Statistical Hypothesis Testing

Author

Michał Janiga

1 Analysis of Mean Values for Medical Data

The file tempciala.txt contains recorded values of heart rate and body temperature for 65 men (gender = 1) and 65 women (gender = 2).

Separately for men and women:

  • estimate the mean and standard deviation of temperature,

  • verify at the significance level \(\alpha = 0.05\) the hypothesis that the mean temperature is equal to 36.6 \(^\circ\)C against the alternative hypothesis that the mean temperature is different, assuming that the temperatures have a normal distribution and the variance of the distribution is unknown.

Since the variance of the distribution is unknown as indicated in the task, it should be estimated based on the sample standard deviation. Student’s t-test provides a wider “width” of the normal distribution, allowing for safer inference with an unknown variance.

The t.test function requires the investigated characteristic to have a normal distribution. This is satisfied by the task description.

show/hide 
# Load data
temp_data <- read.csv("data/tempciala.txt")
names(temp_data) <- c("temperature", "gender", "heart_rate")

# Split into groups
men_temp <- temp_data$temperature[temp_data$gender == 1]
women_temp <- temp_data$temperature[temp_data$gender == 2]

# Parameter estimation
mean_men <- mean(men_temp)
sd_men <- sd(men_temp)
mean_women <- mean(women_temp)
sd_women <- sd(women_temp)

# Display estimation results
data.frame(
  Gender = c("Men", "Women"),
  Mean = c(mean_men, mean_women),
  Std_dev = c(sd_men, sd_women)
)
  Gender     Mean   Std_dev
1    Men 36.72615 0.3882158
2  Women 36.88923 0.4127359
show/hide 
# Hypothesis tests of hypothetical mean body temperature (H0: mu = 36.6)
# By default, a two-sided test is performed
# The confidence level resulting from the task is explicitly provided
men_test <- t.test(men_temp, conf.level = 0.95, mu = 36.6)
women_test <- t.test(women_temp, conf.level = 0.95, mu = 36.6)

men_test

    One Sample t-test

data:  men_temp
t = 2.6199, df = 64, p-value = 0.01097
alternative hypothesis: true mean is not equal to 36.6
95 percent confidence interval:
 36.62996 36.82235
sample estimates:
mean of x 
 36.72615
show/hide 
women_test

    One Sample t-test

data:  women_temp
t = 5.6497, df = 64, p-value = 3.985e-07
alternative hypothesis: true mean is not equal to 36.6
95 percent confidence interval:
 36.78696 36.99150
sample estimates:
mean of x 
 36.88923

1.1 Conclusions

For men, the estimated mean temperature is approx. 36.73 \(^\circ\)C, and the standard deviation is approx. 0.39 \(^\circ\)C. The p-value (0.011) is less than \(\alpha=0.05\), so we reject the null hypothesis \(H_0\) in favor of the alternative. The mean temperature in men differs significantly from 36.6 \(^\circ\)C.

For women, the estimated mean temperature is approx. 36.89 \(^\circ\)C, and the standard deviation is approx. 0.41 \(^\circ\)C. The p-value (3.985e-07) is significantly less than \(\alpha=0.05\), so we reject the null hypothesis \(H_0\) in favor of the alternative. The mean temperature in women also differs significantly from 36.6 \(^\circ\)C.

2 Investigation of Seasonality of Phenomena

The table below contains data regarding the number of suicides in the United States in 1970, broken down by individual months.

Month Number of suicides Number of days
January 1867 31
February 1789 28
March 1944 31
April 2094 30
May 2097 31
June 1981 30
July 1887 31
August 2024 31
September 1928 30
October 2032 31
November 1978 30
December 1859 31

Verify at the significance level \(\alpha = 0.05\) whether the presented data indicate a constant intensity of the investigated phenomenon, or rather suggest seasonal variation in the number of suicides. Assume that in the case of a constant intensity, the number of suicides in a given month is proportional to the number of days in that month.

show/hide 
# Data
suicides <- c(1867, 1789, 1944, 2094, 2097, 1981, 1887, 2024, 1928, 2032, 1978, 1859)
days_in_month <- c(31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31)

# Hypothesis: number of suicides proportional to the number of days
# Theoretical probabilities
theoretical_probs <- days_in_month / sum(days_in_month)

# Chi-squared goodness-of-fit test
seasonality_test <- chisq.test(x = suicides, p = theoretical_probs)
seasonality_test

    Chi-squared test for given probabilities

data:  suicides
X-squared = 47.365, df = 11, p-value = 1.852e-06

2.1 Conclusions

The p-value (1.852011e-06) is significantly less than the significance level \(\alpha=0.05\). This means we reject the hypothesis of a constant intensity of suicides proportional to the number of days in the month. The data indicate the presence of seasonal variation in the number of suicides.

