CI for a Mean - Biostatistics

A researcher calculates a 90% confidence interval for mean reaction time as $(245, 275)$ milliseconds using $n = 10$ subjects. The same researcher wants to be 95% confident that the interval captures the true mean. Assuming the sample statistics remain the same, what must be true about the new interval?
Answer: The new interval will be approximately $(241, 279)$ ms, expanding proportionally based on the ratio of critical values for the two confidence levels When you encounter confidence interval questions involving changes in confidence level, remember that the interval width depends directly on the critical value from the t-distribution (since n = 10 requires t-distribution, not z-distribution). The confidence interval formula is $\bar{x} \pm t_{\alpha/2} \cdot \frac{s}{\sqrt{n}}$ . When you increase confidence from 90% to 95% while keeping sample statistics constant, only the critical value $t_{\alpha/2}$ changes. The margin of error scales proportionally to this change. For $n = 10$ (df = 9): $t_{0.05}$ (90% CI) ≈ 1.833 and $t_{0.025}$ (95% CI) ≈ 2.262. The ratio is $\frac{2.262}{1.833} ≈ 1.234$ . Since the original margin of error was 15 ms (275 - 245 = 30, so ±15), the new margin becomes $15 \times 1.234 ≈ 18.5$ ms. With the same sample mean of 260 ms, the new interval is approximately $(241.5, 278.5)$ or $(241, 279)$ ms. Answer B correctly identifies this proportional expansion based on critical value ratios. Answer A is conceptually right about the interval being wider but doesn't provide the calculation, making it incomplete. Answer C incorrectly suggests the expansion relates to the percentage change (90% to 95%), which isn't how confidence intervals work. Answer D incorrectly uses 1.25 as the expansion factor, confusing the confidence level ratio (95/90 = 1.056) with the critical value ratio. Study tip: Always use the t-distribution critical values for small samples, and remember that confidence interval width changes are determined by the ratio of critical values, not confidence percentages.
A medical device company tests the accuracy of a new blood glucose meter using 28 measurements. They obtain a mean difference of +2.1 mg/dL (compared to a reference standard) with a standard deviation of 4.8 mg/dL. The 95% confidence interval for the mean difference is calculated as (-0.76, +4.96). What conclusion about the meter's accuracy is most appropriate?
Answer: No significant bias can be concluded because the confidence interval includes zero, indicating the true bias could be zero When you encounter confidence intervals in biostatistics, you're looking at a range of plausible values for a population parameter. Here, the key question is whether the blood glucose meter has systematic bias—does it consistently read higher or lower than the true value? The correct interpretation focuses on what the confidence interval tells us about bias. Since the 95% CI (-0.76, +4.96) includes zero, we cannot conclude that significant bias exists. Zero bias means perfect accuracy on average, and since zero falls within our plausible range of values, the true population bias could indeed be zero. This makes answer C correct. Let's examine why the other options miss the mark. Option A commits a common error—confusing the sample statistic (+2.1 mg/dL) with statistical significance. A positive sample mean doesn't automatically indicate significant bias; you need the confidence interval to exclude zero for that conclusion. Option B misunderstands what the confidence interval represents and incorrectly relates interval width to accuracy assessment. The width relative to the standard deviation isn't the relevant comparison here. Option D completely misinterprets confidence intervals—they describe uncertainty about the population mean, not the range where 95% of individual measurements fall (that would be a prediction interval). Study tip: For confidence interval questions, always check whether the interval includes the null hypothesis value (often zero). If it does, you cannot conclude a significant effect exists. The sample statistic alone never determines statistical significance—you need the interval or a p-value.
A researcher reports a 99% confidence interval for mean body temperature as $(97.8, 98.6)°F$ based on a sample of $n = 25$ healthy adults. A colleague argues that a 95% confidence interval would be more appropriate for this research question. If the colleague recalculates using 95% confidence with the same data, which outcome is most likely?
