Paired t-Test - Biostatistics

The mean difference in systolic BP (new drug - placebo) is -12.4 mmHg with a standard error of 3.1 mmHg. When testing the hypothesis that the new drug has no effect, what is the most accurate statement about the degrees of freedom and critical value for a two-tailed test at $\alpha = 0.05$ ?
Answer: Degrees of freedom = 15; critical value ≈ ±2.131, since we're analyzing the differences within each patient When analyzing crossover designs, the key insight is that you're comparing paired differences within the same subjects, not independent groups. This fundamentally changes how you calculate degrees of freedom and approach the statistical test. In this crossover study, each of the 16 patients provides one difference score (new drug BP minus placebo BP). You're essentially performing a one-sample t-test on these 16 difference values to see if the mean difference is significantly different from zero. With 16 difference scores, your degrees of freedom equals $n - 1 = 16 - 1 = 15$ . The critical value from the t-distribution with 15 df at $\alpha = 0.05$ (two-tailed) is approximately ±2.131. Answer C correctly identifies both the degrees of freedom (15) and the appropriate critical value (±2.131) for analyzing paired differences within subjects. Answer A incorrectly uses 14 degrees of freedom, possibly confusing this with other crossover adjustment methods that aren't needed here. Answer B treats this as if you have 30 independent observations (16 × 2), ignoring the paired nature of crossover data and the fact that you're analyzing differences, not raw measurements. This fundamentally misunderstands the design. Answer D uses the normal approximation critical value (±1.960) instead of the t-distribution, which is inappropriate given the small sample size of 16 patients. Remember: in crossover designs, focus on the number of subjects providing difference scores, not the total number of measurements. Each subject contributes one paired difference, making this a one-sample t-test scenario.
A researcher conducts a paired t-test on sleep quality scores (0-10 scale) for 22 patients before and after therapy. The differences have mean = 1.8, standard deviation = 2.4, and the calculated $t = 3.52$ . What is the standard error of the mean difference?
Answer: $SE = 0.51$ When you encounter a paired t-test question that gives you the t-statistic along with sample statistics, you're likely being asked to work backwards to find the standard error. The paired t-test formula connects all these pieces: $t = \frac{\bar{d}}{SE}$ , where $\bar{d}$ is the mean difference and SE is the standard error of the mean difference. Given that $t = 3.52$ and the mean difference $\bar{d} = 1.8$ , you can solve for the standard error by rearranging: $SE = \frac{\bar{d}}{t} = \frac{1.8}{3.52} = 0.511$ , which rounds to 0.51. You can verify this using the standard error formula: $SE = \frac{s}{\sqrt{n}} = \frac{2.4}{\sqrt{22}} = \frac{2.4}{4.69} = 0.511$ . Looking at the wrong answers: Option B (0.55) likely results from a calculation error or slight rounding differences. Option C (2.4) is actually the standard deviation of the differences, not the standard error—a common confusion since both measure variability. Option D (0.11) appears to be a significant calculation error, possibly from incorrectly squaring something in the denominator. Remember that standard error is always smaller than standard deviation because you're dividing by $\sqrt{n}$ . When given a t-statistic and mean difference in paired t-test problems, you can always find SE using $SE = \frac{\text{mean difference}}{t}$ . This relationship is fundamental to understanding how the t-statistic quantifies how many standard errors the mean difference is from zero.
A researcher conducts a paired t-test on reaction time data from 20 participants tested before and after consuming caffeine. The analysis yields $t = -3.26$ with $df = 19$ and $p = 0.004$ for a two-tailed test. If the researcher had instead conducted an independent samples t-test (incorrectly ignoring the pairing), which outcome would be most likely?
