One & Two-Sample t-Tests - Biostatistics

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?
Answer: No error; the conclusion is correct because |t| = 2.14 > 2.048 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 $\alpha = 0.05$ and critical values of $\pm 2.048$ . The calculated t-statistic is $-2.14$ . Since $|{-2.14}| = 2.14 > 2.048$ , the test statistic falls in the rejection region, so rejecting $H_0$ 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 $H_0$ 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 $H_0$ 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 $H_0$ was rejected, not accepted. Choice D incorrectly suggests using a one-tailed test, but the hypothesis $H_1: \mu_1 \neq \mu_2$ 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 $|t| >$ critical value, reject $H_0$ —the decision-making process here was executed properly.
In a two-sample t-test comparing mean recovery times between two surgical procedures, the researcher obtains a 95% confidence interval for the difference in means (μ₁ - μ₂) of (-2.3, 8.7) days. Based on this interval, what can be concluded about the corresponding two-tailed hypothesis test at α = 0.05?
Answer: The null hypothesis H₀: μ₁ = μ₂ would not be rejected because the interval contains zero Since the 95% confidence interval (-2.3, 8.7) contains zero, this indicates that μ₁ - μ₂ = 0 is a plausible value for the difference in population means. This directly corresponds to failing to reject H₀: μ₁ = μ₂ at α = 0.05. Option A incorrectly states the interval contains only positive values (it includes negative values from -2.3 to 0). Option B incorrectly focuses on interval width, which doesn't determine significance. Option D is wrong because the confidence interval provides complete information about the hypothesis test conclusion.
A researcher performs a one-sample t-test with H₀: μ = 50 versus H₁: μ ≠ 50. The sample yields $\bar{x} = 53.2$ , s = 8.1, n = 16. If the researcher had instead tested H₀: μ = 52 versus H₁: μ ≠ 52 with the same sample data, how would the results compare?
Answer: The new test would have a smaller t-statistic and larger p-value than the original test The t-statistic formula is $t = \frac{\bar{x} - \mu_0}{s/\sqrt{n}}$ . For the original test: t = (53.2-50)/(8.1/4) = 1.58. For the new test: t = (53.2-52)/(8.1/4) = 0.59. Since the sample mean (53.2) is closer to the new null value (52) than the original (50), the t-statistic is smaller in absolute value, resulting in a larger p-value. Option B reverses this relationship. Option C incorrectly suggests the t-statistics would be equal. Option D is wrong because the α level doesn't affect the relationship between the test statistics.
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?
Answer: Power would increase because the standard error decreases with larger sample size 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 $\frac{s}{\sqrt{n}}$ , so as sample size increases from 16 to 36, the standard error decreases from $\frac{45}{\sqrt{16}} = 11.25$ to $\frac{45}{\sqrt{36}} = 7.5$ . 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.
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?
Answer: The outliers were inflating the variance estimate, and removing them improved the precision of the test 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.
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?
Answer: -3.27 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: $t = \frac{\bar{x} - \mu_0}{s/\sqrt{n}}$ Here, you're testing if the mean difference equals zero, so $\mu_0 = 0$ . With $\bar{x} = -0.8$ , $s = 1.2$ , and $n = 24$ : $t = \frac{-0.8 - 0}{1.2/\sqrt{24}} = \frac{-0.8}{1.2/4.899} = \frac{-0.8}{0.245} = -3.27$ This confirms answer A is correct. Answer B (-2.67) likely comes from incorrectly using $\sqrt{25}$ instead of $\sqrt{24}$ 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 $n = 15$ . 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.
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?
Answer: 0.63 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: $s_p = \sqrt{\frac{(n_1-1)s_1^2 + (n_2-1)s_2^2}{n_1+n_2-2}}$ . With your data: $s_p = \sqrt{\frac{(19)(2.1)^2 + (17)(1.8)^2}{36}} = \sqrt{\frac{83.79 + 54.36}{36}} = \sqrt{3.84} = 1.96$ The standard error of the difference is: $SE = s_p\sqrt{\frac{1}{n_1} + \frac{1}{n_2}} = 1.96\sqrt{\frac{1}{20} + \frac{1}{18}} = 1.96\sqrt{0.106} = 1.96(0.325) = 0.63$ 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.
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?
Answer: The p-value is between 0.01 and 0.02 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 $df = n - 1 = 24$ 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.
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?
Answer: Fail to reject H₀ because |t| = 1.52 < 2.032 (critical value with df = 34) 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 $\alpha = 0.05$ . The degrees of freedom for a pooled t-test is $df = n_1 + n_2 - 2 = 20 + 16 - 2 = 34$ . With $df = 34$ and $\alpha = 0.05$ (two-tailed), the critical t-value is approximately 2.032. Since your calculated t-statistic of 1.52 falls within the acceptance region ( $|1.52| < 2.032$ ), 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.
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?
Answer: The sample data must follow a normal distribution approximately 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.
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?
Answer: The conclusion would change from reject H₀ to fail to reject H₀; the p-value would remain 0.03 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.
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?