3 Advanced Normality Tests of Financial Time Series

For a selected company listed on the Warsaw Stock Exchange (GPW), load data from the stooq.pl website, and then

  • calculate the percentage changes of the lowest prices on individual days over the last year, plot their histogram, and draw the probability density function of a normal distribution with parameters estimated based on their values,

  • applying various tests discussed in the examples, verify at the significance level \(\alpha = 0.05\) the hypothesis that the percentage changes of the lowest prices on individual days over the last year have a normal distribution.

show/hide 
library(moments)
library(nortest)

# Selected company: KGHM
if (file.exists(".env")) readRenviron(".env")
apikey <- Sys.getenv("STOOQ_API_KEY")
ticker <- 'KGH'
url <- paste0('https://stooq.pl/q/d/l/?s=', ticker, 
                 '&i=d', '&apikey=', apikey)
file_path <- paste0("data/", ticker, ".csv")
if (!file.exists(file_path)) {
  download.file(url, file_path, quiet = TRUE)
}
kgh_df <- read.csv(file_path)
names(kgh_df)[names(kgh_df) == "Data"] <- "Date"
names(kgh_df)[names(kgh_df) == "Najnizszy"] <- "Low"
kgh_df$Date <- as.Date(kgh_df$Date)
kgh_year_df <- kgh_df[which(kgh_df$Date >= (Sys.Date() - 365)),]
low_prices <- kgh_year_df$Low

# Calculation of percentage changes (vector method avoiding NA)
pct_changes <- diff(low_prices) / low_prices[-length(low_prices)] * 100

# Parameters for normal distribution
mu <- mean(pct_changes)
sigma <- sd(pct_changes)

# Histogram and density
hist(pct_changes, breaks = 30, probability = TRUE, 
     main = paste("Histogram of percentage changes:", ticker),
     xlab = "Percentage change [%]", ylab = "Density", col = "lightblue")
curve(dnorm(x, mean = mu, sd = sigma), add = TRUE, col = "red", lwd = 2)

show/hide 
# Normality tests
shapiro_test <- shapiro.test(pct_changes)
ks_test <- ks.test(pct_changes, "pnorm", mean = mu, sd = sigma)
# Tests from the "moments" library
ans_test <- anscombe.test(pct_changes)
ago_test <- agostino.test(pct_changes)
jb_test <- jarque.test(pct_changes)
lillie_test <- lillie.test(pct_changes)

shapiro_test

    Shapiro-Wilk normality test

data:  pct_changes
W = 0.96276, p-value = 0.01876
show/hide 
ks_test

    Exact one-sample Kolmogorov-Smirnov test

data:  pct_changes
D = 0.080729, p-value = 0.637
alternative hypothesis: two-sided
show/hide 
ans_test

    Anscombe-Glynn kurtosis test

data:  pct_changes
kurt = 5.4126, z = 2.9517, p-value = 0.003161
alternative hypothesis: kurtosis is not equal to 3
show/hide 
ago_test

    D'Agostino skewness test

data:  pct_changes
skew = -0.085543, z = -0.338092, p-value = 0.7353
alternative hypothesis: data have a skewness
show/hide 
jb_test

    Jarque-Bera Normality Test

data:  pct_changes
JB = 19.743, p-value = 5.163e-05
alternative hypothesis: greater
show/hide 
lillie_test

    Lilliefors (Kolmogorov-Smirnov) normality test

data:  pct_changes
D = 0.080729, p-value = 0.2151

3.1 Conclusions

Statistical analysis of the percentage changes of the KGH stock price exhibits classic characteristics of financial time series. The results of the tests at the significance level \(\alpha = 0.05\) are as follows:

  1. Rejection of normality: The Shapiro-Wilk test (\(p = 1.876e-02\)) and the Jarque-Bera test (\(p = 5.163e-05\)) unambiguously reject the null hypothesis \(H_0\).
  2. Cause of non-normality (leptokurtosis): The Anscombe-Glynn test shows high kurtosis at 5.41. The very low \(p = 3.161e-03\) provides statistical evidence for the presence of “heavy tails”.
  3. Skewness: D’Agostino’s test shows skewness at -0.086. The p-value \(p = 0.7353\) is greater than \(\alpha\), meaning there are no grounds to reject the hypothesis of distribution symmetry.
  4. KS test specificity: The Kolmogorov-Smirnov test (\(p = 0.637\)) is the only one that does not reject \(H_0\). This confirms its lower sensitivity to tail anomalies compared to moment-based tests (JB, Anscombe). The test measures the maximum distance between cumulative distribution functions, which in this case oscillates around the area of highest probability density, explaining the above statement. Furthermore, the KS test could be somewhat overly optimistic due to estimating parameters from the sample – it is less likely to reject the hypothesis of non-normality even when the data are non-normal.
  5. To “update” the KS test, an additional Lilliefors test was conducted using the “nortest” library. This test clearly indicates non-normality since \(p = 2.151e-01\) clearly rejects the null hypothesis \(H_0\).

