t-Distribution - Biostatistics

1. Two research teams study the same drug using identical protocols but different sample sizes. Team A (n=9) reports a 95% CI of (14.2, 21.8) mg/dL, while Team B (n=25) reports (15.1, 20.9) mg/dL. Assuming both teams observed the same sample mean and population standard deviation, what can be concluded about their analyses?
   Answer: Both analyses are consistent; Team B's narrower interval reflects the improved precision from larger sample size and smaller t-critical value When comparing confidence intervals from studies with different sample sizes, you need to understand how sample size affects interval width through the standard error and t-distribution critical values. Both teams observed the same sample mean, so their intervals should have identical midpoints. Team A's interval (14.2, 21.8) has a midpoint of 18.0, and Team B's interval (15.1, 20.9) also centers at 18.0, confirming consistent sample means. The key difference is interval width. Team B's larger sample size (n=25 vs n=9) provides two advantages: the standard error decreases by $\sqrt{n}$, and the t-critical value becomes smaller due to more degrees of freedom (24 vs 8). Both factors make Team B's interval narrower, reflecting improved precision from the larger sample. Answer C correctly identifies this relationship - both analyses are consistent, with Team B's narrower interval reflecting improved precision from larger sample size and smaller t-critical value. Answer A incorrectly suggests Team A's wider interval is "more reliable" and "conservative." While Team A does use a larger t-critical value due to fewer degrees of freedom, this doesn't make it more reliable - it's simply less precise. Answer B wrongly assumes the analyses are inconsistent and suggests Team A used inappropriate values. The wider interval is actually expected given the smaller sample size. Answer D correctly notes that identical means should produce the same midpoint, but incorrectly claims this makes the analyses inconsistent. Different interval widths are expected with different sample sizes. Study tip: Remember that larger sample sizes always produce narrower confidence intervals due to reduced standard error and smaller critical values - this indicates better precision, not inconsistency.
2. Two researchers independently estimate the mean of the same population using t-distributions. Researcher A uses $n_A = 10$ observations, while Researcher B uses $n_B = 20$ observations. Both construct 95% confidence intervals with identical sample means and standard deviations. How do their intervals compare?
   Answer: Researcher A's interval is wider because both the t-value is larger and the standard error $\frac{s}{\sqrt{n}}$ is larger When comparing confidence intervals constructed using t-distributions, you need to consider both components of the margin of error: the critical t-value and the standard error of the mean. The confidence interval formula is: $\bar{x} \pm t_{\alpha/2,df} \cdot \frac{s}{\sqrt{n}}$ Let's examine how sample size affects each component. The critical t-value decreases as degrees of freedom increase: $t_{0.025,9} = 2.262$ versus $t_{0.025,19} = 2.093$. So Researcher A has a larger t-value. More importantly, the standard error $\frac{s}{\sqrt{n}}$ decreases as sample size increases. With identical standard deviations, Researcher A's standard error is $\frac{s}{\sqrt{10}}$ while Researcher B's is $\frac{s}{\sqrt{20}} = \frac{s}{\sqrt{10}} \cdot \frac{1}{\sqrt{2}}$, making A's standard error larger by a factor of $\sqrt{2} \approx 1.41$. Answer C correctly identifies that Researcher A's interval is wider because both the t-value AND the standard error are larger. Answer A incorrectly claims both researchers have the same standard error—this ignores the sample size difference. Answer B has the relationship backwards; larger samples produce narrower intervals due to reduced variability, not wider ones. Answer D ignores that identical confidence levels don't guarantee identical widths when sample sizes differ. Study tip: Remember that confidence interval width depends on both the critical value AND the standard error. Larger samples always reduce interval width through both smaller t-values and smaller standard errors.
3. A researcher wants to test whether the mean weight loss from a diet program differs from 5 pounds. With $n = 14$ participants, she calculates a t-statistic of $t = -2.45$. Using $\alpha = 0.05$ for a two-tailed test, what can she conclude if $t_{0.025,13} = 2.160$?
