Biostatistics Quiz: One And Two Sample T Tests
20 questions · exam conditions
0:00
One And Two Sample T TestsQuestion 1 of 20

A one-sample t-test is performed on cholesterol levels from 16 patients to test H₀: μ = 200 mg/dL versus H₁: μ ≠ 200 mg/dL. The sample mean is 215 mg/dL with a standard deviation of 25 mg/dL. What assumption is most critical to verify before interpreting the results?

The population variance must be known exactly
The sample data must follow a normal distribution approximately
The sample size must be greater than 30 observations
The population mean must be exactly 200 mg/dL
The observations must be collected using simple random sampling
← Back to quizzes

Biostatistics Quiz

Biostatistics Quiz: One And Two Sample T Tests

Practice One And Two Sample T Tests 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 One And Two Sample T Tests, 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

A one-sample t-test is performed on cholesterol levels from 16 patients to test H₀: μ = 200 mg/dL versus H₁: μ ≠ 200 mg/dL. The sample mean is 215 mg/dL with a standard deviation of 25 mg/dL. What assumption is most critical to verify before interpreting the results?

  1. The population variance must be known exactly
  2. The sample data must follow a normal distribution approximately (correct answer)
  3. The sample size must be greater than 30 observations
  4. The population mean must be exactly 200 mg/dL
  5. The observations must be collected using simple random sampling
Explanation: When you encounter questions about t-test assumptions, you're being tested on the foundational requirements that make statistical inference valid. The one-sample t-test relies on specific conditions to produce trustworthy results. The most critical assumption here is that the sample data must follow an approximately normal distribution (B). The t-test's validity depends on this normality assumption because the test statistic follows a t-distribution only when the underlying population is normal or approximately normal. With a sample size of 16, you're below the threshold where the Central Limit Theorem provides robust protection against non-normality, making this assumption especially important to verify through methods like normal probability plots or the Shapiro-Wilk test. Option A is incorrect because the one-sample t-test is specifically designed for situations where the population variance is unknown—that's why we use the sample standard deviation and the t-distribution instead of the z-distribution. Option C represents a common misconception; while n > 30 is a useful rule of thumb for invoking the Central Limit Theorem, it's not a requirement for the t-test itself. The t-test can work with smaller samples if normality is satisfied. Option D misunderstands hypothesis testing fundamentals—we never assume the null hypothesis is true; rather, we test whether the data provides sufficient evidence against it. For biostatistics exams, remember that assumption checking is as important as performing the test itself. When you see small sample sizes (typically n < 30), always prioritize verifying the normality assumption before trusting your t-test results.

Question 2

In a two-sample t-test comparing mean recovery times between two surgical procedures, the degrees of freedom are calculated as 28. If the first group has 12 patients, how many patients are in the second group, assuming equal variances?

  1. 16 patients
  2. 17 patients
  3. 18 patients (correct answer)
  4. 19 patients
  5. 20 patients
Explanation: When you encounter a two-sample t-test question about degrees of freedom, you're dealing with a fundamental relationship that connects sample sizes to the statistical analysis. The degrees of freedom formula for a two-sample t-test assuming equal variances is: df=n1+n2−2df = n_1 + n_2 - 2, where n1n_1 and n2n_2 are the sample sizes of the two groups. Given that the degrees of freedom equal 28 and the first group has 12 patients, you can solve directly: 28=12+n2−228 = 12 + n_2 - 2. Rearranging: 28=10+n228 = 10 + n_2, so n2=18n_2 = 18 patients. This confirms that answer C is correct. Let's examine why the other options are wrong. Answer A (16 patients) would give you df=12+16−2=26df = 12 + 16 - 2 = 26, which is too low. Answer B (17 patients) yields df=12+17−2=27df = 12 + 17 - 2 = 27, still one degree of freedom short. Answer D (19 patients) produces df=12+19−2=29df = 12 + 19 - 2 = 29, which exceeds the given value by one. The key insight here is understanding where the "-2" comes from: you lose one degree of freedom for each group mean you estimate from the sample data. Remember this pattern for biostatistics exams: always double-check your degrees of freedom calculation, as it's often tested directly and affects your critical value lookup in t-tables. A simple algebraic error here cascades through the entire statistical test.

Question 3

A researcher conducts a one-sample t-test with α = 0.05 and obtains a p-value of 0.03. If the same data were analyzed using α = 0.01, what would happen to the test conclusion and the p-value?