Summary: Since key tests (Shapiro-Wilk, Jarque-Bera) returned \(p < \alpha\), we assume that the price change distribution is not normal. The dominant feature is leptokurtosis, which means that the risk of extreme events is higher than in the theoretical model.

4 Comparison of the Lifespan of Industrial Components

The file lozyska.txt contains the operating times (in millions of cycles) until failure of bearings made of two different materials.

  • Perform a test of no difference between the operating times of bearings made of different materials, assuming that the operating time until failure is described by a normal distribution, without assuming equal variances. Assume a significance level \(\alpha = 0.05\).

  • Perform an analogous test, without assuming normal distributions.

  • (optional) Estimate the probability that a bearing made of the first material will operate longer than a bearing made of the second material.

show/hide 
# Load data
bearing_data <- read.csv("data/lozyska.txt", quote = "'")
type1 <- bearing_data$Typ.I
type2 <- bearing_data$Typ.II

# 1. Welch's t-test (assuming normality, not assuming equal variances)
t_test <- t.test(type1, type2, var.equal = FALSE)
t_test

    Welch Two Sample t-test

data:  type1 and type2
t = 2.0723, df = 16.665, p-value = 0.05408
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
 -0.07752643  7.96352643
sample estimates:
mean of x mean of y 
   10.693     6.750
show/hide 
# 2. Wilcoxon test (without assuming normality)
wilcox_test <- wilcox.test(type1, type2)
wilcox_test

    Wilcoxon rank sum exact test

data:  type1 and type2
W = 75, p-value = 0.06301
alternative hypothesis: true location shift is not equal to 0
show/hide 
# 3. Estimating the probability P(X > Y)
# We can do this by checking all pairs (x_i, y_j)
comparison_matrix <- outer(type1, type2, ">")
p_hat <- mean(comparison_matrix)
p_hat
[1] 0.75

4.1 Conclusions

  • The result of Welch’s t-test (p-value = 0.0541) at the significance level \(\alpha=0.05\) provides no grounds to reject the hypothesis of no difference between the mean bearing operating times (p-value > 0.05).
  • The result of the Wilcoxon test (p-value = 0.063) also does not allow rejecting the hypothesis of no difference between the distributions (p-value > 0.05).

In both cases, we assume that there are no statistically significant differences in the operating times of bearings made of different materials.

Additionally: The estimated probability that a bearing made of the first material will operate longer than a bearing made of the second material is approximately 75%. Although Type I bearings operated longer than Type II bearings in 75% of cases, at the significance level \(\alpha = 0.05\), it was not possible to prove that this difference is permanent and reproducible. This suggests that Type I may be better in practice; however, to confirm this thesis with statistical certainty, further study with a larger sample size is required.

5 Analysis of Statistical Independence in Sports

Using data from the pl.fcstats.com website, verify the hypothesis of independence of home team results (wins, draws, and losses) from the country in which the football matches are played. Assume a significance level \(\alpha = 0.05\).

  • Perform the tests based on data regarding the following leagues:

    • German – Bundesliga (German league),

    • Polish – Ekstraklasa (Liga polska),

    • English – Premier League (English league),

    • Spanish – LaLiga (Liga hiszpańska).

  • The data can be found in the “Porównanie lig” (League comparison) -> “Zwycięzcy meczów” (Match winners) tab, in columns (excluding the [%] sign):

    • 1 – home wins, e.g., 125 for the German league (Bundesliga),

    • x – draws, e.g., 86 for the German league,

    • 2 – home losses, e.g., 95 for the German league.

show/hide 
# Preparation of contingency table (2025/2026 season)
# Rows: League, Columns: Result (win, draw, loss)
results <- matrix(c(
  125, 86, 95,   # Bundesliga
  108, 65, 67,   # Ekstraklasa
  193, 96, 91,  # Premier League
  194, 95, 91   # LaLiga
), nrow = 4, byrow = TRUE)

rownames(results) <- c("Bundesliga", "Ekstraklasa", "Premier League", "LaLiga")
colnames(results) <- c("1", "X", "2")

# Display table
results
                 1  X  2
Bundesliga     125 86 95
Ekstraklasa    108 65 67
Premier League 193 96 91
LaLiga         194 95 91
show/hide 
# Chi-squared test of independence
independence_test <- chisq.test(results)
independence_test

    Pearson's Chi-squared test

data:  results
X-squared = 10.325, df = 6, p-value = 0.1116

5.1 Conclusions

The p-value (0.1116) is greater than the significance level \(\alpha=0.05\). This means there are no grounds to reject the hypothesis of the independence of match results (wins, draws, and losses) from the country/league in which the matches are played. The 1, X, 2 statistics in the analyzed leagues are sufficiently close to each other to prevent finding significant differences.