Answer: The 95% interval will be $(97.9, 98.5)°F$ , narrower than the original but with the same center point of 98.2°F When you encounter confidence interval questions, remember that the interval width depends on the confidence level—higher confidence requires wider intervals to ensure the true parameter is captured more often. The correct approach starts with recognizing that both intervals use the same sample data, so they share the same sample mean (center point). The 99% interval $(97.8, 98.6)$ has a center of $\frac{97.8 + 98.6}{2} = 98.2°F$ , so the 95% interval must also center at $98.2°F$ . Since 95% confidence requires less certainty than 99%, you need a smaller margin of error. The confidence interval formula is $\bar{x} \pm t_{\alpha/2} \cdot \frac{s}{\sqrt{n}}$ . With $n = 25$ , the t-value decreases from approximately 2.797 (99%) to 2.064 (95%), creating a narrower interval while maintaining the same center. Answer choice A incorrectly shifts the center point to 98.2°F when calculating the midpoint of $(97.85, 98.55)$ , which gives 98.2°F, but the interval itself is asymmetric around the mean. Answer choice B makes the interval too narrow—the difference between t-values isn't large enough to shrink the interval from width 0.8°F to just 0.5°F. Answer choice C dramatically underestimates the interval width and misunderstands that lower confidence means less precision is acceptable, not required. Answer D correctly maintains the center at 98.2°F with $(97.9, 98.5)$ and shows appropriate narrowing from the reduced t-value. Study tip: Always check that confidence intervals with the same data maintain the same center point—only the width should change with different confidence levels.
A study measures the effectiveness of a new therapy on 24 patients. The improvement scores have a mean of 8.5 points with a standard deviation of 2.2 points. When calculating the 90% confidence interval, a student writes: "The interval is $8.5 ± 1.714 × \frac{2.2}{\sqrt{23}}$ ". What is incorrect about this setup?
Answer: The standard error calculation should use $\sqrt{24}$ in the denominator, not $\sqrt{23}$ When constructing confidence intervals for small samples, you need to carefully distinguish between sample size (n) and degrees of freedom (df), as they affect different parts of the calculation. The correct confidence interval formula is $\bar{x} \pm t_{\alpha/2,df} \times \frac{s}{\sqrt{n}}$ , where the standard error uses the full sample size n in the denominator. With 24 patients, the standard error should be $\frac{2.2}{\sqrt{24}}$ , not $\frac{2.2}{\sqrt{23}}$ . The sample size n always goes in the standard error calculation because you're dividing the sample standard deviation by the square root of how many observations you actually have. Now let's examine why the other options miss the mark. Option A incorrectly suggests the critical value 1.714 is wrong for a 90% confidence level—but actually, $t_{0.05,23} = 1.714$ is correct since 90% confidence means α = 0.10, so α/2 = 0.05. Option B confuses degrees of freedom with sample size; degrees of freedom for a t-test is always n-1, so 23 is correct here. Option C compounds this confusion by suggesting we need $t_{0.10,23}$ , but for a 90% confidence interval, we need the 0.05 critical value (α/2), not 0.10. Remember this key distinction: degrees of freedom (n-1) determines which t-distribution to use for the critical value, while sample size (n) goes in the standard error denominator. Don't let these two concepts get mixed up—they serve different purposes in the same formula.
A study reports that the 95% confidence interval for mean tumor diameter is $(2.4, 5.6)$ cm based on $n = 25$ patients. What is the approximate value of the sample standard deviation, given that the appropriate t-critical value is 2.064?
Answer: $s \approx 3.88$ cm because the margin of error equals the critical value times the standard error The margin of error is (5.6-2.4)/2 = 1.6 cm. Using ME = t×(s/√n): 1.6 = 2.064×(s/√25) = 2.064×(s/5). Solving for s: s = (1.6×5)/2.064 ≈ 3.88 cm. Choice A incorrectly calculates 1.94. Choice B (9.70) confuses interval width with standard deviation. Choice D (0.78) gives the standard error rather than the sample standard deviation.