   Answer: Reject $H_0$ because $|-2.45| = 2.45 > 2.160$, so the mean differs significantly from 5 pounds When you encounter a two-tailed t-test question, focus on whether the absolute value of your test statistic exceeds the critical value. This determines if your result falls in the rejection region. In this weight loss study, you're testing $H_0: \mu = 5$ versus $H_1: \mu \neq 5$ pounds. With a two-tailed test at $\alpha = 0.05$, you reject the null hypothesis if $|t| > t_{0.025,13} = 2.160$. Since $|-2.45| = 2.45 > 2.160$, you reject $H_0$ and conclude the mean weight loss differs significantly from 5 pounds. Looking at the wrong answers: Option B incorrectly assumes that a negative t-statistic alone indicates insufficient evidence. The sign of t only tells you the direction of the difference, not its significance. Option C makes two errors—it uses the wrong comparison ($t < -2.160$ instead of $|t| > 2.160$) and incorrectly interprets what rejecting $H_0$ means about the direction of weight loss. Option D incorrectly describes the acceptance region as $(-2.160, 2.160)$, but since $|-2.45| = 2.45$ exceeds 2.160, the test statistic actually falls in the rejection region. Study tip: For two-tailed tests, always work with absolute values when comparing to critical values. The rejection region consists of both tails: $t < -t_{\alpha/2}$ OR $t > t_{\alpha/2}$, which simplifies to $|t| > t_{\alpha/2}$. Don't let negative test statistics confuse you—focus on the magnitude.
4. A biostatistician is comparing the effectiveness of three different methods for determining confidence intervals for small samples (n = 6): Method A uses t-distribution with df = 5, Method B uses t-distribution with df = 6, and Method C uses normal distribution. For a sample with mean 24.5 and standard deviation 3.2, which method produces the widest 95% confidence interval, and why?
   Answer: Method A produces the widest interval because it correctly uses n-1 degrees of freedom, requiring the largest critical value for small samples Method A uses the correct df = n-1 = 5, giving t₀.₀₂₅ = 2.571. Method B incorrectly uses df = 6, giving t₀.₀₂₅ = 2.447. Method C uses z₀.₀₂₅ = 1.96. Since smaller degrees of freedom lead to larger critical values, Method A (correct approach) produces the widest interval. The t-distribution approaches the normal distribution as df increases, so t-values are always larger than z-values for small samples.
5. A psychologist compares reaction times before and after caffeine consumption in $n = 15$ subjects. The differences (after - before) have mean $\bar{d} = -12.4$ ms and standard deviation $s_d = 18.6$ ms. What type of t-test should be performed and what are the appropriate degrees of freedom?
   Answer: Paired t-test with $df = 14$ because we have matched pairs from the same subjects When analyzing experiments where the same subjects are measured under two different conditions, you need to identify whether you're dealing with independent groups or matched pairs. This distinction determines both your statistical test and degrees of freedom calculation. Answer A is correct because this is a classic paired t-test scenario. You have the same 15 subjects measured before and after caffeine consumption, creating matched pairs. The key indicator is that you're given difference scores ($\bar{d} = -12.4$ ms and $s_d = 18.6$ ms), which is exactly what paired t-tests analyze. With $n = 15$ pairs, your degrees of freedom equal $n - 1 = 14$. Answer B is wrong because while you are comparing two conditions, they're not independent samples. The $df = 28$ would only apply if you had 15 subjects in one group and 15 different subjects in another group. Here, the same people are measured twice, violating the independence assumption of two-sample tests. Answer C is wrong because although you do have one sample of difference scores, this misses the fundamental design. You don't simply have 15 random differences—you have 15 paired differences from a before-after design. Also, $df = 15$ incorrectly suggests you don't lose any degrees of freedom. Answer D is wrong because it correctly identifies needing $df = 14$ but incorrectly calls for a two-sample test, which doesn't apply to paired data. Study tip: Look for keywords like "same subjects," "before and after," or "matched pairs," and check if difference scores are provided—these signal paired t-tests with $df = n - 1$.
