Why statistical analysis ?

• Draw valid conclusions from sample data

• Test significance and quantify uncertainty

• Compare groups and reveal relationships

• Ensure reproducibility and credible communication

Types of Data

flowchart LR
    A("Data") --> B("Quantitative")
    A --> C("Qualitative")
    B --> D("Continuous")
    B --> E("Discrete")
    C --> F("Nominal") 
    C --> G("Ordinal")
    C --> H("Binary")
    
    B:::Sky
    D:::Sky
    E:::Sky
    classDef Sky stroke-width:1px, stroke-dasharray:none, stroke:#374D7C, fill:#ABABAB, color:#374D7C

Types of Data

flowchart LR
    A("Data") --> B("Quantitative")
    A --> C("Qualitative")
    B --> D("Continuous")
    B --> E("Discrete")
    C --> F("Nominal") 
    C --> G("Ordinal")
    C --> H("Binary")
    
    C:::Sky
    F:::Sky
    G:::Sky
    H:::Sky
    classDef Sky stroke-width:1px, stroke-dasharray:none, stroke:#374D7C, fill:#ABABAB, color:#374D7C

Chi-square test for independence

In a chi-square test, the observed frequencies for two or more groups are compared with expected frequencies by chance

\[\chi^2 = \sum \frac {(O - E)^2}{E}\]

Where \(O\) is the observed frequency and \(E\) is the expected frequency

Chi-square test for independence in R

dat <- data.frame(
  "smoke_no" = c(17, 20),
  "smoke_yes" = c(32, 5),
  row.names = c("Athlete", "Non-athlete"),
  stringsAsFactors = FALSE
)
colnames(dat) <- c("Non-smoker", "Smoker")

dat

            Non-smoker Smoker
Athlete             17     32
Non-athlete         20      5

# View data
barplot(as.matrix(dat))

# Perform Chi-square test
chisq.test(dat)

    Pearson's Chi-squared test with Yates' continuity correction

data:  dat
X-squared = 11.84, df = 1, p-value = 0.0005797

Fisher’s Exact Test

Fisher’s exact test is a statistical test for categorical variables, used to determine if there’s a statistically significant association between two categorical variables, especially when dealing with small sample sizes.

dat <- data.frame(
  "smoke_no" = c(7, 0),
  "smoke_yes" = c(2, 5),
  row.names = c("Athlete", "Non-athlete"),
  stringsAsFactors = FALSE
)
colnames(dat) <- c("Non-smoker", "Smoker")

dat

            Non-smoker Smoker
Athlete              7      2
Non-athlete          0      5

# View data
barplot(as.matrix(dat))

# Perform Fisher's Exact Test
fisher.test(dat)

    Fisher's Exact Test for Count Data

data:  dat
p-value = 0.02098
alternative hypothesis: true odds ratio is not equal to 1
95 percent confidence interval:
 1.449481      Inf
sample estimates:
odds ratio 
       Inf