A study of 30 elderly patients measures their walking speed with a mean of 1.2 m/s and standard deviation of 0.3 m/s. The 95% confidence interval is calculated as (1.09, 1.31). If the same study were conducted with patients half this age but the same sample size and variability, and the mean walking speed increased to 1.8 m/s, what would be the new confidence interval?
Answer: (1.69, 1.91) because the margin of error remains the same since sample size and variability are unchanged When you encounter confidence interval questions, focus on what factors actually determine the interval width: sample size, variability (standard deviation), and confidence level. The population being studied doesn't change these mathematical components. A confidence interval has the form: $\text{mean} \pm \text{margin of error}$ . The margin of error depends on the critical value, standard deviation, and sample size: $\text{ME} = t \times \frac{s}{\sqrt{n}}$ . Since the new study has the same sample size (n = 30) and standard deviation (s = 0.3), the margin of error remains identical. From the original interval (1.09, 1.31), you can calculate the margin of error: $(1.31 - 1.09)/2 = 0.11$ . With the new mean of 1.8 m/s, the confidence interval becomes: $1.8 \pm 0.11 = (1.69, 1.91)$ . Answer A correctly identifies that the margin of error stays constant when sample size and variability are unchanged. Answer B incorrectly assumes younger patients create more variability, but the problem states variability remains the same. Answer C suggests the interval width should be proportional to the mean, which isn't how confidence intervals work – they depend on variability, not the mean's magnitude. Answer D mistakenly claims the critical value changes with different demographics, but the critical value depends only on the confidence level and degrees of freedom (which stay constant here). Study tip: Remember that confidence intervals are purely mathematical constructs. The population characteristics (age, demographics) don't affect the calculation unless they change the actual measured variability or sample size.
A medical researcher studying recovery times obtains a 95% confidence interval of (8.2, 11.8) days from a sample of 20 patients. If she wants to reduce the margin of error by half while maintaining the same confidence level, approximately how many patients should she sample?
Answer: 80 patients, because the margin of error is inversely proportional to the square root of sample size When you encounter confidence interval questions about changing sample sizes, focus on the fundamental relationship between margin of error and sample size. The margin of error formula is $ME = t \times \frac{s}{\sqrt{n}}$ , where t is the critical value, s is the sample standard deviation, and n is the sample size. Since the confidence interval is (8.2, 11.8), the current margin of error is (11.8 - 8.2)/2 = 1.8 days. To reduce this to 0.9 days (half the margin of error), you need to understand that margin of error is inversely proportional to the square root of sample size. If you want to reduce the margin of error by a factor of 2, you must increase the sample size by a factor of 2² = 4. Therefore: 20 × 4 = 80 patients. Option A incorrectly assumes a direct inverse relationship between margin of error and sample size, suggesting you only need to double from 20 to 40 patients. Option B mentions critical value changes, but the critical value stays the same when maintaining the same confidence level - this is a distractor that confuses students who think about changing confidence levels. Option D correctly identifies that you need to increase sample size by a factor of four, but miscalculates by thinking 20 × 4 = 100 instead of 80. Remember the "square root rule" for sample size calculations: to reduce margin of error by half, you need four times the sample size. This inverse square root relationship appears frequently in biostatistics, so memorize that cutting margin of error in half always requires quadrupling your sample.
A study of 15 patients finds a mean cholesterol level of 210 mg/dL with a standard deviation of 25 mg/dL. The researchers want to construct a confidence interval but are unsure about the confidence level. If they use $t_{0.01,14} = 2.977$ , what confidence level are they constructing?