6. A pharmaceutical researcher tests whether a new drug reduces tumor size. With 9 patients, she obtains a test statistic of t = -2.15. She concludes the drug is effective at α = 0.05 using a one-tailed test. However, her supervisor argues that a two-tailed test should have been used due to the possibility of tumor growth. What is the most accurate assessment of this situation?
   Answer: The one-tailed conclusion stands because |-2.15| > 1.860, but the two-tailed test would not show significance since |-2.15| < 2.306 With df = 8, the critical values are t₀.₀₅ = 1.860 (one-tailed) and t₀.₀₂₅ = 2.306 (two-tailed). Since |t| = 2.15 > 1.860, the one-tailed test shows significance. However, 2.15 < 2.306, so the two-tailed test would not reach significance at α = 0.05. This illustrates why the choice of one-tailed vs. two-tailed testing affects conclusions and should be made a priori based on research hypotheses.
7. A researcher collects a sample of size $n = 12$ from a normally distributed population with unknown variance. If the sample mean is $\bar{x} = 45.2$ and the sample standard deviation is $s = 6.8$, what is the appropriate distribution to use for constructing a confidence interval for the population mean?
   Answer: t-distribution with $df = 11$ degrees of freedom When you encounter a confidence interval problem involving a sample from a normally distributed population, the key decision is whether to use a normal distribution or t-distribution. This choice depends on whether the population variance is known and the sample size. Since you have a small sample ($n = 12$) from a normally distributed population with unknown variance, you must use the t-distribution. The sample standard deviation $s = 6.8$ is your estimate of the unknown population standard deviation, which introduces additional uncertainty that the t-distribution accounts for through its heavier tails. The degrees of freedom for a one-sample t-test or confidence interval is always $df = n - 1$. With $n = 12$, you get $df = 12 - 1 = 11$ degrees of freedom, making answer B correct. Let's examine why the other options fail: Answer A incorrectly treats the sample standard deviation as if it were the known population standard deviation and uses the normal distribution. Answer C uses the t-distribution correctly but miscalculates the degrees of freedom as $n = 12$ instead of $n - 1 = 11$. Answer D makes the same fundamental error as A by using the normal distribution when the population variance is unknown, even though it correctly calculates the standard error. Study tip: Remember the decision tree: unknown population variance + small sample (typically $n < 30$) from normal population = t-distribution with $df = n - 1$. Only use the normal distribution when the population variance is actually known, not estimated.
8. A medical researcher tests whether a new treatment reduces recovery time. She collects data from $n = 12$ patients and finds $\bar{x} = 8.2$ days with $s = 2.5$ days. For testing $H_0: \mu \geq 10$ versus $H_1: \mu < 10$, what is the p-value if the calculated t-statistic is $t = -2.49$?
   Answer: $P(T_{11} \leq -2.49)$, which is the area to the left of -2.49 in the t-distribution with 11 df When you encounter a hypothesis test with a calculated t-statistic, determining the p-value requires understanding both the alternative hypothesis direction and the sampling distribution properties. This is a one-tailed (left-tailed) test because $H_1: \mu < 10$ specifies a direction. The researcher wants to know if the treatment reduces recovery time below 10 days. With $n = 12$ patients, you have $df = n - 1 = 11$ degrees of freedom for the t-distribution. The p-value represents the probability of observing a test statistic as extreme or more extreme than what was calculated, assuming $H_0$ is true. Since this is a left-tailed test and $t = -2.49$, you need the probability of getting a t-statistic of -2.49 or smaller (more negative). This is exactly $P(T_{11} \leq -2.49)$, making A correct. B is wrong because it uses the right tail, which would be appropriate for a right-tailed test ($H_1: \mu > 10$). C is incorrect because multiplying by 2 creates a two-tailed p-value, but this is a one-tailed test with a directional alternative hypothesis. D makes a degrees-of-freedom error by using the sample size (12) instead of $n-1 = 11$. Study tip: Always match the p-value calculation to your alternative hypothesis direction. Left-tailed tests use the left tail area, right-tailed tests use the right tail, and two-tailed tests double the smaller tail area. Remember that degrees of freedom for a one-sample t-test is always $n-1$.