Answer: A larger p-value and weaker evidence for a difference due to increased variability When you encounter questions comparing paired versus independent t-tests, focus on how the test design affects variability and statistical power. The key insight is that paired tests control for individual differences between participants, while independent tests treat all observations as unrelated. The correct answer is B because an independent samples t-test would indeed produce a larger p-value and weaker evidence. In a paired t-test, you're analyzing the differences within each participant (after caffeine minus before caffeine), which removes individual baseline variations in reaction time. Some people are naturally faster or slower, but this individual variation gets eliminated when you look at each person's change. An independent samples t-test treats the 40 measurements as completely separate observations, including all that individual variation as "noise" in the analysis. This increased variability makes it harder to detect the caffeine effect, resulting in a larger p-value. Choice A is wrong because identical data analyzed differently will produce different results due to the structural differences between these tests. Choice C incorrectly assumes more degrees of freedom automatically means stronger evidence—while an independent test would have $df = 38$ , the increased variability more than offsets this advantage. Choice D fails because although the degrees of freedom would change to 38, the conclusion would likely change too, since the larger p-value might not reach statistical significance. Remember: paired designs are more powerful when you can match observations because they control for confounding variables. Always consider whether pairing is appropriate and beneficial for detecting the effect of interest.
A researcher conducts a paired t-test on weight loss data from 12 participants (before and after a diet intervention). The test statistic is $t = 2.45$ with $p = 0.032$ . However, examination of the differences reveals one extreme outlier. After removing this outlier, the test statistic becomes $t = 1.78$ with $p = 0.102$ . What is the most appropriate interpretation?
Answer: Report both analyses and acknowledge that the significance depends critically on the outlier observation When results are sensitive to outliers, both analyses should be reported with acknowledgment of the influence of the extreme observation. This provides transparency about the robustness of findings. Choice B is correct. Choice A is wrong because outliers don't automatically invalidate the test. Choice C is incorrect because outlier investigation is appropriate, though post-hoc removal requires careful justification. Choice D is wrong because Bonferroni correction doesn't apply to sensitivity analyses.
In a study of 14 patients, researchers measure biomarker levels before and 24 hours after surgery. The differences (post - pre) have mean = 15.2 units, SD = 8.7 units. A paired t-test yields $t = 6.54$ and $p < 0.001$ . The researchers want to determine if the increase is clinically meaningful (defined as >10 units). What additional analysis is most appropriate?
Answer: Conduct a one-sample t-test testing if the mean difference significantly exceeds 10 units using $H_0: \mu_d = 10$ When you encounter a question about clinical meaningfulness after finding statistical significance, you're dealing with the distinction between statistical and practical significance. The researchers already know the biomarker increased significantly (p < 0.001), but now they need to test whether this increase meets their clinical threshold of 10 units. The most appropriate analysis is a one-sample t-test with $H_0: \mu_d = 10$ versus $H_1: \mu_d > 10$ . This directly tests whether the mean difference significantly exceeds the clinical threshold. You'd calculate $t = \frac{15.2 - 10}{8.7/\sqrt{14}} = \frac{5.2}{2.32} = 2.24$ , then compare this to the critical t-value to determine if the increase is both statistically significant AND clinically meaningful. Option A is incorrect because equivalence tests are used to show that two treatments are similar within a specified margin, not to demonstrate superiority over a threshold. Option B has the right idea about confidence intervals, but simply checking if the entire interval exceeds 10 units is too conservative—you only need evidence that the true mean likely exceeds 10, not certainty that it does. Option D is wrong because Cohen's d measures effect size relative to variability, not whether a specific clinical threshold is exceeded. Effect sizes don't directly translate to clinical meaningfulness without context. Remember: when you have a predetermined clinical threshold, test it directly by setting that threshold as your null hypothesis value. This approach provides the most appropriate statistical evidence for clinical decision-making.
A researcher plans to conduct a paired t-test to compare blood pressure before and after treatment in patients with hypertension. Which of the following scenarios would violate the assumptions of this test?
