Paired t-Test Principles, Calculations, and Software Analysis

Foundations and Principles of the Paired t-Test

• Conceptual Definition:

  • A paired $t$-test is a specific application and special case of the single-sample $t$-test.

  • Instead of evaluating two entirely independent groups, the test evaluates two sets of measurements collected from matched samples or paired observations.

  • In most research configurations, measurements are taken on the exact same experimental unit under two distinct conditions or points in time.

  • Common experimental scenarios include:

  • Before-and-after studies (evaluating an intervention on the same subject).

  • Diurnal comparisons (such as comparing daytime measurements against nighttime measurements on the same subject).

• Reduction to a Single-Sample Framework:

  • While frequently treated as an independent statistical procedure, the paired $t$-test mathematically converts paired bivariate data into a single sample of univariate differences.

  • The core variable of interest is not the raw measurement in condition $1$ or condition $2$, but rather the calculated difference ($\text{DIFF}$) between the paired observations for each individual unit.

  • Once these differences are determined, the analysis proceeds identically to a standard one-sample $t$-test performed on the newly derived sample of differences.

• Underlying Assumptions:

  • Random sampling: The pairs of observations must represent a random sample drawn from the target population.

  • Normality of differences: The sample of differences ($\text{DIFF}$) must be normally distributed in the underlying population.

Experimental Design and Stair-Climbing Case Study

• Research Question:

  • The objective is to determine whether physical exertion affects cardiovascular response, specifically asking: Does climbing stairs affect heart rate?

  • Although a directional hypothesis (e.g., climbing stairs increases heart rate) is biologically plausible and corresponds to a one-tailed test, the analysis is initially framed as a non-directional, two-tailed inquiry (does climbing stairs affect heart rate in either direction).

• Experimental Protocol:

  • A cohort of $n = 8$ subjects has their resting heart rates measured before physical activity.

  • Each subject climbs three flights of stairs.

  • Immediately following the stair climb, each subject's heart rate is measured a second time.

• Baseline Data Matrix:

  • Subject $1$ (Bob): Heart rate before climbing = $62$, Heart rate after climbing = $73$.

  • Subject $2$ (Sally): Heart rate before climbing = $58$, Heart rate after climbing = $66$.

  • Subject $3$ (Meghan): Heart rate before climbing = $70$, Heart rate after climbing = $72$.

  • Subject $4$ (Bill): Heart rate before climbing = $75$, Heart rate after climbing = $75$.

  • Subject $5$ (Harriet): Heart rate before climbing = $64$, Heart rate after climbing = $78$.

  • Subject $6$ (Mary): Heart rate before climbing = $60$, Heart rate after climbing = $59$.

  • Subject $7$ (Carl): Heart rate before climbing = $75$, Heart rate after climbing = $81$.

  • Subject $8$ (Sid): Heart rate before climbing = $58$, Heart rate after climbing = $62$.

Formulation of Hypotheses and Test Setup

• Defining the Difference Variable:

  • The difference metric is explicitly defined as:     $\text{DIFF} = (\text{Heart Rate After}) - (\text{Heart Rate Before})$

  • Under this directional subtraction:

  • A positive difference ($\text{DIFF} > 0$) indicates an increase in heart rate following stair climbing.

  • A negative difference ($\text{DIFF} < 0$) indicates a decrease in heart rate following stair climbing.

  • A difference of zero ($\text{DIFF} = 0$) indicates no change in heart rate.

• Statistical Hypotheses:

  • Null Hypothesis ($H_0$): There is no true difference in mean heart rate before and after climbing stairs across the population.     $H_0: \mu_d = 0$

  • Alternative Hypothesis ($H_a$ or $H_1$): There is a statistically significant difference in mean heart rate before and after climbing stairs.     $H_a: \mu_d \neq 0$

  • Here, $\mu_d$ represents the population mean of the paired differences.

• Significance Level and Tail Configuration:

  • The significance threshold is set at $\alpha = 0.05$.

  • Because the alternative hypothesis tests for any change ($\mu_d \neq 0$), the test is two-tailed, splitting the rejection region into two equal tails of area $\frac{\alpha}{2} = 0.025$ each.

• General Forms of Hypothesis Tests for Means:

  • Two-tailed test:     $H_0: \mu = \mu_0$     $H_a: \mu \neq \mu_0$     Rejection regions are placed symmetrically in both tails beyond $-t^*$ and $+t^*$.

• Right-tailed (upper-tail) test:   $H_0: \mu \le \mu_0$   $H_a: \mu > \mu_0$   Rejection region is located entirely in the right tail beyond $+t^*$.