  1. The conclusion would change from reject H₀ to fail to reject H₀; the p-value would remain 0.03 (correct answer)
  2. The conclusion would remain the same; the p-value would change to 0.01
  3. The conclusion would change from reject H₀ to fail to reject H₀; the p-value would change to 0.01
  4. Both the conclusion and p-value would remain unchanged from the original analysis
  5. The conclusion would remain the same; the p-value would remain 0.03
Explanation: When you encounter questions about hypothesis testing and significance levels, focus on the distinction between p-values (which are fixed properties of your data) and alpha levels (which are researcher-chosen decision criteria). The p-value of 0.03 is a characteristic of your data and test statistic—it represents the probability of observing results at least as extreme as yours, assuming the null hypothesis is true. This value doesn't change regardless of what significance level you choose to use for your decision-making. However, your conclusion does depend on comparing the p-value to your chosen alpha level. With α = 0.05, since p = 0.03 < 0.05, you reject H₀. But with the more stringent α = 0.01, since p = 0.03 > 0.01, you fail to reject H₀. The evidence hasn't changed—only your threshold for what constitutes "significant" evidence. Looking at the wrong answers: B incorrectly suggests the p-value would change to match the new alpha level, but p-values are calculated from data, not chosen arbitrarily. C makes the same error about p-values changing while correctly noting the conclusion change. D fails to recognize that a more stringent alpha level (0.01 vs 0.05) changes whether our p-value of 0.03 meets the significance threshold. Study tip: Remember that p-values are fixed once you've collected data and run your test—only your interpretation of their significance changes based on your chosen alpha level. The smaller your alpha, the stronger the evidence needed to reject H₀.

Question 4

A researcher tests H₀: μ₁ = μ₂ versus H₁: μ₁ ≠ μ₂ using a two-sample t-test. With α = 0.05, the calculated t-statistic is -2.14 and the critical values are ±2.048. The researcher concludes there is a significant difference. What error, if any, has been made?

  1. No error; the conclusion is correct because |t| = 2.14 > 2.048 (correct answer)
  2. Type I error; the null hypothesis was incorrectly rejected when it was actually true
  3. Type II error; the null hypothesis was incorrectly accepted when it was actually false
  4. The wrong critical value was used; should have used a one-tailed test instead
  5. The degrees of freedom were calculated incorrectly, leading to wrong critical values
Explanation: When you encounter hypothesis testing questions, focus on the decision rule: compare your test statistic to the critical value, then determine if any errors occurred in the logic. Here, you have a two-tailed test with α=0.05\alpha = 0.05 and critical values of ±2.048\pm 2.048. The calculated t-statistic is −2.14-2.14. Since ∣−2.14∣=2.14>2.048|{-2.14}| = 2.14 > 2.048, the test statistic falls in the rejection region, so rejecting H0H_0 and concluding a significant difference is the correct statistical decision. Looking at each option: Choice A correctly identifies that no error was made because the absolute value of the test statistic (2.14) exceeds the critical value (2.048), making rejection of H0H_0 appropriate. Choice B suggests a Type I error occurred, but you can't determine this from the given information—Type I error means rejecting a true null hypothesis, but you don't know whether H0H_0 is actually true or false in reality. Choice C mentions Type II error (failing to reject a false null), but this doesn't apply since H0H_0 was rejected, not accepted. Choice D incorrectly suggests using a one-tailed test, but the hypothesis H1:μ1≠μ2H_1: \mu_1 \neq \mu_2 clearly indicates a two-tailed test is appropriate. The key insight is that statistical errors (Type I and Type II) refer to the true state of nature, which you never know in practice. You can only evaluate whether the statistical procedure was followed correctly given the evidence. Remember: when ∣t∣>|t| > critical value, reject H0H_0—the decision-making process here was executed properly.

Question 5

A pharmaceutical company tests whether a new drug reduces reaction time compared to placebo. They use a one-sample t-test on the differences (drug - placebo) from 24 paired subjects. The mean difference is -0.8 seconds with standard deviation 1.2 seconds. What is the t-statistic?