Answer: 98% confidence, because $α = 0.02$ corresponds to 98% confidence with two-tailed critical regions When you encounter t-distribution critical values in confidence interval problems, you need to understand the relationship between the subscript notation and the confidence level. The key is recognizing that $t_{α/2,df}$ represents a two-tailed critical value. The notation $t_{0.01,14}$ means you're using a critical value where $α/2 = 0.01$ . Since this represents half of the total alpha level (due to the two-tailed nature of confidence intervals), the full significance level is $α = 2 × 0.01 = 0.02$ . The confidence level is calculated as $1 - α = 1 - 0.02 = 0.98$ or 98%. Answer A correctly identifies this as 98% confidence and properly explains that $α = 0.02$ corresponds to 98% confidence with two-tailed critical regions. Answer B makes a calculation error by stating that $α/2 = 0.01$ leads to 99% confidence, when it actually leads to 98%. Answer C misinterprets the subscript notation, incorrectly assuming $t_{0.01,14}$ means $α = 0.01$ rather than $α/2 = 0.01$ . Answer D simply states a common default without engaging with the given critical value. Remember this pattern: when you see $t_{value,df}$ notation, that subscript value represents $α/2$ , not the full alpha. Always double it to find the total significance level, then subtract from 1 to get your confidence level. This subscript interpretation is crucial for correctly working backwards from critical values to confidence levels.
A researcher reports: "The 90% confidence interval for mean systolic blood pressure is 128.4 ± 4.6 mmHg, based on 14 participants." When reviewing this result, which statement represents the most significant concern about the methodology?
Answer: The reported interval assumes the population is normally distributed, which may not be valid for blood pressure When evaluating confidence intervals, you need to consider the underlying assumptions that make the statistical procedure valid. The most critical assumption for constructing confidence intervals about means is that the data comes from a normally distributed population, especially with small sample sizes. The correct answer is C because blood pressure data often exhibits skewness or non-normal distributions in real populations. With only 14 participants, the Central Limit Theorem hasn't had enough opportunity to normalize the sampling distribution, making the normality assumption crucial but potentially violated. If the underlying distribution isn't normal, the confidence interval may not have the stated coverage probability (90% in this case), making the inference unreliable. Let's examine why the other options are less concerning: A is incorrect because 90% confidence intervals are perfectly acceptable in statistical reporting - there's no universal requirement for 95%. The choice of confidence level depends on the research context and desired precision. B is wrong because with 14 participants, you'd use a t-distribution (not an approximation) - this is exactly what the t-distribution was designed for with small samples. D is incorrect because the margin of error's size depends on the study's variability and sample size; 4.6 mmHg isn't inherently "too large" without knowing the study's specific context and goals. Study tip: Always check distributional assumptions first when evaluating statistical procedures. With small samples, normality becomes critical because you can't rely on the Central Limit Theorem to rescue you from non-normal populations.
A pharmaceutical company analyzes the concentration of an active ingredient in 15 tablets. They find $\bar{x} = 250.4$ mg, $s = 12.6$ mg. The quality control specification requires that they report whichever is wider: a 90% confidence interval or a 95% confidence interval. What should they report?
Answer: The 95% interval (243.4, 257.4) because it is wider and meets the specification requirement When you encounter confidence interval comparison problems, remember that confidence level and interval width have a direct relationship: higher confidence requires wider intervals to capture the true parameter with greater certainty. The question specifically states that the company must report whichever interval is wider. Since both intervals use the same sample data ( $n=15$ , $\bar{x}=250.4$ , $s=12.6$ ), the only difference is the critical t-value. A 95% confidence interval uses $t_{0.025,14} = 2.145$ , while a 90% interval uses $t_{0.05,14} = 1.761$ . The larger critical value for 95% confidence creates a wider interval: $250.4 \pm 2.145 \times \frac{12.6}{\sqrt{15}} = (243.4, 257.4)$ versus $(244.7, 256.1)$ for 90%. Choice B correctly identifies that the 95% interval is wider and meets the specification requirement. This is the only answer that directly addresses what the company was asked to do. Choice A incorrectly suggests reporting the narrower 90% interval, directly contradicting the specification to report the wider one. Choice C wrongly claims the difference is negligible when there's a clear 2 mg difference in width that could matter for quality control. Choice D makes an unsupported claim about pharmaceutical standards and still recommends the wrong (narrower) interval. Study tip: In confidence interval problems, always check what's actually being asked. Don't assume "better precision" (narrower intervals) is always preferred—sometimes regulatory or specification requirements override statistical preferences.