Answer: The differences between paired measurements are strongly skewed and the sample size is 8 When you encounter paired t-test questions, focus on the three key assumptions: the differences between paired observations must be approximately normally distributed, the pairs must be independent, and the data should come from a representative sample. The paired t-test is robust to violations of normality when sample sizes are large (typically n ≥ 30) due to the Central Limit Theorem. However, with small samples, severe departures from normality can invalidate the test results. In option B, you have both a strongly skewed distribution of differences AND a very small sample size (n = 8). This combination violates the normality assumption without sufficient sample size to compensate, making the test results unreliable. Option A is incorrect because paired t-tests don't assume equal variances between the before and after measurements themselves. The test focuses on the distribution of the differences, not the individual group variances. Option C is wrong because having different baseline values is exactly what paired designs are meant to handle - each subject serves as their own control, so varying starting points don't violate any assumptions. Option D is incorrect because while different measurement times might introduce additional variability, this doesn't violate the core statistical assumptions of the paired t-test. Remember this pattern: for paired t-tests with small samples, look for severe violations of normality in the differences. The "magic number" is around 30 - below that, normality matters much more. Above 30, the test becomes quite robust to non-normal distributions.
In a paired t-test comparing pre- and post-treatment cholesterol levels in 20 patients, the calculated t-statistic is 2.31. The researcher initially planned a two-tailed test but, after seeing that all observed differences were negative, decides to conduct a one-tailed test instead. What is the most appropriate conclusion about this analytical approach?
Answer: This approach is inappropriate because the choice of one-tailed versus two-tailed testing must be made before examining the data The decision between one-tailed and two-tailed testing must be made a priori, before examining the data, to maintain the validity of the statistical inference. Choice B is correct. Choice A is wrong because switching after seeing the data direction constitutes data snooping. Choice C is incorrect because no alpha adjustment can fix this fundamental violation of hypothesis testing principles. Choice D is wrong because the treatment design doesn't determine the statistical test choice after data collection.
A researcher measures blood glucose levels in 26 diabetic patients before and after a new medication. The differences (after - before) have a mean of -15.2 mg/dL and standard deviation of 18.4 mg/dL. What is the 99% confidence interval for the mean difference? Use $t_{0.005,25} = 2.787$ .
Answer: $-15.2 \pm 2.787 \times \frac{18.4}{\sqrt{26}} = -15.2 \pm 10.06$ When you encounter a paired samples study like this one, you're dealing with a single sample of differences rather than comparing two separate groups. This calls for a one-sample t-test confidence interval using the differences as your data. The correct formula for a confidence interval is: $\bar{x} \pm t_{\alpha/2} \times \frac{s}{\sqrt{n}}$ , where $\bar{x}$ is the sample mean, $s$ is the sample standard deviation, and $n$ is the sample size. For a 99% confidence interval with 25 degrees of freedom (n-1 = 26-1), you need $t_{0.005,25} = 2.787$ . Answer A correctly applies this formula: $-15.2 \pm 2.787 \times \frac{18.4}{\sqrt{26}} = -15.2 \pm 10.06$ . Answer B uses $\sqrt{25}$ instead of $\sqrt{26}$ in the denominator. This reflects confusion about degrees of freedom versus sample size - while degrees of freedom equal n-1, the standard error formula uses the actual sample size n. Answer C uses the wrong t-value (2.576), which is the z-value for a 99% confidence interval. This ignores that we're working with a small sample requiring the t-distribution. Answer D omits the $\sqrt{n}$ entirely from the denominator, using just the standard deviation rather than the standard error. This fundamental error treats individual observations' variability the same as the mean's variability. Study tip: For paired data, always work with the differences as a single sample. Remember that confidence intervals need the standard error ( $\frac{s}{\sqrt{n}}$ ), not just the standard deviation, and use t-values for small samples.
A study measures cholesterol levels in 18 patients before and after a dietary intervention. The mean difference (after - before) is -12.8 mg/dL. For a paired t-test with $\alpha = 0.05$ , what are the degrees of freedom and the critical value for a two-tailed test?