9. The researchers calculate a 95% confidence interval for mean drug concentration as (12.3, 18.7) μg/mL. If they want to reduce the width of this interval by 25% while maintaining the same confidence level, and assuming the population standard deviation remains constant, approximately how many total subjects would they need?
   Answer: Approximately 19 subjects, since reducing interval width by 25% requires increasing sample size by approximately 78% When you encounter confidence interval problems involving changes to interval width, you're dealing with the fundamental relationship between sample size and precision. The key insight is that confidence interval width is inversely proportional to the square root of sample size. The width of a confidence interval is $2 \times t \times \frac{s}{\sqrt{n}}$. To reduce width by 25% means the new width should be 75% of the original, or multiplied by 0.75. Since width is inversely proportional to $\sqrt{n}$, if you want the width to be 0.75 times smaller, you need $\sqrt{n}$ to be $\frac{1}{0.75} = 1.33$ times larger. Therefore, the new sample size must be $(1.33)^2 = 1.78$ times the original size. With 11 original subjects: $11 \times 1.78 ≈ 19$ subjects. Choice A incorrectly focuses on the t-distribution's degrees of freedom effect, which is minimal and doesn't drive the primary relationship. Choice B falsely assumes a linear relationship between sample size and interval width – this is wrong because the relationship involves a square root. Choice D correctly identifies that you need to square something, but applies it incorrectly to the reduction factor itself rather than the reciprocal. Study tip: Remember the "inverse square root rule" for confidence intervals: to cut the width in half, you need four times the sample size. For any proportional width reduction, divide 1 by that proportion, then square the result to find the sample size multiplier.
10. A quality analyst compares the tensile strength of materials from two suppliers using samples of $n_1 = 7$ and $n_2 = 9$. She assumes equal variances and calculates a pooled t-statistic of $t = 2.85$. For testing equality of means at $\alpha = 0.01$, what should she conclude if $t_{0.005,14} = 2.977$?
    Answer: Fail to reject $H_0$ because $t = 2.85 < t_{0.005,14} = 2.977$, providing insufficient evidence When you encounter a two-sample t-test with pooled variances, you're testing whether two population means are significantly different. The key is properly comparing your calculated t-statistic to the critical value at your chosen significance level. Here's the setup: You have $n_1 = 7$ and $n_2 = 9$ samples, giving you $df = n_1 + n_2 - 2 = 14$ degrees of freedom. For a two-tailed test at $\alpha = 0.01$, you need $t_{0.005,14} = 2.977$ as your critical value (the 0.005 comes from splitting the 0.01 alpha between two tails). Since your calculated $t = 2.85$ is less than the critical value of 2.977, you fail to reject the null hypothesis. The difference isn't statistically significant at the 1% level. Option A incorrectly states that 2.85 > 2.977, which is mathematically wrong. Option C makes the mistake of thinking any positive t-value indicates significance—but statistical significance depends on exceeding the critical value, not just being positive. Option D incorrectly calculates degrees of freedom; for two independent samples, it's always $n_1 + n_2 - 2$, not $n_1 + n_2 - 1$. Study tip: Always double-check your inequality direction when comparing test statistics to critical values. For rejection, your calculated statistic must exceed the critical value in absolute terms. Many students mix this up under exam pressure, so practice writing out "reject if $|t| > t_{critical}$" until it's automatic.
11. A pharmaceutical company tests a new drug on $n = 9$ patients. The mean reduction in blood pressure is $\bar{x} = 12.6$ mmHg with standard deviation $s = 4.3$ mmHg. To test $H_0: \mu = 10$ versus $H_1: \mu > 10$, what is the calculated t-statistic?