A researcher wants to estimate mean hemoglobin levels and needs the margin of error to be no more than 0.5 g/dL with 95% confidence. A pilot study with $n = 12$ patients yielded $s = 1.8$ g/dL. Approximately how many additional patients should be recruited beyond the original 12?
Answer: 42 additional patients, accounting for the t-distribution throughout the iterative calculation process Using ME = t×(s/√n) ≤ 0.5, we need t×(1.8/√n) ≤ 0.5. Starting with z ≈ 1.96: n ≥ (1.96×1.8/0.5)² ≈ 49.8. But we need t-values. For n = 50, df = 49, t ≈ 2.01, giving n ≥ (2.01×1.8/0.5)² ≈ 52.2. Trying n = 54: t ≈ 2.006, n ≥ 52.0. So approximately 54 total needed, minus 12 existing = 42 additional. Choice A uses z incorrectly. Choice C ignores the t-distribution. Choice D uses an incorrect approach to the calculation.
A medical researcher calculates a 95% confidence interval for mean glucose level and obtains $(105.2, 118.8)$ mg/dL. Due to budget constraints, the study could only recruit $n = 12$ patients instead of the planned $n = 20$ . If the researcher had achieved the original target sample size with the same sample mean and standard deviation, how would the confidence interval have changed?
Answer: The interval would be approximately $(106.9, 117.1)$ mg/dL, moderately narrower with the same center point maintained Current: n=12, center=112, ME=6.8. For df=11: t₀.₀₂₅,₁₁ ≈ 2.201. So s/√12 = 6.8/2.201 ≈ 3.09, giving s ≈ 10.7. For n=20: t₀.₀₂₅,₁₉ ≈ 2.093, new ME = 2.093×(10.7/√20) ≈ 2.093×2.39 ≈ 5.0. New interval ≈ (107.0, 117.0). Choice C is closest. Choice A gives ME≈4.2 (too small). Choice B gives ME≈3.5 (much too small). Choice D gives ME≈2.9 (far too small) and incorrectly mentions 'exponential' improvement.
A researcher studying enzyme activity levels obtains the following sample statistics: n = 19, $\bar{x} = 45.2$ units, s = 6.8 units. She wants to construct a 98% confidence interval but accidentally looks up $t_{0.02,18} = 2.214$ instead of the correct critical value. How will this error affect her confidence interval?
Answer: The interval will be too narrow, providing less than 98% confidence, because she should have used $t_{0.01,18} = 2.552$ When constructing confidence intervals with small samples, you must carefully match the critical value to your desired confidence level. For a t-distribution, the critical value depends on both the degrees of freedom and the area in the tails. For a 98% confidence interval, you want 98% of the area in the middle, leaving 2% in the tails (1% in each tail). With df = n - 1 = 18, you need $t_{0.01,18} = 2.552$ . The researcher mistakenly used $t_{0.02,18} = 2.214$ , which corresponds to 96% confidence (leaving 4% in the tails, or 2% in each tail). Since confidence interval width equals $2t \times \frac{s}{\sqrt{n}}$ , using a smaller critical value (2.214 instead of 2.552) creates a narrower interval. This narrower interval captures the true population mean less frequently than the intended 98% of the time—specifically, only 96% of the time. Answer D correctly identifies this: the interval is too narrow and provides less than 98% confidence because she should have used $t_{0.01,18} = 2.552$ . Answer A makes the right conclusion but confuses the alpha values—she used $α = 0.04$ when she needed $α = 0.02$ , not the reverse. Answer B incorrectly states the interval is too wide when it's actually too narrow. Answer C is wrong because $t_{0.02,18}$ gives 96% confidence, not 98%. Study tip: Remember that for confidence intervals, the subscript on the t-value equals $\frac{α}{2}$ , where $α = 1 - \text{confidence level}$ . For 98% confidence, $α = 0.02$ , so you need $t_{0.01}$ .