  1. -3.27 (correct answer)
  2. -2.67
  3. -1.96
  4. -1.33
  5. -0.67
Explanation: When you encounter a paired t-test scenario, you're looking at before-and-after or matched-pairs data where each subject serves as their own control. The key is recognizing that you're testing whether the mean difference significantly differs from zero. The t-statistic formula for a one-sample t-test is: t=xˉ−μ0s/nt = \frac{\bar{x} - \mu_0}{s/\sqrt{n}} Here, you're testing if the mean difference equals zero, so μ0=0\mu_0 = 0. With xˉ=−0.8\bar{x} = -0.8, s=1.2s = 1.2, and n=24n = 24: t=−0.8−01.2/24=−0.81.2/4.899=−0.80.245=−3.27t = \frac{-0.8 - 0}{1.2/\sqrt{24}} = \frac{-0.8}{1.2/4.899} = \frac{-0.8}{0.245} = -3.27 This confirms answer A is correct. Answer B (-2.67) likely comes from incorrectly using 25\sqrt{25} instead of 24\sqrt{24} in the denominator, a common arithmetic error. Answer C (-1.96) might result from confusing this with a z-critical value or using an incorrect sample size like n=15n = 15. Answer D (-1.33) could stem from major calculation errors, such as using the wrong standard deviation value or significantly miscalculating the standard error. Remember that in paired t-tests, your sample size is the number of pairs (24), not individual observations (48). The negative t-statistic makes sense here since the drug is reducing reaction time, creating negative differences. Always double-check your arithmetic with the standard error calculation—it's where most errors occur in t-test problems.

Question 6

A researcher performs a one-sample t-test with n = 9 and obtains t = 2.1. For a two-tailed test, what range contains the p-value?

  1. p < 0.01
  2. 0.01 < p < 0.02
  3. 0.02 < p < 0.05
  4. 0.05 < p < 0.10 (correct answer)
  5. p > 0.10
Explanation: When you encounter a t-test question asking for p-value ranges, you need to use the t-distribution table to find where your calculated t-statistic falls relative to critical values. With n = 9, your degrees of freedom = n - 1 = 8. For a two-tailed test with t = 2.1, you need to find where this value sits among the critical t-values for df = 8. Looking at a t-table for df = 8: t = 1.860 corresponds to α = 0.10 (two-tailed), and t = 2.306 corresponds to α = 0.05 (two-tailed). Since your calculated t = 2.1 falls between 1.860 and 2.306, your p-value must fall between 0.05 and 0.10. This makes D correct. Choice A (p < 0.01) is wrong because t = 2.1 isn't nearly large enough to reach this significance level, which would require t > 3.355 for df = 8. Choice B (0.01 < p < 0.02) is incorrect for the same reason—this range would require t-values around 2.9-3.4. Choice C (0.02 < p < 0.05) is wrong because while closer, t = 2.1 doesn't quite reach the t = 2.306 threshold needed for p = 0.05. Remember: for t-test p-value questions, always identify your degrees of freedom first, then locate your calculated t-statistic within the appropriate column ranges on the t-table. The p-value will fall between the α levels that bracket your t-statistic. Practice reading t-tables in both directions—from α to t, and from t to α.

Question 7

When comparing mean blood glucose levels between diabetic and non-diabetic patients using a two-sample t-test, which statement about the assumptions is most accurate?

  1. The sample sizes must be equal for the test to be valid
  2. The populations must have exactly normal distributions regardless of sample size
  3. Independence of observations is more critical than normality when sample sizes are large (correct answer)
  4. Equal population variances must be verified using an F-test before proceeding
  5. The central limit theorem eliminates all distributional assumptions when n > 30
Explanation: When evaluating two-sample t-test assumptions, you need to understand the relative importance and robustness of each requirement, especially as sample size changes. The correct answer is C because independence of observations is a fundamental, non-negotiable assumption that cannot be violated regardless of sample size. When you have large samples (typically n > 30 per group), the Central Limit Theorem makes the t-test quite robust to departures from normality, but independence violations can severely bias your results and invalidate statistical inference entirely. Let's examine why the other options are incorrect: Option A is wrong because equal sample sizes, while sometimes beneficial for power, are not required for validity. The two-sample t-test works perfectly well with unequal group sizes using appropriate formulas for degrees of freedom. Option B misses a crucial point about robustness. While normality is important, the t-test becomes increasingly robust to normality violations as sample size increases due to the Central Limit Theorem. Requiring "exactly normal distributions regardless of sample size" is unnecessarily restrictive. Option D represents poor statistical practice. While equal variances help, you shouldn't routinely pre-test with an F-test because this creates multiple testing problems and the two-sample t-test has variants (like Welch's t-test) that handle unequal variances well. Study tip: Remember the hierarchy of t-test assumptions by criticality: Independence is non-negotiable, normality becomes less important with larger samples, and equal variances can be addressed with test modifications. Focus most on ensuring your observations are truly independent.

Question 8

A two-sample t-test yields t = 1.87 with 34 degrees of freedom. The researchers report this as "marginally significant" at α = 0.05. What is the most likely explanation for their interpretation?