    Answer: $t = \frac{12.6 - 10}{\frac{4.3}{\sqrt{9}}} = 1.814$ When you encounter a one-sample t-test question, you're testing whether a sample mean differs significantly from a hypothesized population mean. The key is recognizing that you need the standard error of the mean, not just the sample standard deviation. The t-statistic formula for a one-sample test is: $t = \frac{\bar{x} - \mu_0}{s/\sqrt{n}}$, where $\bar{x}$ is the sample mean, $\mu_0$ is the hypothesized mean, $s$ is the sample standard deviation, and $n$ is the sample size. Here, you calculate: $t = \frac{12.6 - 10}{4.3/\sqrt{9}} = \frac{2.6}{4.3/3} = \frac{2.6}{1.433} = 1.814$. This matches answer choice B. Choice A incorrectly uses the sample standard deviation (4.3) directly in the denominator instead of the standard error. This is a common mistake that ignores sampling variability. Choice C uses $\sqrt{8}$ instead of $\sqrt{9}$ in the denominator. This error stems from confusing the sample size (n = 9) with degrees of freedom (df = n-1 = 8). While degrees of freedom matter for the critical value, the standard error calculation always uses the actual sample size. Choice D squares the standard deviation in the denominator, which has no statistical basis and dramatically inflates the denominator. Study tip: Remember the pattern: for one-sample t-tests, always divide the sample standard deviation by $\sqrt{n}$ to get the standard error. The degrees of freedom (n-1) only come into play when finding critical values from the t-table, not when calculating the test statistic itself.
12. A food scientist compares the protein content of two brands of cereal using independent samples. Brand A: $n_1 = 8$, $\bar{x}_1 = 12.4$ g, $s_1 = 2.1$ g. Brand B: $n_2 = 12$, $\bar{x}_2 = 10.8$ g, $s_2 = 1.9$ g. Assuming equal population variances, what is the pooled standard deviation?
    Answer: $s_p = \sqrt{\frac{(8-1)(2.1)^2 + (12-1)(1.9)^2}{8+12-2}} = \sqrt{\frac{30.87 + 39.71}{18}} = 1.98$ When comparing means from two independent samples with assumed equal variances, you need the pooled standard deviation to estimate the common population standard deviation. This combines information from both samples to get a more reliable estimate than using either sample alone. The pooled standard deviation formula is $s_p = \sqrt{\frac{(n_1-1)s_1^2 + (n_2-1)s_2^2}{n_1+n_2-2}}$. Notice that it weights each sample's variance by its degrees of freedom ($n-1$), not just the sample size. This gives you an unbiased estimate of the population variance. Choice A correctly applies this formula: $s_p = \sqrt{\frac{(8-1)(2.1)^2 + (12-1)(1.9)^2}{8+12-2}} = \sqrt{\frac{7(4.41) + 11(3.61)}{18}} = \sqrt{\frac{30.87 + 39.71}{18}} = 1.98$. Choice B incorrectly weights by sample sizes (8 and 12) rather than degrees of freedom (7 and 11), which would bias the estimate. Choice C makes two errors: it weights by sample sizes AND averages the standard deviations directly instead of variances, completely ignoring that variances (not standard deviations) are additive. Choice D simply averages the two variances without any weighting, throwing away valuable information about sample sizes. Remember: pooled variance calculations always use degrees of freedom as weights, and you must work with variances ($s^2$) in the calculation, then take the square root at the end to get back to standard deviation units.
13. A biostatistician receives conflicting advice about analyzing data from 8 subjects. Advisor A recommends using df = 7 for the t-test because "you always use n-1 degrees of freedom." Advisor B suggests using df = 6 because "with such a small sample, you need to be more conservative." Advisor C proposes df = 8 because "the sample size itself determines degrees of freedom." Which advisor gives the most appropriate guidance for a standard one-sample t-test?