A nutritionist measures the vitamin C content (mg) in 12 orange juice samples. The sample mean is 58.3 mg and the sample standard deviation is 4.2 mg. When computing the 95% confidence interval, what is the margin of error?
Answer: $2.201 × \frac{4.2}{\sqrt{12}} = 2.67$ When calculating confidence intervals for small samples with unknown population standard deviation, you need three key components: the critical value, the sample standard deviation, and the standard error formula. Since you have a small sample (n=12) and don't know the population standard deviation, you must use the t-distribution rather than the normal distribution. With 12 samples, your degrees of freedom = n - 1 = 11. For a 95% confidence interval with 11 degrees of freedom, the critical t-value is 2.201. The margin of error formula is: $t_{critical} × \frac{s}{\sqrt{n}}$ Substituting the values: $2.201 × \frac{4.2}{\sqrt{12}} = 2.201 × 1.213 = 2.67$ This confirms answer A is correct. Answer B uses the wrong denominator in the standard error calculation. It shows $\sqrt{11}$ instead of $\sqrt{12}$ . While 11 is the degrees of freedom, the sample size n=12 goes in the denominator for standard error. Answer C uses 1.96 as the critical value, which is the z-score for 95% confidence with large samples. Since this is a small sample with unknown population standard deviation, you need the t-distribution, not the normal distribution. Answer D combines two errors: it uses both the wrong critical value (should be from t-table, not shown correctly) and the wrong denominator ( $\sqrt{11}$ instead of $\sqrt{12}$ ). Study tip: For small samples (n<30) with unknown σ, always use t-distribution. Remember: degrees of freedom for the critical value, but actual sample size n for the standard error denominator.
A quality control manager samples 16 items and finds the mean weight is 50.2 grams with a standard deviation of 2.8 grams. She constructs a 99% confidence interval and gets (48.1, 52.3). Her colleague argues they should use a z-interval instead of a t-interval. If they used $z_{0.005} = 2.576$ , what would the new interval be?
Answer: (48.4, 52.0) because the z-critical value is smaller than the t-critical value, creating a narrower interval When you encounter confidence interval problems involving small samples, you need to decide between t-intervals and z-intervals. The key distinction is that t-intervals use the sample standard deviation (unknown population σ), while z-intervals assume the population standard deviation is known. Let's calculate what happens when switching from t to z. The original t-interval is (48.1, 52.3) with mean 50.2. The margin of error is 50.2 - 48.1 = 2.1. For a t-interval with n=16, we use df=15, and the 99% critical value is approximately 2.947. This gives us: margin of error = 2.947 × (2.8/√16) = 2.947 × 0.7 = 2.06 ≈ 2.1. Using the z-critical value instead: margin of error = 2.576 × 0.7 = 1.80. The new interval becomes 50.2 ± 1.80 = (48.4, 52.0). Answer A is correct because the z-critical value (2.576) is indeed smaller than the t-critical value (≈2.947), creating a narrower interval. Answer B incorrectly suggests z-intervals are more accurate for small samples—actually, t-intervals are more appropriate when σ is unknown. Answer C gets the concept right but produces the wrong interval calculation. Answer D mentions population standard deviation correctly but suggests an adjustment that makes the interval wider, which contradicts the mathematical relationship. Study tip: Remember that t-critical values are always larger than z-critical values for the same confidence level, so switching from t to z always produces narrower intervals. For small samples with unknown σ, stick with t-intervals.
A researcher collects data on the daily sodium intake (mg) for 16 adults and calculates a sample mean of 2400 mg with a sample standard deviation of 320 mg. When constructing a 95% confidence interval for the population mean, which of the following critical values should be used?