  1. They used a one-tailed test when a two-tailed test was more appropriate
  2. They used a two-tailed test and obtained p ≈ 0.07, which is close to but exceeds 0.05 (correct answer)
  3. They incorrectly calculated the degrees of freedom and should have used a larger value
  4. They applied a Bonferroni correction for multiple comparisons inappropriately
  5. They confused the t-statistic with the p-value in their interpretation
Explanation: When you encounter a question about interpreting statistical significance, focus on the relationship between test statistics, p-values, and significance thresholds. With t = 1.87 and 34 degrees of freedom, you can determine the p-value by consulting a t-table or using statistical software. For a two-tailed test, this t-statistic corresponds to a p-value of approximately 0.07. Since 0.07 > 0.05, the result is technically not significant at α = 0.05, but it's close enough that researchers often describe it as "marginally significant" or "approaching significance." This explains why option B is correct. Option A is incorrect because using a one-tailed test when a two-tailed test was appropriate would actually make the p-value smaller (about 0.035), which would be significant, not marginal. Option C is wrong because the degrees of freedom calculation for a two-sample t-test follows established formulas based on sample sizes and variance assumptions—there's no indication these were calculated incorrectly, and larger df would only make significance more likely. Option D doesn't fit because a Bonferroni correction would make the significance threshold more stringent (like α = 0.025 or lower), and with p ≈ 0.07, the result would be clearly non-significant, not marginally so. Study tip: When you see "marginally significant" in biostatistics, it almost always means the p-value is slightly above 0.05 (typically 0.05-0.10). Remember that statistical significance is a binary threshold, but researchers often acknowledge results that nearly reach significance.

Question 9

In a two-sample t-test with equal variances, Group A (n₁ = 20) has x̄₁ = 85 and s₁ = 15, while Group B (n₂ = 16) has x̄₂ = 78 and s₂ = 12. If the calculated t-statistic is 1.52, what conclusion should be drawn using α = 0.05 for a two-tailed test?

  1. Reject H₀ because the sample means differ by 7 units, which is practically significant
  2. Fail to reject H₀ because |t| = 1.52 < 2.032 (critical value with df = 34) (correct answer)
  3. Reject H₀ because the pooled standard deviation indicates high variability between groups
  4. Fail to reject H₀ because the sample sizes are unequal, violating test assumptions
  5. The test is inconclusive because the t-statistic falls in the critical region boundary
Explanation: When you encounter a two-sample t-test question, focus on the decision rule: compare your calculated t-statistic to the critical value at your chosen significance level. This determines whether you have sufficient evidence to reject the null hypothesis. Here, you need to determine the critical value for a two-tailed test with α=0.05\alpha = 0.05. The degrees of freedom for a pooled t-test is df=n1+n2−2=20+16−2=34df = n_1 + n_2 - 2 = 20 + 16 - 2 = 34. With df=34df = 34 and α=0.05\alpha = 0.05 (two-tailed), the critical t-value is approximately 2.032. Since your calculated t-statistic of 1.52 falls within the acceptance region (∣1.52∣<2.032|1.52| < 2.032), you fail to reject the null hypothesis. Option A incorrectly focuses on practical significance rather than statistical significance. A 7-unit difference between means tells you nothing about statistical significance without considering variability and sample size. Option C misunderstands what the pooled standard deviation represents—it estimates within-group variability, not between-group differences. A larger pooled standard deviation would actually make it harder to detect differences between groups. Option D incorrectly suggests that unequal sample sizes violate t-test assumptions. The two-sample t-test can handle unequal sample sizes; the key assumptions are normality and equal variances (which the question states). Study tip: Always compare your calculated test statistic to the critical value, not to practical differences or descriptive statistics. The formal decision rule—not intuition about effect size—determines statistical significance.

Question 10

A researcher wants to test if the mean reaction time for a cognitive task differs from 500 ms. From 16 subjects, the sample mean is 485 ms with s = 45 ms. What would happen to the power of the test if the sample size were increased to 36 subjects, assuming the same effect size and variability?

