Biostatistics Quiz: Paired T Test

20 questions · exam conditions

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Paired T TestQuestion 1 of 20

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?

df = 17

, critical value =

\pm 2.110

df = 18

, critical value =

\pm 2.101

df = 34

, critical value =

\pm 2.032

df = 16

, critical value =

\pm 2.120

df = 17

, critical value =

\pm 1.740

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Biostatistics Quiz

Biostatistics Quiz: Paired T Test

Practice Paired T Test in Biostatistics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Paired T Test, giving you a quick way to practice the rules, question types, and explanations that matter most for Biostatistics.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

$df = 17$ , critical value = $\pm 2.110$ (correct answer)
$df = 18$ , critical value = $\pm 2.101$
$df = 34$ , critical value = $\pm 2.032$
$df = 16$ , critical value = $\pm 2.120$
$df = 17$ , critical value = $\pm 1.740$

Explanation: 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.

Question 2

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?

Reject $H_0$ because $t = 2.08 > 1.96$ which exceeds the standard critical value
Fail to reject $H_0$ because $t = 2.08 < 3.499$ and the test statistic is not extreme enough (correct answer)
Reject $H_0$ because $t = 2.08$ is positive, indicating a significant difference
Cannot determine without calculating the exact p-value for $t = 2.08$
Fail to reject $H_0$ because the sample size is too small for reliable inference

Explanation: 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.

Question 3

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 $t_{0.025,19} = 2.093$ ?

$-3.2 \pm 2.093 \times \frac{4.8}{\sqrt{20}} = -3.2 \pm 2.25$ (correct answer)
$-3.2 \pm 2.093 \times \frac{4.8}{\sqrt{19}} = -3.2 \pm 2.30$
$-3.2 \pm 2.093 \times 4.8 = -3.2 \pm 10.05$
$-3.2 \pm 1.96 \times \frac{4.8}{\sqrt{20}} = -3.2 \pm 2.10$
$-3.2 \pm 2.093 \times \frac{4.8}{20} = -3.2 \pm 0.50$

Explanation: 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.

Question 4

A paired t-test is conducted on weight loss data from 25 participants. The mean difference is -4.2 kg with $t = -2.89$ and $p = 0.008$ . If this was originally planned as a one-tailed test (expecting weight loss), but the researcher now wants to report it as a two-tailed test, what should be done?

Double the p-value to get $p = 0.016$ and use the same t-statistic of $-2.89$ (correct answer)
Keep the p-value at $p = 0.008$ since the direction matches the expected outcome
Recalculate the t-statistic by dividing by 2 to get $t = -1.445$ and double the p-value
Use the absolute value $|t| = 2.89$ and keep $p = 0.008$ for the two-tailed test
The test cannot be converted from one-tailed to two-tailed without recollecting the data

Explanation: When converting between one-tailed and two-tailed tests, you're essentially changing how you interpret the same statistical evidence. The key insight is that your calculated t-statistic represents the actual difference you observed in your data - this never changes regardless of how you choose to test it. The correct approach is option A: double the p-value to get

p = 0.016

while keeping

t = -2.89

. Here's why: a one-tailed test examines probability in only one direction of the distribution, while a two-tailed test splits that probability across both tails. When you originally conducted a one-tailed test expecting weight loss, you found

p = 0.008

in the left tail. For a two-tailed test, you must account for the possibility of extreme results in either direction, so you double this probability. Option B incorrectly suggests keeping the original p-value, which would underestimate the true two-tailed probability. Option C makes the fundamental error of changing the t-statistic itself - your observed data and calculated t-value don't change just because you decide to test it differently. Option D commits a similar error by taking the absolute value of t, plus it fails to adjust the p-value appropriately for two-tailed testing. Study tip: Remember that the t-statistic is a property of your data and never changes when converting between one- and two-tailed tests. Only the p-value interpretation changes - and for most cases, you simply double the one-tailed p-value to get the two-tailed equivalent.

Question 5

Before conducting a paired t-test on reaction time differences, a researcher should verify that the differences follow approximately which distribution?

The individual reaction times before treatment should follow a normal distribution
The individual reaction times after treatment should follow a normal distribution
The differences (after - before) should follow approximately a normal distribution (correct answer)
Both the before and after measurements should follow normal distributions separately
The variances of the before and after measurements should be approximately equal

Explanation: 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.