    Answer: Advisor A is correct; degrees of freedom equal n-1 because one parameter (sample mean) is estimated from the data, reducing available information In a one-sample t-test, degrees of freedom = n-1 because we estimate one parameter (the population mean) using the sample mean. This estimation 'uses up' one degree of freedom, leaving n-1 for the test statistic. This is a fundamental property of the t-distribution derivation, not a matter of conservatism or preference. Advisor B's approach would be overly conservative without theoretical justification, and Advisor C's approach would underestimate the uncertainty in the test.
14. As the degrees of freedom in a t-distribution increase from 5 to 30, which statement best describes what happens to the distribution?
    Answer: The distribution becomes less peaked and approaches the standard normal distribution with thinner tails When analyzing how t-distributions behave as degrees of freedom change, you're examining one of the most important convergence properties in statistics. The t-distribution was specifically designed to handle situations with small sample sizes, and its shape depends entirely on the degrees of freedom parameter. As degrees of freedom increase from 5 to 30, the t-distribution undergoes a predictable transformation. With more degrees of freedom, the distribution becomes less peaked (lower kurtosis) and its tails become progressively thinner. Most importantly, it approaches the standard normal distribution as a limiting case. This happens because larger degrees of freedom indicate more information in your sample, reducing the uncertainty that creates the t-distribution's characteristic heavy tails. Answer C correctly captures this behavior - the distribution becomes less peaked and approaches normality with thinner tails. Answer A is wrong because the t-distribution always remains symmetric (never skewed) and has no relationship to the chi-square distribution. Answer B incorrectly states the distribution becomes more peaked with thicker tails - this is exactly backwards from what actually occurs. Answer D is wrong because t-distributions don't shift laterally (they remain centered at zero) and while variance does decrease with more degrees of freedom, the distribution doesn't shift right. Remember this key principle: as degrees of freedom increase, t-distributions always converge toward the standard normal. By about 30 degrees of freedom, they're practically indistinguishable from $N(0,1)$. This is why many statistical tables switch from t to z at around 30 df.
15. A clinical researcher analyzes data from 7 patients and calculates a test statistic t = 2.95 for testing H₀: μ = 100 vs H₁: μ ≠ 100. She reports a p-value of 0.026. A colleague suggests that with such a small sample, the normality assumption should be verified before trusting the t-test results. What is the most appropriate response to this concern?
    Answer: The concern is valid; with n = 7, the t-test requires strict normality, and the significant result should be interpreted cautiously without normality verification When you encounter t-test questions involving small sample sizes, you need to carefully consider the underlying assumptions, particularly normality of the population distribution. The t-test relies on the assumption that either the population is normally distributed OR the sample size is large enough for the Central Limit Theorem to apply. With only 7 patients, this is a small sample where the CLT hasn't kicked in yet, making the normality assumption critical for valid results. Answer D correctly identifies that with n = 7, strict normality is required for the t-test to be valid. Without verifying this assumption, the calculated p-value of 0.026 could be misleading, and any conclusion about rejecting H₀ should be made cautiously. Small samples are particularly sensitive to departures from normality, which can substantially affect the actual Type I error rate. Answer A incorrectly assumes the CLT applies regardless of sample size when using t-tests. The CLT typically requires n ≥ 30 for non-normal populations, and even then, it applies to the sampling distribution shape, not the validity of the t-distribution assumption. Answer B dismisses a legitimate concern. The t-test is NOT robust to non-normality with such small samples, and treating the p-value as reliable without checking assumptions is poor statistical practice. Answer C mentions the wrong issue. The problem isn't about adjusting critical values for non-normal data, but about whether the t-distribution is even appropriate given potential assumption violations. Study tip: For t-tests, remember the "rule of 30" - below n = 30, always verify normality assumptions. Above n = 30, the CLT typically provides protection against moderate non-normality.
16. A nutritionist measures vitamin C content in $n = 8$ orange samples. The data follows a normal distribution with $\bar{x} = 65.4$ mg and $s = 7.2$ mg. If she wants to test $H_0: \mu = 60$ mg versus $H_1: \mu \neq 60$ mg at $\alpha = 0.01$, what critical value(s) should she use?