  1. Power would decrease because the degrees of freedom increase, making the critical value larger
  2. Power would remain the same because the effect size and population variance are unchanged
  3. Power would increase because the standard error decreases with larger sample size (correct answer)
  4. Power would increase initially but then decrease due to diminishing returns from larger samples
  5. The change in power cannot be determined without knowing the true population mean
Explanation: When you encounter statistical power questions, focus on the core relationship: power increases when you can better distinguish between the null hypothesis and alternative hypothesis. Power depends on effect size, sample size, significance level, and population variability. The key insight here is understanding how sample size affects the standard error. The standard error of the mean equals sn\frac{s}{\sqrt{n}}, so as sample size increases from 16 to 36, the standard error decreases from 4516=11.25\frac{45}{\sqrt{16}} = 11.25 to 4536=7.5\frac{45}{\sqrt{36}} = 7.5. A smaller standard error means your sample mean is more precisely estimated, making it easier to detect when the true mean actually differs from 500 ms. This increased precision directly translates to higher power. Option A incorrectly suggests that larger degrees of freedom hurt power by increasing critical values. Actually, as degrees of freedom increase, t-critical values approach the smaller z-critical values, which slightly helps rather than hurts power. Option B reflects a common misconception that power only depends on effect size and variability. While these matter, sample size is equally crucial because it affects how precisely you can estimate the population parameter. Option D incorrectly implies diminishing returns eventually decrease power - while the rate of power increase does slow with very large samples, power itself never decreases from adding more subjects. Remember this pattern: larger sample size → smaller standard error → more precise estimates → higher power. This fundamental relationship appears frequently in biostatistics, so master the connection between sample size and standard error.

Question 11

A researcher tests whether the mean systolic blood pressure in a sample of 25 hypertensive patients differs from the population mean of 140 mmHg. The sample mean is 148 mmHg with a standard deviation of 15 mmHg. If the calculated t-statistic is 2.67, what can be concluded about the p-value for this two-tailed test?

  1. The p-value is less than 0.01
  2. The p-value is between 0.01 and 0.02 (correct answer)
  3. The p-value is between 0.02 and 0.05
  4. The p-value is between 0.05 and 0.10
  5. The p-value is greater than 0.10
Explanation: When you encounter t-test questions involving p-values, you need to connect the calculated t-statistic to probability using the t-distribution and degrees of freedom. With a sample of 25 patients, you have df=n−1=24df = n - 1 = 24 degrees of freedom. The calculated t-statistic is 2.67 for this two-tailed test. To find the p-value range, you need to locate where 2.67 falls on the t-distribution table for 24 degrees of freedom. Looking at critical values for a two-tailed test with 24 df: the critical value for α = 0.02 is approximately 2.492, and for α = 0.01 is approximately 2.797. Since your calculated t-statistic of 2.67 falls between these two values (2.492 < 2.67 < 2.797), the p-value falls between 0.01 and 0.02. Answer choice A (p < 0.01) is incorrect because 2.67 is less than the critical value of 2.797 needed for p < 0.01. Answer choice C (0.02 < p < 0.05) is wrong because 2.67 exceeds the critical value of 2.492 for α = 0.02, making the p-value smaller than 0.02. Answer choice D (0.05 < p < 0.10) is incorrect because 2.67 is much larger than the critical value needed for α = 0.05 (approximately 2.064). Study tip: Always memorize or have access to key critical values for common degrees of freedom. Practice reading t-tables in both directions – from α to critical values and from t-statistics back to p-value ranges.

Question 12

A one-sample t-test yields a 95% confidence interval for the mean of (12.3, 18.7). If the null hypothesis was H₀: μ = 15, what can be concluded about the two-tailed test at α = 0.05?

  1. Reject H₀ because the sample mean (15.5) differs from the hypothesized mean
  2. Fail to reject H₀ because 15 falls within the confidence interval (correct answer)
  3. Reject H₀ because the confidence interval is too wide, indicating high variability
  4. Fail to reject H₀ because the margin of error exceeds 3 units
  5. The test is inconclusive because the confidence interval includes the null value
Explanation: When you encounter a question linking confidence intervals to hypothesis testing, remember that these two statistical methods are mathematically equivalent for two-tailed tests at the same significance level. A 95% confidence interval corresponds directly to a two-tailed test at α = 0.05. The key principle is this: if the null hypothesis value falls within the confidence interval, you fail to reject the null hypothesis. Since the 95% confidence interval is (12.3, 18.7) and the null hypothesis states μ = 15, we can see that 15 falls within this interval (12.3 < 15 < 18.7). Therefore, we fail to reject H₀. Option A is incorrect because it focuses on comparing the sample mean to the hypothesized mean without considering variability. The sample mean is xˉ=12.3+18.72=15.5\bar{x} = \frac{12.3 + 18.7}{2} = 15.5, but the small difference from 15 isn't significant given the sampling variability reflected in the confidence interval. Option C is wrong because the width of the confidence interval doesn't determine rejection or non-rejection of the null hypothesis. High variability (wide intervals) simply indicates less precision, not grounds for rejection. Option D incorrectly suggests that the margin of error size (3.2 units) is relevant to the hypothesis test decision. The margin of error's magnitude doesn't directly influence whether we reject or fail to reject H₀. Study tip: Always check whether the null hypothesis value falls within the confidence interval. If it does, fail to reject H₀; if it doesn't, reject H₀. This creates a quick verification method for two-tailed tests.