Question 6

A paired t-test is performed on cognitive test scores for 30 elderly patients before and after a memory training program. The mean improvement is 2.3 points with a 95% confidence interval of (0.8, 3.8). What can be concluded about a two-tailed test at $\alpha = 0.05$ ?

We would reject $H_0$ because the entire confidence interval is above zero, indicating a significant difference (correct answer)
We would fail to reject $H_0$ because the confidence interval includes values less than 2.3
We cannot determine the test result without knowing the exact p-value from the confidence interval
We would fail to reject $H_0$ because the confidence interval is too wide, indicating high variability
We would reject $H_0$ only if this were a one-tailed test, but not for a two-tailed test

Explanation: When you encounter a paired t-test with a confidence interval, you're looking at the relationship between confidence intervals and hypothesis testing. For a two-tailed test at

\alpha = 0.05

, the 95% confidence interval tells you directly whether to reject the null hypothesis. The key insight is that if the 95% confidence interval for the mean difference doesn't include zero, you can reject

H_0

(no difference) at the 0.05 significance level. Here, the confidence interval is (0.8, 3.8), which lies entirely above zero. This means we can be 95% confident that the true mean improvement is between 0.8 and 3.8 points - all positive values. Since zero (representing no improvement) falls outside this interval, we reject

H_0

and conclude there's a statistically significant improvement. Answer A correctly identifies this reasoning. Answer B is wrong because it misunderstands what matters - the confidence interval excluding zero is what's important, not whether it includes values different from the sample mean of 2.3. Answer C is incorrect because confidence intervals and hypothesis tests are mathematically equivalent; you don't need the exact p-value when you have the confidence interval. Answer D misinterprets interval width - while a wide interval indicates more variability, the critical factor for hypothesis testing is whether the interval contains zero, not its width. Study tip: Remember this shortcut: for a 95% confidence interval and

\alpha = 0.05

, if zero isn't in the interval, reject

H_0

. This direct relationship saves time and reduces errors on exams.

Question 7

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 $t_{0.005,25} = 2.787$ .

$-15.2 \pm 2.787 \times \frac{18.4}{\sqrt{26}} = -15.2 \pm 10.06$ (correct answer)
$-15.2 \pm 2.787 \times \frac{18.4}{\sqrt{25}} = -15.2 \pm 10.26$
$-15.2 \pm 2.576 \times \frac{18.4}{\sqrt{26}} = -15.2 \pm 9.29$
$-15.2 \pm 2.787 \times 18.4 = -15.2 \pm 51.28$
$-15.2 \pm 2.787 \times \frac{18.4}{26} = -15.2 \pm 1.97$

Explanation: 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.

Question 8

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?

The before and after measurements have different variances (heteroscedasticity)
The differences between paired measurements are strongly skewed and the sample size is 8 (correct answer)
Some patients have higher baseline blood pressure than others (different starting points)
The blood pressure measurements are taken at different times of day for different patients
The treatment effect varies between patients (some benefit more than others)

Explanation: 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.

Question 9

A pharmaceutical company tests a new pain medication by measuring pain scores (0-10 scale) before and 2 hours after administration in 24 patients. The mean reduction in pain is 2.3 points with $t = 4.15$ and $p = 0.0004$ . The company wants to claim the medication provides 'significant pain relief.' Which statement about this claim is most appropriate?

The claim is supported because $p < 0.05$ indicates the medication significantly reduces pain scores
The claim is questionable because statistical significance doesn't necessarily mean clinically meaningful relief (correct answer)
The claim is invalid because a 2.3-point reduction is too small to be considered significant pain relief
The claim is supported because the large t-statistic (4.15) indicates a strong treatment effect
The claim cannot be evaluated without knowing the baseline pain scores of the patients