Answer: $t_{0.025,15} = 2.131$ When you encounter confidence interval problems with small sample sizes and unknown population standard deviation, you need to determine whether to use the t-distribution or z-distribution, and identify the correct degrees of freedom and alpha level. Since you have a sample of 16 adults with an unknown population standard deviation (only the sample standard deviation of 320 mg is given), you must use the t-distribution rather than the normal distribution. The t-distribution accounts for the additional uncertainty introduced by estimating the population standard deviation from sample data. For a 95% confidence interval, your confidence level is 0.95, making α = 0.05. Since you're constructing a two-tailed interval, you need the critical value that leaves α/2 = 0.025 in each tail. The degrees of freedom equal n - 1 = 16 - 1 = 15. Choice A is correct: $t_{0.025,15} = 2.131$ uses the proper α/2 = 0.025 and correct degrees of freedom of 15. Choice B is wrong because $t_{0.05,15} = 1.753$ uses α = 0.05 instead of α/2 = 0.025, which would give you a 90% confidence interval, not 95%. Choice C is incorrect because $t_{0.025,16} = 2.120$ uses 16 degrees of freedom instead of 15. The degrees of freedom should always be n - 1, not n. Choice D is wrong because $z_{0.025} = 1.96$ uses the normal distribution instead of the t-distribution. The z-distribution is only appropriate when the population standard deviation is known or when sample sizes are large (typically n ≥ 30). Study tip: Remember the t-distribution formula: use n - 1 degrees of freedom when the population standard deviation is unknown, and always use α/2 for two-tailed confidence intervals.
A researcher calculates a 95% confidence interval for mean reaction time as (0.82, 1.18) seconds from a sample of 20 subjects. She then realizes that all her measurements were recorded in seconds, but she needs to report the results in milliseconds. After converting the data, what will be the new confidence interval?
Answer: (820, 1180) milliseconds, obtained by multiplying both endpoints by 1000 When you encounter questions about transforming confidence intervals, remember that confidence intervals behave predictably under linear transformations. The key insight is that if you transform your original data by multiplying each value by a constant, the confidence interval endpoints transform in exactly the same way. Since there are 1000 milliseconds in 1 second, converting from seconds to milliseconds requires multiplying each measurement by 1000. When you multiply all the original data points by 1000, the sample mean gets multiplied by 1000, and consequently, both endpoints of the confidence interval must also be multiplied by 1000. Therefore: (0.82 × 1000, 1.18 × 1000) = (820, 1180) milliseconds, making A correct. Looking at the wrong answers: B shows (0.00082, 0.00118), which would result from dividing by 1000 instead of multiplying—this converts milliseconds to seconds, the opposite direction. C gives (82, 118), which comes from multiplying by 100 rather than 1000, confusing the conversion factor between seconds and milliseconds. D shows (8.2, 11.8), obtained by dividing by 100, which both uses the wrong conversion factor and goes in the wrong direction. Study tip: For any linear transformation of data (adding/subtracting constants, multiplying/dividing by constants), confidence intervals transform using the exact same mathematical operation. Always double-check your unit conversions—1000 milliseconds = 1 second, so going from seconds to milliseconds means multiplying by 1000.
A clinical trial measures pain reduction scores for 22 patients. The sample mean is 6.8 points with a standard deviation of 2.1 points. When constructing a 95% confidence interval, the researcher uses the formula $\bar{x} \pm t_{0.025,21} \cdot \frac{s}{\sqrt{n-1}}$ . What error has been made?
Answer: The standard error formula should use $\sqrt{n}$ in the denominator, not $\sqrt{n-1}$ When constructing confidence intervals for small samples with unknown population standard deviation, you need to use the t-distribution and get the standard error formula exactly right. The correct formula for a confidence interval is $\bar{x} \pm t_{\alpha/2,df} \cdot \frac{s}{\sqrt{n}}$ . The standard error of the mean is always $\frac{s}{\sqrt{n}}$ , where n is the sample size. In this case, with 22 patients, the denominator should be $\sqrt{22}$ , not $\sqrt{21}$ . The researcher incorrectly used $\sqrt{n-1}$ instead of $\sqrt{n}$ in the standard error calculation, making answer B correct. Let's examine why the other options are wrong: A) The degrees of freedom is correctly 21, calculated as n-1 = 22-1 = 21 when using the t-distribution. C) For a 95% confidence interval, you need $t_{0.025,21}$ because you want 2.5% in each tail of the two-tailed distribution (0.05/2 = 0.025). D) Using the sample standard deviation s is appropriate when the population standard deviation σ is unknown, which is typical in real-world scenarios. The confusion likely stems from mixing up the degrees of freedom formula (n-1) with the standard error formula (which uses n). Remember: degrees of freedom uses n-1, but standard error always uses the full sample size n in the denominator. This distinction is crucial for accurate confidence interval calculations.