Question 13

In comparing two independent groups using a t-test, Group A has n₁ = 8 subjects and Group B has n₂ = 12 subjects. If the assumption of equal variances is violated, which modification is most appropriate?

  1. Use the larger sample standard deviation for both groups in calculations
  2. Apply Welch's t-test with adjusted degrees of freedom using the Satterthwaite approximation (correct answer)
  3. Increase both sample sizes to n > 30 and proceed with the standard t-test
  4. Use the pooled variance but apply a Bonferroni correction to the significance level
  5. Transform the data using logarithms and then apply the standard equal-variance t-test
Explanation: When you encounter questions about comparing two independent groups with unequal variances, you're dealing with one of the key assumptions of the standard two-sample t-test. The standard t-test assumes equal population variances (homoscedasticity), but real data often violates this assumption. When variances are unequal, Welch's t-test provides the most appropriate solution. This modification uses separate variance estimates for each group rather than pooling them, and crucially, it adjusts the degrees of freedom using the Satterthwaite approximation. The formula accounts for the different sample sizes and variances, giving you: df≈(s12/n1+s22/n2)2(s12/n1)2n1−1+(s22/n2)2n2−1df \approx \frac{(s_1^2/n_1 + s_2^2/n_2)^2}{\frac{(s_1^2/n_1)^2}{n_1-1} + \frac{(s_2^2/n_2)^2}{n_2-1}} Option A is incorrect because using the larger standard deviation for both groups artificially inflates the variance estimate for one group, creating bias. Option C misses the point entirely—even with larger samples, unequal variances remain problematic, and the "n > 30 rule" doesn't magically fix variance inequality. Option D makes no sense because pooled variance assumes equal variances (the very assumption that's violated), and Bonferroni correction addresses multiple comparisons, not variance heterogeneity. Remember this pattern: when you see "unequal variances" in a t-test question, immediately think Welch's t-test. It's more robust than the standard t-test and is actually the default in most statistical software precisely because it handles both equal and unequal variance situations effectively.

Question 14

A two-sample t-test compares mean protein levels between diseased (n = 20, x̄ = 8.5, s = 2.1) and healthy (n = 18, x̄ = 7.2, s = 1.8) subjects. Assuming equal variances, what is the standard error of the difference in means?

  1. 0.48
  2. 0.63 (correct answer)
  3. 0.89
  4. 1.25
  5. 1.95
Explanation: When you encounter a two-sample t-test problem asking for the standard error of the difference in means, you're dealing with a fundamental concept in comparing groups. The standard error quantifies the uncertainty in estimating how different the population means truly are. For equal variances, you first calculate the pooled standard deviation: sp=(n1−1)s12+(n2−1)s22n1+n2−2s_p = \sqrt{\frac{(n_1-1)s_1^2 + (n_2-1)s_2^2}{n_1+n_2-2}}. With your data: sp=(19)(2.1)2+(17)(1.8)236=83.79+54.3636=3.84=1.96s_p = \sqrt{\frac{(19)(2.1)^2 + (17)(1.8)^2}{36}} = \sqrt{\frac{83.79 + 54.36}{36}} = \sqrt{3.84} = 1.96 The standard error of the difference is: SE=sp1n1+1n2=1.96120+118=1.960.106=1.96(0.325)=0.63SE = s_p\sqrt{\frac{1}{n_1} + \frac{1}{n_2}} = 1.96\sqrt{\frac{1}{20} + \frac{1}{18}} = 1.96\sqrt{0.106} = 1.96(0.325) = 0.63 This confirms answer B is correct. Answer A (0.48) likely results from calculation errors in the pooled variance or forgetting to include both sample size terms. Answer C (0.89) might come from incorrectly using the individual standard deviations without proper pooling. Answer D (1.25) could result from using an unpooled approach or major computational mistakes. Study tip: Always remember the two-step process: first pool the variances when assuming equal variances, then apply the standard error formula with both sample sizes. Practice this formula until it becomes automatic—it's essential for understanding confidence intervals and hypothesis tests in biostatistics.

