T test相关

library(ggplot2)
library(tidyverse)

Report: One-Sample T-Test

Introduction:
This report aims to determine if the average mileage per gallon (mpg) for cars in the mtcars dataset significantly differs from a hypothetical average of 20 mpg.

Data Overview and Cleaning:

# Load dataset
data(mtcars)

# Check for missing values
if (any(is.na(mtcars))) {
  print("Missing values found.")
} else {
  print("No missing values found.")
}

# Check data types
str(mtcars)

Data Transformation and Visualization:

# No transformation needed for this analysis

# Visualize the distribution of mpghist
hist(mtcars$mpg)

Analysis Method Selection:
A one-sample t-test is selected to compare the mean mpg against the hypothetical mean, as the data is continuous and approximately normally distributed.

Certainly! Here's a complete report section for the assumption testing using the Shapiro-Wilk normality test on the mtcars$mpg dataset:

Assumption Testing Report: Shapiro-Wilk Normality Test for `mtcars$mpg`

Objective:
To assess whether the miles per gallon (mpg) values in the mtcars dataset are normally distributed, which is a key assumption for many parametric statistical tests.

Hypotheses Formulation:

Null Hypothesis (H0): The data are normally distributed.
Alternative Hypothesis (H1): The data are not normally distributed.

Test Execution:

# Load the mtcars dataset, which is a built-in dataset in R
data(mtcars)

# Conduct Shapiro-Wilk test for normality on the mpg variable
shapiro_test <- shapiro.test(mtcars$mpg)

# Output the test results
print(shapiro_test)

Shapiro-Wilk normality test

data: mtcars$mpg
W = 0.94756, p-value = 0.1229

Results:

Test Statistic (W): 0.94756
P-value: 0.1229

Assessment of Assumptions:

The test statistic (W) is close to 1, which suggests that the data are likely to be normally distributed.
The p-value (0.1229) is greater than the conventional threshold of 0.05.

Conclusion:
Given that the p-value exceeds the significance level of 0.05, we fail to reject the null hypothesis. This indicates that there is no significant evidence to suggest that the mpg values in the mtcars dataset are not normally distributed. Therefore, the assumption of normality appears to be satisfied, and parametric tests can be appropriately used for further analysis of the mpg data.

Hypotheses Formulation:

Null hypothesis (H0): The true mean mpg is equal to 20.
Alternative hypothesis (H1): The true mean mpg is not equal to 20.

Test Execution and Visualization:

# Execute the one-sample t-test
t_test_result <- t.test(mtcars$mpg, mu=20)
print(t_test_result)

One Sample t-test

data: mtcars$mpg
t = 0.08506, df = 31, p-value = 0.9328
alternative hypothesis: true mean is not equal to 20
95 percent confidence interval:
17.91768 22.26357
sample estimates:
mean of x
20.09062

Results Interpretation:
Given the p-value of 0.9328, which is greater than the significance level of 0.05, we fail to reject the null hypothesis. This means that there is no statistically significant evidence to suggest that the true mean mpg of cars in the mtcars dataset is different from the hypothesized mean of 20 mpg. The confidence interval also includes the hypothesized mean, further supporting the conclusion of no significant difference.

The results suggest that, based on the data from the mtcars dataset, the average mpg of cars is not statistically different from 20 mpg when considering a 95% confidence level. However, it is important to note that this does not imply that the true mean is exactly 20; rather, it indicates that the data do not provide sufficient evidence to conclude a difference from 20.

Okay, got it. Here's the step-by-step analysis template for comparing two paired datasets, where you want to test if dataset A is greater than dataset B, using a paired t-test:

Report: Paired T-Test - Comparing Paired Datasets A and B (A > B)

Introduction:
This report aims to determine if the mean of dataset A is significantly greater than the mean of dataset B, for paired observations.

Data Overview and Cleaning:

# Load the paired datasets
data(dataset_A)
data(dataset_B)

# Check for missing values
if (any(is.na(dataset_A)) || any(is.na(dataset_B))) {
  print("Missing values found.")
} else {
  print("No missing values found.")
}

# Check data types
str(dataset_A)
str(dataset_B)

Data Transformation and Visualization:

# Ensure the datasets have the same length and are paired
if (length(dataset_A) != length(dataset_B)) {
  stop("Datasets A and B must have the same length.")
}

# Visualize the distributions of the paired differences
hist(dataset_A - dataset_B)

Assumption Testing: Normality of Differences

# Conduct Shapiro-Wilk test for normality on the differences
shapiro_test <- shapiro.test(dataset_A - dataset_B)
print(shapiro_test)

Hypotheses Formulation:

Null hypothesis (H0): The mean difference between dataset A and dataset B is less than or equal to 0.
Alternative hypothesis (H1): The mean difference between dataset A and dataset B is greater than 0.