Explanation: When you encounter questions about research claims and statistical significance, remember that statistical significance and clinical significance are two distinct concepts that don't always align. The key issue here is distinguishing between what the statistics tell us versus what matters in practice. The study shows clear statistical significance (

p = 0.0004

, well below 0.05), meaning we can confidently say the medication reduces pain more than placebo or baseline. However, statistical significance doesn't automatically translate to meaningful clinical benefit. A 2.3-point reduction on a 10-point scale might be statistically reliable but may not represent relief that patients would notice or care about in their daily lives. This is why option B correctly identifies the claim as questionable. Option A falls into the common trap of equating statistical significance with clinical meaningfulness. Yes,

p < 0.05

confirms the effect is real, but it doesn't tell us if the effect matters to patients. Option C goes too far by declaring the claim "invalid" - we can't definitively say 2.3 points is too small without additional context about what patients consider meaningful relief. Option D misinterprets the t-statistic, which reflects the reliability of the finding relative to sample variability, not necessarily the practical importance of the effect size. Study tip: In biostatistics questions about research interpretation, always ask yourself two questions: "Is the effect statistically significant?" and "Is the effect clinically meaningful?" These are separate issues, and distinguishing them is crucial for proper study interpretation.

Question 10

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?

The same t-statistic and p-value since the data are identical
A larger p-value and weaker evidence for a difference due to increased variability (correct answer)
A smaller p-value and stronger evidence because of increased degrees of freedom
The same conclusion but with $df = 38$ instead of $df = 19$
An invalid test because the independence assumption would be violated

Explanation: 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.

Question 11

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?

$SE = 0.51$ (correct answer)
$SE = 0.55$
$SE = 2.4$
$SE = 0.11$
$SE = 1.02$

Explanation: 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.

Question 12

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?

Reject $H_0$ because $|-1.89| > |-1.734|$ and conclude the technique significantly reduces stress
Fail to reject $H_0$ because $-1.89 < -1.734$ and the test statistic is not extreme enough
Reject $H_0$ because $-1.89 < -1.734$ and conclude the technique significantly reduces stress (correct answer)
Fail to reject $H_0$ because we need the positive critical value $+1.734$ for comparison
Cannot determine without knowing whether this is an upper or lower tail test

Explanation: 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!

Question 13

A researcher studies the effect of a new medication on blood pressure by measuring systolic blood pressure in 12 patients before and after treatment. The mean difference (after - before) is -8.5 mmHg with a standard deviation of differences of 12.5 mmHg. What is the test statistic for the paired t-test?

$t = -4.75$
$t = -1.37$
$t = -2.36$ (correct answer)
$t = -0.68$
$t = -0.73$

Explanation: When you encounter a paired t-test scenario, you're dealing with before-and-after measurements on the same subjects. The key is recognizing that you already have the differences calculated, so you can jump straight to the t-test formula. For a paired t-test, the test statistic is:

t = \frac{\bar{d}}{s_d/\sqrt{n}}

, where

\bar{d}

is the mean difference,

s_d

is the standard deviation of differences, and

n

is the sample size. Plugging in the given values:

t = \frac{-8.5}{12.5/\sqrt{12}} = \frac{-8.5}{12.5/3.464} = \frac{-8.5}{3.608} = -2.36

This confirms answer C is correct. Looking at the wrong answers: Answer A (-4.75) likely comes from incorrectly using

\sqrt{n}

instead of dividing by it, or confusing this with a different statistical test. Answer B (-1.37) might result from using the wrong denominator formula, perhaps forgetting to divide the standard deviation by

\sqrt{n}

. Answer D (-0.68) could come from accidentally using

n

instead of

\sqrt{n}

in the denominator, making it much larger than it should be. Study tip: Always double-check your denominator calculation in t-tests. The standard error formula is

s/\sqrt{n}

, and many students make arithmetic errors here. Practice calculating

\sqrt{12} \approx 3.464

and similar square roots, as they appear frequently in biostatistics problems with common sample sizes.

Question 14

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?

$H_0: \mu_{after} = \mu_{before}$ vs. $H_1: \mu_{after} < \mu_{before}$
$H_0: \mu_d = 0$ vs. $H_1: \mu_d < 0$ where $\mu_d = \mu_{after} - \mu_{before}$ (correct answer)
$H_0: \mu_d = 0$ vs. $H_1: \mu_d > 0$ where $\mu_d = \mu_{after} - \mu_{before}$
$H_0: \mu_{after} - \mu_{before} = 0$ vs. $H_1: \mu_{after} - \mu_{before} \neq 0$
$H_0: \bar{d} = 0$ vs. $H_1: \bar{d} < 0$ where $\bar{d} = \bar{x}_{after} - \bar{x}_{before}$

Explanation: 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).