Question 15

A one-sample t-test is conducted on 25 patients' cholesterol reduction after treatment. The sample mean reduction is 18 mg/dL with s = 12 mg/dL. Testing H₀: μ = 15 versus H₁: μ ≠ 15, what is the 95% confidence interval for the mean reduction?

  1. (13.08, 22.92)
  2. (13.27, 22.73)
  3. (13.05, 22.95) (correct answer)
  4. (13.52, 22.48)
  5. (14.18, 21.82)
Explanation: When you encounter confidence interval questions involving small samples (n < 30), you need to use the t-distribution rather than the normal distribution. This accounts for the additional uncertainty from estimating the population standard deviation with the sample standard deviation. For a 95% confidence interval with the t-distribution, the formula is: xˉ±tα/2×sn\bar{x} \pm t_{\alpha/2} \times \frac{s}{\sqrt{n}} Here, you have xˉ=18\bar{x} = 18, s=12s = 12, and n=25n = 25. With df = 24, the critical t-value for 95% confidence is approximately 2.064. The margin of error is: 2.064×1225=2.064×2.4=4.952.064 \times \frac{12}{\sqrt{25}} = 2.064 \times 2.4 = 4.95 This gives you the interval: 18±4.95=(13.05,22.95)18 \pm 4.95 = (13.05, 22.95), which is answer C. Answer A (13.08, 22.92) uses an incorrect t-value, likely rounded incorrectly or taken from the wrong degrees of freedom. Answer B (13.27, 22.73) appears to use the normal distribution z-value (1.96) instead of the t-value, which would give a margin of error of 4.73. Answer D (13.52, 22.48) uses an even smaller margin of error, possibly from using a 90% confidence level instead of 95%. Remember that small sample confidence intervals always require the t-distribution. Also note that the confidence interval calculation is independent of the null hypothesis value (15 mg/dL) - it's based solely on your sample statistics.

Question 16

Two independent groups of patients receive different treatments for anxiety. Group 1 (n₁ = 12) has a mean anxiety score of 45 with s₁ = 8, and Group 2 (n₂ = 15) has a mean score of 38 with s₂ = 10. Assuming equal variances, what is the pooled standard deviation for the two-sample t-test?

  1. 8.65
  2. 9.15 (correct answer)
  3. 9.32
  4. 9.78
  5. 10.24
Explanation: When you encounter a two-sample t-test with the assumption of equal variances, you need to calculate the pooled standard deviation, which combines the variability from both groups into a single estimate. This pooled estimate provides more statistical power than using separate standard deviations. The pooled standard deviation formula is: sp=(n1−1)s12+(n2−1)s22n1+n2−2s_p = \sqrt{\frac{(n_1-1)s_1^2 + (n_2-1)s_2^2}{n_1+n_2-2}} Substituting the given values: sp=(12−1)(8)2+(15−1)(10)212+15−2=11(64)+14(100)25=704+140025=210425=84.16=9.15s_p = \sqrt{\frac{(12-1)(8)^2 + (15-1)(10)^2}{12+15-2}} = \sqrt{\frac{11(64) + 14(100)}{25}} = \sqrt{\frac{704 + 1400}{25}} = \sqrt{\frac{2104}{25}} = \sqrt{84.16} = 9.15 The correct answer is B) 9.15. Answer A (8.65) likely results from incorrectly using the simple average of the two standard deviations: (8+10)/2 = 9, then making a calculation error. Answer C (9.32) might come from forgetting to subtract 1 from each sample size in the numerator, using n₁ and n₂ instead of (n₁-1) and (n₂-1). Answer D (9.78) could result from using the wrong denominator, perhaps just adding the sample sizes without subtracting 2. Remember that the pooled standard deviation will typically fall between the two individual standard deviations, but closer to the one from the larger sample. Always double-check that you're using (n-1) terms in both the numerator and denominator calculations.

Question 17

In a clinical study, researchers compare pain scores between two treatments using a two-sample t-test. They report t = -2.34 with df = 58 and p = 0.023. However, they realize that 3 patients in each group had measurement errors and should be excluded. If the recalculated test statistic becomes t = -2.89 with the reduced sample, what can be concluded about the impact of removing these outliers?