• Left-tailed (lower-tail) test:   $H_0: \mu \ge \mu_0$   $H_a: \mu < \mu_0$   Rejection region is located entirely in the left tail beyond $-t^*$.

Critical Values, Decision Rules, and Sampling Distribution

• Degrees of Freedom:

  • The degrees of freedom ($df$) for a paired $t$-test are determined by the number of paired differences minus one:     $df = n - 1 = 8 - 1 = 7$

• Critical Value Identification:

  • Evaluating the Student's $t$-distribution table with $df = 7$ and an upper tail probability of $0.025$ (corresponding to a two-tailed confidence level of $95\%$ or $\alpha = 0.05$):     $t^* = 2.365$

  • Because the test is two-tailed, the two critical thresholds are:     $-t^* = -2.365$     $+t^* = +2.365$

• Statistical Decision Criteria:

  • If the calculated test statistic satisfies $t < -2.365$ or $t > 2.365$, the test statistic falls into the rejection region.

  • Falling in the critical region indicates that the probability of observing such extreme differences under the assumption of a true null hypothesis is less than $5\%$ ($p < 0.05$).

  • Decision rule: Reject $H_0$ in favor of $H_a$ if $|t| > 2.365$. Fail to reject $H_0$ if $-2.365 \le t \le 2.365$.

Manual Step-by-Step Mathematical Calculation

• Test Statistic Formula:

  • The general test statistic for a sample of differences is:     $t = \frac{\bar{d} - \mu_0}{\frac{s_d}{\sqrt{n}}}$

  • Where:

  • $\bar{d}$ is the sample mean of the differences.

  • $\mu_0$ is the hypothesized population difference under the null hypothesis (typically $\mu_0 = 0$).

  • $s_d$ is the sample standard deviation of the differences.

  • $n$ is the total number of paired units ($n = 8$).

  • $\frac{s_d}{\sqrt{n}}$ represents the estimated standard error of the mean difference ($SE_{\bar{d}}$).

  • Setting $\mu_0 = 0$, the formula reduces to:     $t = \frac{\bar{d}}{\frac{s_d}{\sqrt{n}}}$

• Subject-Level Differences:

  • Bob: $\text{DIFF} = 73 - 62 = 9$ (recorded manually as $9$)

  • Sally: $\text{DIFF} = 66 - 58 = 8$

  • Meghan: $\text{DIFF} = 72 - 70 = 2$

  • Bill: $\text{DIFF} = 75 - 75 = 0$

  • Harriet: $\text{DIFF} = 78 - 64 = 14$

  • Mary: $\text{DIFF} = 59 - 60 = -1$

  • Carl: $\text{DIFF} = 81 - 75 = 6$

  • Sid: $\text{DIFF} = 62 - 58 = 4$

• Manual Sample Summary Statistics:

  • Sample size: $n = 8$

  • Sample mean difference: $\bar{d} = 5.25$

  • Sample standard deviation of differences: $s_d = 5.04$

• Evaluation of the Test Statistic:

  • Standard error denominator:     $SE_{\bar{d}} = \frac{5.04}{\sqrt{8}}$

  • Substitution into the $t$-ratio:     $t = \frac{5.25}{\frac{5.04}{\sqrt{8}}} = 2.94$

• Hypothesis Decision:

  • Comparing the calculated statistic to the critical value:     $t = 2.94 > 2.365$

  • The test statistic falls deep within the upper rejection tail.

  • Conclusion: Reject $H_0$. There is statistically significant evidence at the $\alpha = 0.05$ level that climbing stairs affects heart rate.

Implementation and Analysis in Microsoft Excel

• Data Arrangement:

  • Column A: Subject identifiers (Name), rows $2$ through $9$.

  • Column B: Baseline measurements (Before), rows $2$ through $9$.

  • Column C: Post-intervention measurements (After), rows $2$ through $9$.

• Software Navigation:

  • Navigate through the Excel ribbon to the Data tab.

  • Within the Analysis group, select Data Analysis.

  • From the list of analysis tools, choose t-Test: Paired Two Sample for Means and click OK.

• Parameter Configuration:

  • Variable 1 Range: $C$1:$C$9 (corresponding to the After column, including header label).

  • Variable 2 Range: $B$1:$B$9 (corresponding to the Before column, including header label).

  • Hypothesized Mean Difference: leave blank or enter 0.

  • Labels: select the checkbox (indicating row $1$ contains column headers).