Answer: $df = 17$ , critical value = $\pm 2.110$ When you encounter a paired t-test question, you're dealing with before-and-after measurements on the same subjects, which requires specific calculations for degrees of freedom and critical values. For a paired t-test, the degrees of freedom equals $n - 1$ , where $n$ is the number of paired observations (not the total number of measurements). Since we have 18 patients with before-and-after measurements, $df = 18 - 1 = 17$ . With $df = 17$ and $\alpha = 0.05$ for a two-tailed test, the critical value from the t-distribution table is $\pm 2.110$ . Answer A correctly identifies both $df = 17$ and the critical value $\pm 2.110$ . Answer B uses the wrong degrees of freedom (18 instead of 17) and consequently has an incorrect critical value of $\pm 2.101$ , which corresponds to $df = 18$ . Answer C makes the common error of treating this as an independent samples t-test by using $df = 34$ (which would be $18 + 18 - 2$ ) and gives the critical value $\pm 2.032$ for that incorrect degree of freedom. Answer D uses $df = 16$ , which has no logical basis in this context, and provides the corresponding but irrelevant critical value. Remember: paired t-tests always use $df = n - 1$ where $n$ is the number of pairs, not individual observations. Don't confuse paired designs with independent samples designs, which would double your degrees of freedom calculation.
In a paired t-test comparing pain scores before and after treatment in 14 patients, the researcher finds $t = -2.45$ with $p = 0.029$ . Using $\alpha = 0.05$ , which interpretation is correct for this two-tailed test?
Answer: The result is statistically significant, providing evidence that treatment affects pain scores (direction unspecified) When interpreting paired t-test results, you need to understand what statistical significance means and how two-tailed tests work. A paired t-test compares the mean difference between paired observations, and the p-value tells you the probability of observing your result (or more extreme) if there's truly no difference. Here, $p = 0.029 < 0.05$ , so the result is statistically significant. This means you can reject the null hypothesis that there's no difference between before and after treatment scores. The negative t-statistic ( $t = -2.45$ ) indicates the direction of the difference, but in a two-tailed test, you're testing whether treatment has any effect, not specifically whether it increases or decreases pain. Option A is wrong because the sign of the t-statistic doesn't determine significance—only the p-value compared to $\alpha$ does that. A negative t-statistic simply indicates the direction of the difference. Option B incorrectly makes a directional claim. While the negative t-statistic suggests pain scores decreased, a two-tailed test doesn't allow you to conclude a specific direction—that would require a one-tailed test with appropriate hypotheses. Option D misunderstands two-tailed testing. You don't compare $p$ to $\alpha/2$ directly; you compare the full p-value to your chosen $\alpha$ . The p-value already accounts for both tails of the distribution. Study tip: Remember that two-tailed tests detect any difference (bidirectional), while one-tailed tests detect differences in a specific direction. Your conclusion should match your test design.
A study compares pain scores (0-10 scale) before and after treatment in 15 patients. The differences (after - before) have a mean of -2.1 and standard deviation of 2.8. If the 95% confidence interval for the mean difference is (-3.6, -0.6), what can be concluded about a two-tailed paired t-test at $\alpha = 0.05$ ?
Answer: The test would reject the null hypothesis because the confidence interval excludes zero and the mean difference is negative Since the 95% confidence interval (-3.6, -0.6) does not contain zero, we can reject the null hypothesis of no difference at α = 0.05. Choice A is correct. Choice B is wrong because the width of the CI doesn't determine significance; it's whether zero is included. Choice C is incorrect because the normality assumption doesn't change the conclusion given this CI. Choice D is imprecise because the key factor is excluding zero, not just having the same signs.
A researcher wants to test if a new exercise program reduces resting heart rate. Heart rates are measured before and after the program in 16 participants. Which of the following correctly states the null and alternative hypotheses for this paired t-test?