    Answer: $\pm t_{0.005,7} = \pm 3.499$ because it's a two-tailed test with $\alpha = 0.01$ and $df = 7$ When you encounter hypothesis testing with small samples and unknown population standard deviation, you need to determine the correct distribution and critical values. This scenario calls for a t-test since we have $n = 8$ (small sample) and only know the sample standard deviation $s = 7.2$ mg. The correct approach uses $\pm t_{0.005,7} = \pm 3.499$. Here's why: The alternative hypothesis $H_1: \mu \neq 60$ indicates a two-tailed test, so you split $\alpha = 0.01$ between both tails, giving $\alpha/2 = 0.005$ in each tail. The degrees of freedom equal $n - 1 = 8 - 1 = 7$. Therefore, you need $t_{0.005,7} = 3.499$, and since it's two-tailed, the critical values are $\pm 3.499$. Option B incorrectly suggests this is a one-tailed test. The $\neq$ symbol in $H_1$ always indicates two tails, not one. Option C makes two errors: it uses the wrong degrees of freedom (8 instead of 7) and fails to split alpha properly for a two-tailed test, using $t_{0.01,8}$ instead of $t_{0.005,7}$. Option D incorrectly applies the z-distribution. Even though the population is normally distributed, when $\sigma$ is unknown and estimated by $s$, you must use the t-distribution regardless of the population's distribution. Study tip: Remember the decision tree: unknown $\sigma$ = t-test, known $\sigma$ = z-test. For two-tailed tests, always divide your alpha by 2 when finding critical values, and degrees of freedom for one-sample t-tests always equal $n - 1$.
17. In a bioequivalence study with 10 subjects, the 90% confidence interval for the ratio of test to reference drug concentrations is calculated as (0.87, 1.19) using log-transformed data and back-transforming the results. If the regulatory acceptance range is (0.80, 1.25), what can be concluded about the analysis and its implications?
    Answer: Bioequivalence is established since the entire confidence interval falls within acceptance limits, and the t-distribution was appropriately used for the small sample In bioequivalence testing, 90% confidence intervals are standard (equivalent to two one-sided tests at α = 0.05). The interval (0.87, 1.19) falls entirely within the acceptance range (0.80, 1.25), so bioequivalence is established. With n = 10, df = 9, using t-distribution is correct for small samples. The 90% CI is appropriate for bioequivalence regardless of sample size.
18. A pharmacist tests the dissolution time of $n = 9$ tablets and finds $\bar{x} = 18.6$ minutes with $s = 3.2$ minutes. She constructs a 95% confidence interval and gets $(16.19, 21.01)$. If one additional tablet is tested and shows a dissolution time that makes the new sample mean $\bar{x}_{new} = 19.1$ minutes, how will this affect the confidence interval width?
    Answer: The width will decrease because the larger sample size reduces both the t-value and the standard error When you encounter confidence interval questions involving sample size changes, focus on how the width formula responds: $\text{Width} = 2 \times t_{\alpha/2} \times \frac{s}{\sqrt{n}}$ Adding one tablet increases the sample size from $n = 9$ to $n = 10$. This creates two beneficial effects that both reduce the interval width. First, the degrees of freedom increase from 8 to 9, which decreases the critical t-value (from approximately 2.306 to 2.262 for 95% confidence). Second, and more importantly, the standard error $\frac{s}{\sqrt{n}}$ decreases because you're dividing by $\sqrt{10}$ instead of $\sqrt{9}$. Even if the sample standard deviation $s$ changes slightly, the $\sqrt{n}$ effect dominates. Answer A correctly identifies that both the t-value and standard error decrease with larger sample size, narrowing the interval. Answer B incorrectly assumes the standard deviation will increase significantly—one additional data point rarely causes dramatic changes in $s$, and even if it did, the $\sqrt{n}$ increase would likely offset this. Answer C misses that going from $n = 9$ to $n = 10$ represents an 11% sample size increase, which produces noticeable effects on confidence interval width. Answer D confuses confidence interval center movement with width changes—the new mean of 19.1 is well within the original interval (16.19, 21.01), and interval width depends on variability measures, not mean location. Remember: Larger sample sizes almost always produce narrower confidence intervals due to reduced standard error, regardless of minor changes in the sample mean or standard deviation.
