histogram of blood sugar level (FBS)", 15)

dataset_sbp$FBS %>% mean

## [1] 98.86443

dataset_sbp$FBS %>% sd

## [1] 22.9813

dataset_sbp$FBS %>% median

## [1] 94

So, if we use z-score to guess the 2.5% percentile’s FBS level,

(dataset_sbp$FBS %>% mean) - 2 * (dataset_sbp$FBS %>% sd)

## [1] 52.90183

However, in actual database,

quantile(dataset_sbp$FBS, 0.025)

## 2.5% 
##   74

Thee z-score cannot represent the actual dataset.

Bibliography

## Computing. R Foundation for Statistical Computing, Vienna, Austria. <https://www.R-project.org/>. We have invested a lot of time and effort in creating R, please cite it when using it for data analysis. See also 'citation("pkgname")' for citing R packages.
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## R. version 0.5.0. Buffalo, New York. http://github.com/trinker/pacman

COD_week4_2_MGK_BTE3207

Minsik Kim

2024-09-27