Answer: $H_0: \mu_d = 0$ vs. $H_1: \mu_d < 0$ where $\mu_d = \mu_{after} - \mu_{before}$ When you encounter paired data in biostatistics—like before/after measurements on the same subjects—you need to focus on the differences between paired observations, not the individual group means. This is the foundation of paired t-tests. The correct approach is option B because it properly frames the hypotheses in terms of the mean difference ( $\mu_d$ ). Since we want to test if the exercise program reduces heart rate, we're looking for evidence that post-exercise heart rates are lower than pre-exercise rates. With $\mu_d = \mu_{after} - \mu_{before}$ , a reduction would produce negative differences, so our alternative hypothesis should be $H_1: \mu_d < 0$ . Option A is incorrect because it treats this as a two-sample t-test comparing independent groups, but we have paired data from the same 16 participants. This ignores the correlation between before/after measurements on the same person. Option C uses the correct paired format but has the wrong directional hypothesis. With $\mu_d = \mu_{after} - \mu_{before}$ , the alternative $\mu_d > 0$ would test if heart rates increased after exercise, which contradicts the research question. Option D technically represents the null hypothesis correctly ( $\mu_{after} - \mu_{before} = 0$ is equivalent to $\mu_d = 0$ ) but uses a two-tailed alternative hypothesis ( $\neq$ ) when we have a specific directional prediction that heart rates should decrease. Study tip: For paired t-tests, always define your difference variable clearly and make sure your alternative hypothesis direction matches what you're trying to prove. "Reduces" means you expect negative differences when calculated as (after - before).
A researcher conducts a paired t-test on cognitive test scores before and after an intervention in 22 elderly patients. The assumption of normality for the differences appears questionable based on a Shapiro-Wilk test ( $p = 0.02$ ). The paired t-test gives $t = 2.89$ , $p = 0.009$ . What is the most appropriate next step?
Answer: Perform a Wilcoxon signed-rank test as a non-parametric alternative and compare the conclusions When you encounter assumption violations in statistical tests, the key decision is whether to modify your approach or switch to a more robust alternative that doesn't rely on the violated assumption. The Shapiro-Wilk test with $p = 0.02$ indicates significant departure from normality in the difference scores, violating a key assumption of the paired t-test. In this situation, the Wilcoxon signed-rank test is the appropriate non-parametric alternative because it tests the same research question (whether there's a systematic difference between paired observations) without requiring normality. Comparing conclusions from both tests helps you assess whether the assumption violation meaningfully affects your results. Let's examine why the other options fall short. Choice A incorrectly assumes that $n = 22$ is large enough for the central limit theorem to override normality concerns—while the CLT helps with larger samples (typically $n ≥ 30$ ), it doesn't eliminate the need to address clear assumption violations in smaller samples. Choice B misunderstands how significance levels work—changing $\alpha$ doesn't fix assumption violations; it only changes your threshold for statistical significance. Choice C suggests data transformation, which could work but is less straightforward than using a test designed for non-normal data, and may complicate interpretation of results. Study tip: When you see assumption violations in biostatistics problems, your first instinct should be to consider the corresponding non-parametric test rather than trying to force the original test to work. Learn the parametric/non-parametric pairs: t-test → Wilcoxon, ANOVA → Kruskal-Wallis, correlation → Spearman's rank.
A study examines the effect of a relaxation technique on stress scores (measured on a 1-50 scale) in 19 participants. The researcher wants to test if the technique reduces stress using a one-tailed paired t-test at $\alpha = 0.05$ . The critical value is $t_{0.05,18} = -1.734$ . If the calculated test statistic is $t = -1.89$ , what conclusion should be drawn?
Answer: Reject $H_0$ because $-1.89 < -1.734$ and conclude the technique significantly reduces stress When you encounter a one-tailed paired t-test, the key is understanding both the direction of your hypothesis and how to properly compare your test statistic to the critical value. In this study, you're testing whether the relaxation technique reduces stress, so your alternative hypothesis is $H_1: \mu_d < 0$ (where $d$ represents the difference "after - before"). This creates a left-tailed test, meaning you reject $H_0$ when your test statistic falls in the left tail beyond the critical value. The correct conclusion is C: Reject $H_0$ because $-1.89 < -1.734$ . Since $-1.89$ is more negative than $-1.734$ , your test statistic falls further into the rejection region, providing sufficient evidence that the technique significantly reduces stress. Let's examine why the other options miss the mark. Option A makes a critical error by using absolute values ( $|-1.89| > |-1.734|$ ), which ignores the directional nature of this one-tailed test. Option B incorrectly interprets the comparison $-1.89 < -1.734$ , failing to recognize that being "more negative" means the test statistic is extreme enough for this left-tailed test. Option D shows confusion about critical values—in a left-tailed test, you use the negative critical value, not the positive one. Study tip: For one-tailed tests, always visualize the rejection region. In left-tailed tests, you reject when your test statistic is more negative than the negative critical value. Don't let absolute values confuse you—direction matters!