19. A quality control engineer takes a random sample of $n = 7$ measurements from a normally distributed process. If $\bar{x} = 24.8$ and $s = 2.1$, what is the 90% confidence interval for the population mean using $t_{0.05,6} = 1.943$?
    Answer: $24.8 \pm 1.943 \times \frac{2.1}{\sqrt{7}} = 24.8 \pm 1.541$ or $(23.26, 26.34)$ When you encounter a confidence interval problem with a small sample from a normal distribution, you need to use the t-distribution formula: $\bar{x} \pm t_{\alpha/2,df} \times \frac{s}{\sqrt{n}}$, where the degrees of freedom equal $n-1$. Let's work through this systematically. You have $n=7$ measurements, so $df = 7-1 = 6$. For a 90% confidence interval, you need $t_{0.05,6} = 1.943$ (given). The standard error of the mean is $\frac{s}{\sqrt{n}} = \frac{2.1}{\sqrt{7}} = 0.794$. Therefore: $24.8 \pm 1.943 \times 0.794 = 24.8 \pm 1.541$, giving you the interval $(23.26, 26.34)$. Answer A correctly applies this formula. Answer B makes a common error by using $\sqrt{6}$ instead of $\sqrt{7}$ in the denominator—perhaps confusing the degrees of freedom (6) with the sample size (7) in the standard error calculation. Answer C omits the square root entirely, using just $s$ instead of $\frac{s}{\sqrt{n}}$, which dramatically overestimates the margin of error. Answer D incorrectly squares the standard deviation and divides by $n$ without taking the square root, creating a completely wrong formula. Study tip: Always double-check that you're using $\sqrt{n}$ (not $n$ or $\sqrt{df}$) in the denominator of the standard error, and remember that degrees of freedom for a single sample is always $n-1$, not $n$.
20. Two independent samples from normal populations have $n_1 = 6$ and $n_2 = 10$ observations. If the population variances are unknown and unequal, what is the approximate degrees of freedom for Welch's t-test using $s_1^2 = 16$ and $s_2^2 = 25$?
    Answer: $df = \frac{(\frac{16}{6} + \frac{25}{10})^2}{\frac{(16/6)^2}{5} + \frac{(25/10)^2}{9}} = \frac{(5.167)^2}{1.421 + 0.694} = 12.7$ When you encounter a two-sample t-test with unequal variances, you need Welch's t-test, which uses a special formula to approximate the degrees of freedom. This situation arises when you can't assume equal population variances, making the standard pooled t-test inappropriate. The correct approach uses the Welch-Satterthwaite equation: $df = \frac{(\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2})^2}{\frac{(s_1^2/n_1)^2}{n_1-1} + \frac{(s_2^2/n_2)^2}{n_2-1}}$ Substituting our values: $df = \frac{(\frac{16}{6} + \frac{25}{10})^2}{\frac{(16/6)^2}{5} + \frac{(25/10)^2}{9}} = \frac{(2.667 + 2.5)^2}{\frac{7.111}{5} + \frac{6.25}{9}} = \frac{26.69}{1.422 + 0.694} = 12.6$ This matches option A's calculation and result of 12.7. Option B incorrectly uses $n_1$ and $n_2$ in the denominator instead of $(n_1-1)$ and $(n_2-1)$, which inflates the degrees of freedom. Option C gives you the degrees of freedom for a pooled t-test assuming equal variances—this would be wrong when variances are clearly unequal. Option D represents an overly conservative approach that some statisticians might use, but it's not the standard Welch approximation. Study tip: Always check whether the problem states "equal variances" or "unequal variances." If unequal (or unknown and you suspect they're different), use Welch's formula with $(n-1)$ terms in the denominator, not the sample sizes themselves.

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