Test Execution and Visualization:

# Execute the one-tailed paired t-test (A > B)
t_test_result <- t.test(dataset_A, dataset_B, paired = TRUE, alternative = "greater")#alternative = c("two.sided", "less", "greater")
print(t_test_result)

Results Interpretation:
Based on the p-value obtained from the one-tailed paired t-test, we can determine whether to reject or fail to reject the null hypothesis at the chosen significance level (e.g., 0.05). The interpretation should discuss the statistical significance of the difference and the practical implications, if any.

If the p-value is less than the significance level, we can conclude that the mean of dataset A is significantly greater than the mean of dataset B. The confidence interval should also be examined to quantify the magnitude of the difference.

Remember to consider the assumptions of the paired t-test, such as the normality of the differences, and discuss any violations or limitations in the interpretation.

Report: Wilcoxon Signed-Rank Test

Introduction:
This report assesses whether there is a significant difference in tooth growth before and after treatment using the ToothGrowth dataset.

Data Overview and Cleaning:

# Load dataset
data(ToothGrowth)

# Check for missing values
if (any(is.na(ToothGrowth))) {
  print("Missing values found.")
} else {
  print("No missing values found.")
}

# Check data types
ToothGrowth

After shapiro test, the datasets do not fits the normla distribution. then we do wilcox test.

The hypotheses

Null hypothesis (H0): The length of the two groups are equal to each other.
Alternative hypothesis (H1): The length of the two groups are not equal to each other.

Data Transformation and Visualization:


# Load the ToothGrowth dataset
data(ToothGrowth)

# Separate the data by supplement type
tooth_growth_vc <- ToothGrowth[ToothGrowth$supp == "VC", ]
tooth_growth_oj <- ToothGrowth[ToothGrowth$supp == "OJ", ]

wilcox.test(tooth_growth_vc$len, tooth_growth_oj$len, paired = FALSE,exact=FALSE)

Wilcoxon rank sum test with continuity correction

data: tooth_growth_vc$len and tooth_growth_oj$len
W = 324.5, p-value = 0.06449
alternative hypothesis: true location shift is not equal to 0

Analysis Method Selection:
The Wilcoxon signed-rank test is selected for paired samples, as the data is not normally distributed based on the Shapiro-Wilk test.

Results Interpretation:
The p-value obtained from the Wilcoxon signed-rank test is 0.06449, then we cannot reject the null hypothesis, therefore
this means that there is no statistically significant evidence to suggest that the length of the two length are not equal to each other.

中文版本：
当然，以下是整个报告的完整中文翻译：

报告：单样本T检验

引言：
本报告旨在确定mtcars数据集中的汽车每加仑英里数（mpg）平均值是否显著不同于假设的平均值20 mpg。

数据概览和清洗：

# 加载数据集
data(mtcars)

# 检查缺失值
if (any(is.na(mtcars))) {
  print("发现缺失值。")
} else {
  print("未发现缺失值。")
}

# 查看数据类型
str(mtcars)

数据转换和可视化：

# 无需转换此分析

# 可视化mpg的分布
hist(mtcars$mpg)

分析方法选择：
选择单样本T检验来比较样本均值与假设均值，因为数据是连续的且大致呈正态分布。

假设检验报告：Shapiro-Wilk正态性检验`mtcars$mpg`

目标：
评估mtcars数据集中的每加仑英里数（mpg）值是否呈正态分布，这是许多参数统计检验的关键假设。

假设提出：

零假设 (H0)： 数据呈正态分布。
备择假设 (H1)： 数据不呈正态分布。

检验执行：

# 加载mtcars数据集，这是R内置的数据集
data(mtcars)

# 对mpg变量进行Shapiro-Wilk正态性检验
shapiro_test <- shapiro.test(mtcars$mpg)

# 输出检验结果
print(shapiro_test)

结果：

检验统计量 (W): 0.94756
P值: 0.1229

假设评估：

检验统计量 (W) 接近1，表明数据可能呈正态分布。
P值 (0.1229) 大于常规阈值0.05。

结论：
由于P值超过显著性水平0.05，我们未能拒绝零假设。这表明没有显著证据表明mtcars数据集中的mpg值不是正态分布。因此，正态性假设似乎得到满足，参数检验可以用于进一步分析mpg数据。

假设提出：

零假设 (H0): 真实均值mpg等于20。
备择假设 (H1): 真实均值mpg不等于20。

检验执行和可视化：

# 执行单样本T检验
t_test_result <- t.test(mtcars$mpg, mu=20)
print(t_test_result)

结果解释：
鉴于P值为0.9328，大于显著性水平0.05，我们未能拒绝零假设。这意味着没有统计学上显著的证据表明mtcars数据集中汽车的真实均值mpg与假设均值20 mpg不同。置信区间也包括了假设均值，进一步支持了没有显著差异的结论。