  • Alpha: 0.05.

  • Output options: select New Worksheet Ply and title it Paired t-test results.

• Excel Computational Output:

  • Mean: After = $70.75$, Before = $65.25$

  • Variance: After = $59.92857$, Before = $51.07143$

  • Observations: After = $8$, Before = $8$

  • Pearson Correlation: $0.75014$

  • Hypothesized Mean Difference: $0$

  • df: $7$

  • t Stat: $2.939874$

  • P(T<=t) one-tail: $0.010859$

  • t Critical one-tail: $1.894579$

  • P(T<=t) two-tail: $0.021717$

  • t Critical two-tail: $2.364624$

• Interpretation of Excel Output:

  • The two-tailed $p$-value is $0.021717$.

  • Because $p = 0.021717 < 0.05$, the null hypothesis is rejected.

  • The calculated $t$-statistic of $2.939874$ exceeds the critical threshold of $2.364624$.

Implementation and Analysis in Minitab

• Worksheet Setup:

  • Column C1-T: Name (Bob, Sally, Meghan, Bill, Harriet, Mary, Carl, Sid).

  • Column C2: Before ($62$, $58$, $70$, $75$, $64$, $60$, $75$, $58$).

  • Column C3: After ($73$, $66$, $72$, $75$, $78$, $59$, $81$, $62$).

• Menu Pathway and Dialog Options:

  • Menu path: Stat -> Basic Statistics -> Paired t....

  • In the primary dialog window:

  • Choose the radio button Samples in columns.

  • Set First sample: After.

  • Set Second sample: Before.

  • (Minitab evaluates the paired test as first sample minus second sample).

  • Click the Options... sub-menu:

  • Confidence level: 95.0.

  • Test mean: 0.0.

  • Alternative: select not equal.

  • Confirm by selecting OK in both dialogs.

• Minitab Session Window Output:

  • Paired $T$ for After - Before:

  • After: $N = 8$, $\text{Mean} = 70.7500$, $\text{StDev} = 7.7414$, $\text{SE Mean} = 2.7370$

  • Before: $N = 8$, $\text{Mean} = 65.2500$, $\text{StDev} = 7.1464$, $\text{SE Mean} = 2.5266$

  • Difference: $N = 8$, $\text{Mean} = 5.50000$, $\text{StDev} = 5.29150$, $\text{SE Mean} = 1.87083$

  • $95\%$ Confidence Interval for mean difference:     $(1.07619, 9.92381)$

  • $T$-Test of mean difference = $0$ (vs not = $0$):

  • $T\text{-Value} = 2.94$

  • $P\text{-Value} = 0.022$

Graphical Interpretation and Final Conclusions

• Visual Analysis of the Boxplot of Differences:

  • The boxplot displays the distribution of the eight difference scores along an axis spanning from $-2$ to $14$.

  • The median difference lies near $5$.

  • The sample mean difference is marked at $\bar{X} = 5.5$.

  • The horizontal blue interval bracket denotes the $95\%$ $t$-confidence interval for the true mean difference, spanning from $1.08$ to $9.92$.

  • The null hypothesis value ($H_0 = 0$) is plotted as a red crossed circle at $0$.

  • Because the null value ($0$) falls completely outside the $95\%$ confidence interval boundary ($1.08, 9.92$), the null hypothesis is rejected at the $\alpha = 0.05$ level.

• Synthesis of Test Results Across Methods:

  • Manual calculation: $t = 2.94$, exceeding the critical value $t^* = 2.365$.

  • Excel analysis: $t = 2.939874$, critical $t = 2.364624$, $p = 0.021717$.

  • Minitab analysis: $t = 2.94$, $p = 0.022$, $95\%$ CI: $(1.07619, 9.92381)$.

  • All computational methods provide consistent and conclusive statistical evidence rejecting the null hypothesis, confirming that climbing stairs produces a statistically significant change in heart rate.

Lecture

Paired t-test looks at two sets of measurements on MATCHED samples; meaning its the same measurements; kinda like a before and after, daytime and nighttime. Single sample of DIFFERENCES.

Calculating t:

• $t=\frac{\overline{x_{diff}}-\mu_{0 diff}}{\frac{s_{diff}}{\sqrt{n}}}$

• $\mu$ will usually be 0, so it can be dropped from the equation.

For Minitab

• Go to Basic Statistics

• Choose Paired T-test

• It will subtract second sample from first sample; First sample should be for “AFTER” data, and the second sample for “BEFORE” data

• Go into options to determine the parameters and what type of tailed test