A paired t-test is conducted on weight change data from 12 participants in a diet study. The mean weight change is -3.4 kg with $s_d = 4.2$ kg. The researcher reports a 90% confidence interval. Which t-value should be used?
Answer: $t_{0.05,11} = 1.796$ When you encounter paired t-test confidence interval questions, you need to identify two key components: the correct degrees of freedom and the appropriate alpha level for your confidence interval. For a 90% confidence interval, you're leaving 10% for the tails (5% in each tail), so you need the t-value corresponding to α/2 = 0.05. The degrees of freedom for a paired t-test equals n - 1, where n is the number of paired observations. With 12 participants, you have df = 12 - 1 = 11. Therefore, you need $t_{0.05,11} = 1.796$ , which is option B. This t-value will be used in the formula: $\bar{d} \pm t_{\alpha/2,df} \times \frac{s_d}{\sqrt{n}}$ to construct your 90% confidence interval. Let's examine why the other options are incorrect. Option A uses $t_{0.10,11}$ , which corresponds to α = 0.10 (not α/2 = 0.05) — this would be for a one-tailed test, not a two-tailed confidence interval. Option C uses the correct alpha level (0.05) but incorrect degrees of freedom (12 instead of 11) — a common mistake when students forget to subtract 1 from the sample size. Option D combines both errors: wrong alpha level (0.10) and wrong degrees of freedom (12). Study tip: For confidence intervals, always remember the "halving rule" — a 90% CI uses α/2 = 0.05, a 95% CI uses α/2 = 0.025, and so on. Also, degrees of freedom for paired t-tests is always n - 1, never n.
A study compares pre- and post-treatment anxiety scores in 20 patients. The differences (post - pre) have a mean of -3.2 and standard deviation of 4.8. Which of the following represents the correct 95% confidence interval for the mean difference, given $t_{0.025,19} = 2.093$ ?
Answer: $-3.2 \pm 2.093 \times \frac{4.8}{\sqrt{20}} = -3.2 \pm 2.25$ When you encounter paired data like pre- and post-treatment measurements, you're dealing with a one-sample t-test on the differences. The key insight is that you're analyzing the 20 difference scores as a single sample, not comparing two separate groups. The confidence interval formula for a mean is: $\bar{x} \pm t_{\alpha/2,df} \times \frac{s}{\sqrt{n}}$ , where the standard error is $\frac{s}{\sqrt{n}}$ . Here, you have 20 patients, so n = 20, and the degrees of freedom are n - 1 = 19, which matches the given t-value. Answer A correctly applies this formula: $-3.2 \pm 2.093 \times \frac{4.8}{\sqrt{20}} = -3.2 \pm 2.25$ . This uses the right sample size (20) in the denominator and the correct t-value. Answer B makes a subtle but critical error by using $\sqrt{19}$ instead of $\sqrt{20}$ in the denominator. While the degrees of freedom are 19, the sample size for calculating standard error is still 20. Answer C omits the square root entirely, using just the standard deviation (4.8) instead of the standard error. This inflates the margin of error dramatically to ±10.05. Answer D uses the z-value (1.96) instead of the t-value. With only 20 observations, you must use the t-distribution, not the normal distribution, because you're estimating the population standard deviation from a small sample. Remember: for paired data, always use n (not n-1) in the standard error calculation, but use df = n-1 for the t-value. Small samples require t-distributions, not z-distributions.
A researcher compares reaction times before and after caffeine consumption in 20 subjects. The correlation between before and after measurements is r = 0.85. If the researcher had instead used an independent samples t-test (incorrectly treating the data as unpaired), how would this affect the analysis compared to the appropriate paired t-test?
Answer: The independent samples test would have lower power due to increased standard error but would have higher degrees of freedom An independent samples t-test would have increased standard error (losing the benefit of pairing with r = 0.85) but more degrees of freedom (38 vs 19). The high correlation means the paired test is much more efficient, so despite fewer df, the paired test has higher power. Choice D correctly identifies both effects. Choice A is wrong about df. Choice B incorrectly suggests independence violation rather than efficiency loss. Choice C is wrong because the standard errors differ substantially.