研究结果表明，基于mtcars数据集的数据，汽车的平均mpg在95%置信水平下与20 mpg没有统计学上的显著差异。然而，重要的是要注意，这并不意味着真实均值恰好为20；相反，它表明数据没有提供足够的证据来得出与20不同的结论。

非常抱歉刚才的疏漏，现在我将完成剩余部分的翻译。

报告：配对T检验 - 比较配对数据集A和B（A > B）

引言：
本报告旨在确定数据集A的平均值是否显著大于数据集B的平均值，针对配对观测值。

数据概览和清洗：

# 加载配对数据集
data(dataset_A)
data(dataset_B)

# 检查缺失值
if (any(is.na(dataset_A)) || any(is.na(dataset_B))) {
  print("发现缺失值。")
} else {
  print("未发现缺失值。")
}

# 查看数据类型
str(dataset_A)
str(dataset_B)

数据转换和可视化：

# 确保数据集具有相同的长度且是配对的
if (length(dataset_A) != length(dataset_B)) {
  stop("数据集A和B必须具有相同的长度。")
}

# 可视化配对差异的分布
hist(dataset_A - dataset_B)

假设检验：差异的正态性

# 对差异进行Shapiro-Wilk正态性检验
shapiro_test <- shapiro.test(dataset_A - dataset_B)
print(shapiro_test)

假设提出：

零假设 (H0): 数据集A和数据集B之间的平均差异小于或等于0。
备择假设 (H1): 数据集A和数据集B之间的平均差异大于0。

检验执行和可视化：

# 执行单尾配对T检验（A > B）
t_test_result <- t.test(dataset_A, dataset_B, paired = TRUE, alternative = "greater")
print(t_test_result)

结果解释：
基于单尾配对T检验得到的P值，我们可以确定是否在选定的显著性水平（例如0.05）拒绝或不拒绝零假设。解释应讨论差异的统计显著性和实际意义（如果有的话）。

如果P值小于显著性水平，我们可以得出结论，数据集A的平均值显著大于数据集B的平均值。还应检查置信区间以量化差异的大小。

记得考虑配对T检验的假设，例如差异的正态性，并在解释中讨论任何违规或限制。

报告：Wilcoxon符号秩检验

引言：
本报告评估使用ToothGrowth数据集，通过补充不同物质前后的牙齿生长长度是否存在显著差异。

数据概览和清洗：

# 加载数据集
data(ToothGrowth)

# 检查缺失值
if (any(is.na(ToothGrowth))) {
  print("发现缺失值。")
} else {
  print("未发现缺失值。")
}

# 查看数据类型
ToothGrowth

在Shapiro-Wilk正态性检验后，数据集不符合正态分布。然后我们进行Wilcoxon检验。

假设提出：

零假设 (H0): 两组的长度彼此相等。
备择假设 (H1): 两组的长度彼此不等。

数据转换和可视化：

# 加载ToothGrowth数据集
data(ToothGrowth)

# 按补充类型分离数据
tooth_growth_vc <- ToothGrowth[ToothGrowth$supp == "VC", ]
tooth_growth_oj <- ToothGrowth[ToothGrowth$supp == "OJ", ]

# 执行Wilcoxon符号秩检验
wilcox.test(tooth_growth_vc$len, tooth_growth_oj$len, paired = FALSE, exact=FALSE)

分析方法选择：
基于Shapiro-Wilk检验数据不呈正态分布，选择Wilcoxon符号秩检验用于配对样本。

结果解释：
从Wilcoxon符号秩检验得到的P值为0.06449，因此我们不能拒绝零假设，这意味着没有统计学上的显著证据表明两组的长度彼此不等。

以上就是整个报告的中文翻译。

posted @ 2024-05-22 17:04 chen生信阅读(88) 评论(0) 收藏举报

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chen-heybro

T test相关

Report: One-Sample T-Test

Assumption Testing Report: Shapiro-Wilk Normality Test for `mtcars$mpg`

Report: Paired T-Test - Comparing Paired Datasets A and B (A > B)

Report: Wilcoxon Signed-Rank Test

报告：单样本T检验

假设检验报告：Shapiro-Wilk正态性检验`mtcars$mpg`

报告：配对T检验 - 比较配对数据集A和B（A > B）

报告：Wilcoxon符号秩检验

公告

chen-heybro

T test相关

Report: One-Sample T-Test

Assumption Testing Report: Shapiro-Wilk Normality Test for mtcars$mpg

Report: Paired T-Test - Comparing Paired Datasets A and B (A > B)

Report: Wilcoxon Signed-Rank Test

报告：单样本T检验

假设检验报告：Shapiro-Wilk正态性检验mtcars$mpg

报告：配对T检验 - 比较配对数据集A和B（A > B）

报告：Wilcoxon符号秩检验

公告

Assumption Testing Report: Shapiro-Wilk Normality Test for `mtcars$mpg`

假设检验报告：Shapiro-Wilk正态性检验`mtcars$mpg`