  1. The outliers were primarily in the control group, making the original difference appear smaller
  2. The outliers were reducing the apparent treatment effect, and removing them strengthened the evidence
  3. The outliers were equally distributed between groups but were increasing measurement noise throughout
  4. The outliers were inflating the variance estimate, and removing them improved the precision of the test (correct answer)
Explanation: When analyzing changes in t-test results after removing data points, focus on how the test statistic formula responds to changes in sample size, means, and variance. The t-statistic equals the difference in means divided by the pooled standard error, so any change that increases the numerator or decreases the denominator will increase the absolute value of t. Here, removing 6 patients (3 from each group) changed t from -2.34 to -2.89, making it more extreme. Since equal numbers were removed from both groups, the difference in means likely remained similar or increased slightly. However, the most significant impact comes from how outliers affect variance calculations. Measurement errors typically increase within-group variability, inflating the standard error in the denominator. When these error-prone measurements were removed, the pooled variance decreased substantially, reducing the standard error and making the test statistic larger in magnitude. Option A incorrectly assumes unequal distribution of outliers between groups. Option B uses imprecise language—the "treatment effect" (difference in means) may not have changed much; rather, our ability to detect it improved. Option C mentions equal distribution correctly but focuses on "measurement noise" generally rather than the specific statistical impact on variance estimation. Option D correctly identifies that outliers inflated the variance estimate. Removing them improved test precision by reducing the standard error, making the t-statistic more extreme and the p-value smaller. Remember: when outliers are removed equally from both groups in a t-test, look for variance reduction as the primary driver of statistical changes.

Question 18

A researcher is conducting a one-sample t-test to determine if the mean systolic blood pressure in a population of diabetic patients differs from the general population mean of 120 mmHg. The sample of 25 diabetic patients has a mean of 128 mmHg and standard deviation of 15 mmHg. If the researcher incorrectly uses a z-test instead of a t-test, how will this affect the p-value compared to the correct t-test approach?

  1. The p-value will be smaller because the z-distribution has thinner tails than the t-distribution (correct answer)
  2. The p-value will be larger because the z-distribution has thicker tails than the t-distribution
  3. The p-value will be essentially the same since n = 25 is large enough for normal approximation
  4. The p-value cannot be determined without knowing the population standard deviation for the z-test
Explanation: When using a z-test instead of the appropriate t-test with df = 24, the p-value will be smaller because the z-distribution has thinner tails than the t-distribution. The t-distribution accounts for additional uncertainty from estimating the population standard deviation, making it more conservative. Option B incorrectly states the relationship between tail thickness. Option C is wrong because even with n = 25, there's still a meaningful difference between t and z distributions. Option D misses the point that the researcher is using the sample standard deviation as if it were the population parameter.

Question 19

Two independent samples are being compared using a two-sample t-test. Group A has n1=20n_1 = 20, xˉ1=85\bar{x}_1 = 85, s1=12s_1 = 12. Group B has n2=25n_2 = 25, xˉ2=78\bar{x}_2 = 78, s2=15s_2 = 15. If the pooled variance approach is used, what assumption is most critical for the validity of this test that cannot be directly verified from the given information?

  1. The assumption that both population variances are equal, which requires additional testing or prior knowledge
  2. The assumption that both samples are normally distributed, which requires examining residual plots
  3. The assumption that the samples are independent, which requires knowledge of the sampling methodology (correct answer)
  4. The assumption that the population means are different, which is what the test is designed to evaluate
Explanation: The independence assumption is most critical and cannot be verified from the numerical data alone - it requires knowledge of how the samples were collected. While equal variances (A) is important for pooled variance, it can be assessed using F-tests or variance ratios. Normality (B) can be evaluated with the given data using various tests. Option D is incorrect because we don't assume the means are different; that's what we're testing. Independence violations can completely invalidate the test regardless of other assumptions being met.

Question 20

A clinical trial compares two treatments using a two-sample t-test. The result shows t = 2.15 with df = 38 and p = 0.037. The researcher concludes that Treatment A is superior to Treatment B. Which of the following represents the most significant limitation of this conclusion?

  1. The p-value is too close to 0.05, indicating borderline significance that requires replication studies
  2. Statistical significance does not necessarily imply clinical significance or practical importance of the difference (correct answer)
  3. The degrees of freedom suggest unequal sample sizes, which violates the assumptions of the t-test
  4. A two-tailed test should have been used instead of the implied one-tailed test for superiority claims
Explanation: The most significant limitation is that statistical significance (p < 0.05) doesn't automatically mean the difference is clinically meaningful or practically important. Option A incorrectly suggests that p-values near 0.05 are inherently problematic. Option C is wrong because unequal sample sizes don't violate t-test assumptions (df = 38 could come from various sample size combinations). Option D incorrectly assumes a one-tailed test was used - the researcher's conclusion of 'superiority' could be based on a two-tailed test showing a significant difference in a particular direction.