Question 15

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?

$t_{0.10,11} = 1.363$
$t_{0.05,11} = 1.796$ (correct answer)
$t_{0.05,12} = 1.782$
$t_{0.10,12} = 1.356$
$t_{0.01,11} = 2.718$

Explanation: 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.

Question 16

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?

The result is not statistically significant because the test statistic is negative
The result is statistically significant, providing evidence that treatment reduces pain scores
The result is statistically significant, providing evidence that treatment affects pain scores (direction unspecified) (correct answer)
The result is not statistically significant because $p > 0.025$ for a two-tailed test
The result is inconclusive because we need to know the actual pain score differences

Explanation: 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

\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.

Question 17

In a paired t-test with 15 subjects, the critical value for a two-tailed test at $\alpha = 0.05$ is approximately $t_{0.025,14} = 2.145$ . If the calculated test statistic is $t = -2.67$ , which conclusion is most appropriate?

Reject the null hypothesis because $|-2.67| > 2.145$ and conclude there is insufficient evidence of a difference
Fail to reject the null hypothesis because $t = -2.67 < 0$ and negative values cannot be significant
Reject the null hypothesis because $|-2.67| > 2.145$ and conclude there is sufficient evidence of a difference (correct answer)
Fail to reject the null hypothesis because $-2.67 < 2.145$ and the test statistic must exceed the critical value
Cannot determine significance because the test statistic is negative and we need the positive critical value

Explanation: When analyzing paired t-test results, you need to understand how two-tailed tests work with critical values. In a two-tailed test, you're testing whether there's a significant difference in either direction, so both extremely positive and extremely negative test statistics can lead to rejection of the null hypothesis. The decision rule for a two-tailed test is: reject

H_0

|t| > t_{\alpha/2, df}

. Here, you compare the absolute value of your test statistic to the critical value. Since

|-2.67| = 2.67 > 2.145

, you reject the null hypothesis and conclude there is sufficient evidence of a difference between the paired observations. Option A makes the correct comparison (

|-2.67| > 2.145

) but draws the wrong conclusion, confusing "reject

H_0

" with "insufficient evidence." When you reject the null hypothesis, you're concluding there IS sufficient evidence of a difference. Option B reflects a fundamental misunderstanding that negative test statistics cannot be significant. In two-tailed tests, the direction doesn't matter—only the magnitude. A

t = -2.67

is just as significant as

t = +2.67

. Option D incorrectly compares

-2.67

directly to

2.145

without taking the absolute value. This ignores how two-tailed tests work—you must consider the absolute value of the test statistic. Study tip: For two-tailed t-tests, always remember the absolute value rule:

|t| > t_{critical}

means reject

H_0

, regardless of whether your test statistic is positive or negative. The sign only tells you the direction of the difference.

Question 18

A pharmaceutical company tests a new antihypertensive medication using a randomized crossover design. Each of 16 patients receives both the new drug and a placebo in random order, with a 2-week washout period between treatments. Systolic blood pressure is measured after each treatment period.

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 $\alpha = 0.05$ ?

Degrees of freedom = 14; critical value ≈ ±2.145, after adjusting for the crossover design structure
Degrees of freedom = 30; critical value ≈ ±2.042, since we have 16 patients with 2 measurements each
Degrees of freedom = 15; critical value ≈ ±2.131, since we're analyzing the differences within each patient (correct answer)
Degrees of freedom = 15; critical value ≈ ±1.960, since the sample size is large enough for normal approximation

Explanation: 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.

Question 19

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?

This approach is acceptable since the data clearly supports a directional hypothesis and increases statistical power
This approach is inappropriate because the choice of one-tailed versus two-tailed testing must be made before examining the data (correct answer)
This approach is acceptable provided the researcher adjusts the alpha level to account for the post-hoc decision
This approach is inappropriate unless the researcher can demonstrate that the treatment was specifically designed to decrease cholesterol

Explanation: 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.

Question 20

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?

Report the result without the outlier since outliers violate the normality assumption of the paired t-test
Report both analyses and acknowledge that the significance depends critically on the outlier observation (correct answer)
Report the original result since removing data points after analysis constitutes inappropriate data manipulation
Report the result without the outlier and adjust the p-value using a Bonferroni correction for multiple testing

Explanation: 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.