Detailed Notes on Paired Samples T-test and Related Statistical Concepts
PH102: Biostatistics in Public Health - Paired Samples T-test
One Sample T Test Review
Definition: The One Sample T Test is utilized to assess whether the mean of a continuous random variable (R.V.) from one population is equal to a hypothesized value, denoted as $$.
Example of One Sample T Test
Question: Is the mean juice consumption in 1997 the same as the known mean juice consumption of 1975 with $ = 4.5$ oz/day?
Comparing Mean Between Two Populations
Test Selection Criteria: The choice of statistical test depends on:
Relationship between the populations:
Are the two populations related (dependent)?
Are the two populations unrelated (independent)?
Nature of the variable:
Is the question regarding a continuous variable (means) or a binary variable (proportions)?
Population Standard Deviation Status:
Independent means lead to the use of:
Z Test for a population mean if standard deviation is known.
One Sample t-Test if standard deviation is not known.
Dependent means lead to:
Paired Sample t-Test if samples are paired.
Two Independent Samples t-Test if samples are independent.
Paired Samples Design
Characteristics: Pre- and post-intervention studies involve measuring a random variable before and after treatment on the same individuals, creating effective paired designs.
Example:
Measure plasma cholesterol of the same random sample of individuals before and after a specified treatment, such as eating avocados for six months.
Statistical Power: Paired samples are considered statistically powerful since utilizing the same participants minimizes variability caused by extraneous factors.
Paired Samples Design Hypothesis Set-Up
Example of Pre- vs. Post-Treatment Studies: The hypothesis posits that treatment alters a continuous variable outcome:
Sample 1: Pre-Treatment
Sample 2: Post-Treatment
Null Hypothesis Overview: Can be stated in three equivalent ways:
$H0: {pre-treatment} = _{post-treatment}$
$H0: {post-treatment} - _{pre-treatment} = 0$
$H0: d = 0$ where $d = {post-treatment} - _{pre-treatment}$
Alternative Hypothesis (Ha): Opposite statements to the null hypothesis based on the defined parameters.
Paired Samples t Test Procedure
Utilizes the same operational approach as a One Sample t Test:
Null Hypothesis: $H0: d = 0$ vs $Ha: d
eq 0$
A new variable, $d$, can be created as $( ext{post-treatment} – ext{pre-treatment})$, upon which the One Sample t Test is run.
Practical Example: Hypertension Study
Study Overview:
Dataset: Hypertension.sav with $n=409$ subjects diagnosed and treated for hypertension.
Objective: To compare systolic blood pressure before and after treatment with baseline SBP and follow-up SBP representing Sample 1 and Sample 2 respectively.
Null Hypothesis: No change in mean SBP between baseline and follow-up:
$H0: {baseline} = {follow-up}$ vs $Ha: {baseline} eq {follow-up}$
Equivalently: $H0: d = 0$ vs. $Ha: d
eq 0$ with $d = {follow-up} - _{baseline}$
Data Presentation: Summary Statistics for Hypertension Study
Descriptive Statistics Output Overview:
Variables:
SBP.Baseline (409 samples, Mean: 144.70, SD: 23.467)
SBP.Followup (409 samples, Mean: 135.74, SD: 20.240)
SBP.Change (Mean: -8.95, SD: 23.23938)
Mean SBP Change Calculation: ($ ext{follow-up} - ext{baseline} = -8.95$ mmHg).
Testing Procedures
Null Hypothesis: Mean change in SBP equals zero:
$H0: d = 0$ vs $Ha: d
eq 0$, where $d = {baseline}-_{follow-up}$.
Confidence Interval Calculation
95% Confidence Interval for $_d$: Based on our analysis, the 95% CI for mean change in SBP was calculated as: $(-11.21, -6.69)$ mmHg, indicating that since zero is not included in the interval, the Null Hypothesis is rejected, concluding a significant decrease in SBP from baseline to follow-up.
Calculating the T-Statistic
T-statistic Formula: $ ext{t}_{stat}$:
Formula: $ ext{t}{stat} = rac{ar{x} - 0}{ rac{s_d}{ ext{sqrt}(n)}}$ where:
$ar{x}$ is the mean difference.
$s_d$ is standard deviation of the differences.
$n$ is the number of observations.
Example Calculation with Data
For Hypertension Study:
$ar{x} = -8.95$ mmHg, $s_d = 23.24$ mmHg, $n = 409$
Calculate: $ ext{t}_{stat} = rac{-8.95 - 0}{ rac{23.24}{ ext{sqrt}(409)}} = -7.79$.
Critical Value Determination
Determining Critical Values: Determine critical t-values from the t-distribution table for your specific $ ext{df}$ (degrees of freedom) and p-value criteria.
P-value Assessment
After obtaining the t-statistic, the associated p-value must be determined to conclude whether to reject the Null Hypothesis or not, based on standard thresholds (e.g., $ ext{p} < 0.05$ denotes significance).
Summary of Assumptions for Paired Samples t-Test
The random variable under evaluation must be continuous.
Normality: The distribution of differences between pairs should be approximately normal; alternatively, if the sample size is large enough (n > 30), the Central Limit Theorem allows for normality.
Related Groups: Samples must consist of matched pairs or be related in some way.
Using SPSS for Paired Samples T-Test
SPSS Process Overview:
One-sample t-test for differences or paired sample t-test selections.
Calculating differences via the Transform function to create a new variable linking SBP changes.
Analyzing results through Analyze > Compare Means for either perspective.
Exercise Proposal
Task: Assume normality in a dataset; conduct Paired Sample t-Test using SPSS and perform hand calculation verifications for hypothesis testing in comparable drug efficacy for sleep induction.