In a paired t-test with 8 participants, the differences have mean $\bar{d} = 2.5$ and standard deviation $s_d = 3.4$ . If the researcher conducts a two-tailed test at $\alpha = 0.01$ and finds $t_{0.005,7} = 3.499$ , what is the conclusion?
Answer: Fail to reject $H_0$ because $t = 2.08 < 3.499$ and the test statistic is not extreme enough When you encounter a paired t-test question, you're testing whether the mean difference between paired observations is significantly different from zero. The key is calculating the test statistic correctly and comparing it to the appropriate critical value. First, calculate the test statistic using the paired t-test formula: $t = \frac{\bar{d}}{s_d/\sqrt{n}} = \frac{2.5}{3.4/\sqrt{8}} = \frac{2.5}{1.202} = 2.08$ For a two-tailed test at $\alpha = 0.01$ with 7 degrees of freedom (n-1 = 8-1 = 7), you need $|t| > 3.499$ to reject the null hypothesis. Since $|2.08| < 3.499$ , you fail to reject $H_0$ . Option A incorrectly uses the standard normal critical value (1.96) instead of the t-distribution critical value. With small samples like this, you must use the t-distribution, which has wider tails and higher critical values than the standard normal. Option C commits a fundamental error by assuming any positive test statistic indicates significance. The sign only tells you the direction of the difference, not whether it's statistically significant. Option D is wrong because you don't need the exact p-value when you have the critical value. Since $|t| = 2.08 < 3.499$ , you know $p > 0.01$ without calculating the precise p-value. Remember: in hypothesis testing, always use the correct distribution (t for small samples) and compare your test statistic to the appropriate critical value, not just its sign or magnitude alone.
A researcher measures blood pressure in 25 patients before and after a 6-week exercise intervention. The mean difference (after - before) is -8.2 mmHg with a standard deviation of differences of 12.4 mmHg. When conducting a paired t-test at $\alpha = 0.05$ , which of the following represents the most critical assumption that, if violated, would invalidate the test results?
Answer: The differences between paired observations must be approximately normally distributed in the population For a paired t-test, the critical assumption is that the differences (not the original measurements) are approximately normally distributed. Choice A is correct. Choice B is incorrect because normality of individual measurements is not required. Choice C is wrong because equal variances of the original measurements is not an assumption of the paired t-test. Choice D is incorrect because while pairing is most efficient with positive correlation, it's not a formal assumption for validity.
Before conducting a paired t-test on reaction time differences, a researcher should verify that the differences follow approximately which distribution?
Answer: The differences (after - before) should follow approximately a normal distribution When you encounter questions about paired t-tests, focus on what data the test actually analyzes. A paired t-test doesn't work with the original measurements—it works with the differences between paired observations. The correct answer is C because the paired t-test specifically tests whether the mean of the differences equals zero (or some hypothesized value). The test statistic $t = \frac{\bar{d} - \mu_0}{s_d/\sqrt{n}}$ is based on the sample mean of differences ( $\bar{d}$ ) and the standard deviation of differences ( $s_d$ ). For this t-statistic to follow a t-distribution, the underlying differences must be approximately normally distributed. Option A is wrong because the normality of pre-treatment measurements alone doesn't guarantee the differences will be normal. Two normal distributions can produce non-normal differences depending on their correlation. Option B falls into the same trap—post-treatment normality doesn't ensure difference normality. Option D represents a common misconception that both original datasets need to be normal. While this might seem logical, it's unnecessary and insufficient for the paired t-test assumption. This is a crucial distinction: even if both before and after measurements are perfectly normal, their differences might not be normal if there are outliers in the pairing or unusual correlation patterns. Conversely, the differences could be approximately normal even when the original measurements aren't. Study tip: For any paired test question, always ask "what is the test actually analyzing?" Paired tests work with difference scores, so the assumptions apply to those differences, not the original measurements.

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