A researcher calculates a 95% confidence interval for mean daily calorie intake as (1847, 2153) calories based on 25 subjects. Later, she discovers that one extreme outlier was incorrectly included in the analysis. After removing this outlier, she recalculates with 24 subjects and finds the sample mean is now 1950 calories with a standard deviation of 180 calories. What is the new 95% confidence interval?
Answer: (1874.2, 2025.8) because the smaller sample size increases the critical value from 2.064 to 2.069 When analyzing confidence intervals with changing sample conditions, you need to carefully consider how sample size affects the critical value and how outliers impact both the mean and standard deviation. To find the new 95% confidence interval, use the formula: $\bar{x} \pm t_{\alpha/2} \cdot \frac{s}{\sqrt{n}}$ With n = 24, df = 23, so the critical t-value is 2.069. The margin of error is: $2.069 \times \frac{180}{\sqrt{24}} = 2.069 \times 36.74 = 76.0$ This gives the interval: $1950 \pm 76.0 = (1874.0, 2026.0)$ The closest match is answer B at (1874.2, 2025.8), which correctly identifies that the critical value increased from 2.064 (for n=25) to 2.069 (for n=24). Answer A incorrectly assumes the mean decreased significantly, but we're told it's now 1950 calories, which is actually near the center of the original interval (1847, 2153). Answer C wrongly suggests that reduced variability "more than compensates" for smaller sample size, but the confidence interval width depends on both factors working together—the smaller sample size still increases the critical value. Answer D exaggerates the impact, claiming the critical value increases "substantially," when the change from 2.064 to 2.069 is relatively modest. Study tip: When sample size changes in confidence interval problems, always recalculate the degrees of freedom and corresponding t-value. Small changes in sample size near n=25 produce only modest changes in critical values, so be wary of answers claiming dramatic effects.
A researcher studying sleep quality measures 35 participants and calculates a 99% confidence interval for mean sleep duration as (6.8, 8.4) hours. A colleague suggests that since the sample size is greater than 30, they should use a z-interval instead of a t-interval. If they recalculate using $z_{0.005} = 2.576$ , which statement best describes the expected change?
Answer: The change cannot be determined without knowing whether the population standard deviation is known or estimated When deciding between t-intervals and z-intervals for confidence intervals, you need to consider both sample size and whether the population standard deviation is known. While the "n > 30" rule is commonly taught, it oversimplifies the decision process. The key insight here is that we're using the sample standard deviation (not the population standard deviation), which means a t-interval is theoretically correct regardless of sample size. However, you're right that t-values do approach z-values as sample size increases. With n = 35, we have 34 degrees of freedom, and $t_{0.005,34} = 2.728$ compared to $z_{0.005} = 2.576$ . Since the z-critical value is smaller, switching to a z-interval would make the interval narrower: (6.8, 8.4) would become approximately (6.9, 8.3) hours. This creates the illusion of improved precision, but it's actually less accurate because we're ignoring the additional uncertainty from estimating the population standard deviation. Looking at the wrong answers: Answer A correctly identifies that the z-interval will be narrower but incorrectly suggests this "improves precision" when it actually underestimates uncertainty. Answer B gets the direction wrong—z-intervals are narrower, not wider. Answer C overstates how similar the intervals are; while they're closer than with small samples, the difference is still meaningful. Answer D incorrectly claims z-intervals are more accurate for large samples when using sample standard deviations. Study tip: Always use t-intervals when working with sample standard deviations, regardless of sample size. The "n > 30" rule applies mainly when you're forced to approximate and don't have statistical software available.

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