Answer: 9.15 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: $s_p = \sqrt{\frac{(n_1-1)s_1^2 + (n_2-1)s_2^2}{n_1+n_2-2}}$ Substituting the given values: $s_p = \sqrt{\frac{(12-1)(8)^2 + (15-1)(10)^2}{12+15-2}} = \sqrt{\frac{11(64) + 14(100)}{25}} = \sqrt{\frac{704 + 1400}{25}} = \sqrt{\frac{2104}{25}} = \sqrt{84.16} = 9.15$ 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.
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?
Answer: Statistical significance does not necessarily imply clinical significance or practical importance of the difference 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.
A researcher conducts a two-sample t-test and reports: 'The mean difference was 4.8 units (95% CI: 1.2 to 8.4), t(48) = 2.67, p = 0.010.' A colleague questions whether the equal variances assumption was reasonable. If this assumption was violated and Welch's t-test should have been used instead, which result would most likely change?
Answer: The degrees of freedom would likely decrease and the p-value would likely increase When switching from pooled variance t-test to Welch's t-test (unequal variances), the degrees of freedom typically decrease from n₁+n₂-2 to the Welch approximation, which is usually smaller. The smaller df makes the critical t-value larger, generally resulting in a larger p-value and wider confidence interval (more conservative test). Option A incorrectly suggests the CI would narrow. Option C is wrong because while the standard error calculation changes, the t-statistic often changes only modestly. Option D is incorrect because both tests estimate the difference in means the same way ( $\bar{x}_1 - \bar{x}_2$ ).
Two research groups study the same research question using identical methodology but different sample sizes. Group 1 uses n₁ = 25 and obtains t = 2.201, p = 0.037. Group 2 uses n₂ = 100 and obtains the same sample mean and standard deviation as Group 1. Approximately what t-statistic would Group 2 obtain?
Answer: t ≈ 4.402, because the standard error decreases by a factor of 2 when sample size quadruples The t-statistic is $t = \frac{\bar{x} - \mu_0}{s/\sqrt{n}}$ . When n increases from 25 to 100 (a factor of 4), the standard error $s/\sqrt{n}$ decreases by $\sqrt{4} = 2$ . Since the numerator stays the same, the t-statistic doubles: 2.201 × 2 ≈ 4.402. Option A ignores the effect of sample size on standard error. Option B gives the right answer but wrong reasoning about proportionality. Option D incorrectly suggests the t-statistic would quadruple, confusing the relationship between sample size and standard error.
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?
Answer: The p-value will be smaller because the z-distribution has thinner tails than the t-distribution 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.
A medical researcher wants to determine the minimum sample size needed to detect a difference of 5 units between two groups using a two-sample t-test with 80% power and α = 0.05. If the common standard deviation is estimated as 12 units, and she calculates that n = 92 per group is needed, what would happen to the required sample size if she decided to use α = 0.01 instead?
Answer: The required sample size would increase because a smaller α requires stronger evidence to reject H₀ When you encounter sample size calculations for hypothesis testing, remember that four key factors are interconnected: the effect size you want to detect, the variability in your data, your desired power, and your significance level (α). Changing α from 0.05 to 0.01 makes your test more conservative—you're requiring stronger evidence before you'll reject the null hypothesis. Think of it this way: with α = 0.01, you need your test statistic to be more extreme to reach statistical significance compared to α = 0.05. To achieve the same 80% power (ability to detect a true difference when it exists), you need more data to generate that stronger evidence. Therefore, the required sample size increases. Let's examine why the other options miss the mark. Choice A incorrectly suggests the sample size would decrease and confuses stringency with power—a more stringent test (smaller α) actually requires more data, not less. Choice B brings up one-tailed versus two-tailed tests, but this distinction doesn't change the fundamental relationship between α and sample size; regardless of test type, smaller α values require larger samples. Choice C contains a critical misconception—while α directly controls Type I error, it absolutely affects the sample size needed to maintain a specified power level. Here's your key takeaway: α and sample size move in opposite directions. When α decreases (more conservative), sample size must increase to maintain the same power. When α increases (less conservative), you can get away with smaller samples. This inverse relationship is fundamental to study design.
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?
Answer: 18 patients 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: $df = n_1 + n_2 - 2$ , where $n_1$ and $n_2$ 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: $28 = 12 + n_2 - 2$ . Rearranging: $28 = 10 + n_2$ , so $n_2 = 18$ patients. This confirms that answer C is correct. Let's examine why the other options are wrong. Answer A (16 patients) would give you $df = 12 + 16 - 2 = 26$ , which is too low. Answer B (17 patients) yields $df = 12 + 17 - 2 = 27$ , still one degree of freedom short. Answer D (19 patients) produces $df = 12 + 19 - 2 = 29$ , 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.
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?
Answer: (13.05, 22.95) 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: $\bar{x} \pm t_{\alpha/2} \times \frac{s}{\sqrt{n}}$ Here, you have $\bar{x} = 18$ , $s = 12$ , and $n = 25$ . With df = 24, the critical t-value for 95% confidence is approximately 2.064. The margin of error is: $2.064 \times \frac{12}{\sqrt{25}} = 2.064 \times 2.4 = 4.95$ This gives you the interval: $18 \pm 4.95 = (13.05, 22.95)$ , 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.
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?
Answer: Fail to reject H₀ because 15 falls within the confidence interval 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 $\bar{x} = \frac{12.3 + 18.7}{2} = 15.5$ , 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.

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