Biostatistics Quiz: Ci For A Mean

20 questions · exam conditions

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Ci For A MeanQuestion 1 of 20

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?

37 additional patients, using the z-critical value since the target sample will be large enough

42 additional patients, accounting for the t-distribution throughout the iterative calculation process

49 additional patients, based on the standard formula with z-critical value for large samples

35 additional patients, using the initial t-critical value and adjusting for the existing sample size

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Biostatistics Quiz

Biostatistics Quiz: Ci For A Mean

Practice Ci For A Mean in Biostatistics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Ci For A Mean, giving you a quick way to practice the rules, question types, and explanations that matter most for Biostatistics.

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

All questions

Question 1

37 additional patients, using the z-critical value since the target sample will be large enough
42 additional patients, accounting for the t-distribution throughout the iterative calculation process (correct answer)
49 additional patients, based on the standard formula with z-critical value for large samples
35 additional patients, using the initial t-critical value and adjusting for the existing sample size

Explanation: 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.

Question 2

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?

$t_{0.025,15} = 2.131$ (correct answer)
$t_{0.05,15} = 1.753$
$t_{0.025,16} = 2.120$
$z_{0.025} = 1.96$
$t_{0.05,16} = 1.746$

Explanation: 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.

Question 3

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?

$2.201 × \frac{4.2}{\sqrt{12}} = 2.67$ (correct answer)
$2.179 × \frac{4.2}{\sqrt{11}} = 2.76$
$1.96 × \frac{4.2}{\sqrt{12}} = 2.38$
$2.201 × \frac{4.2}{\sqrt{11}} = 2.80$
$2.228 × \frac{4.2}{\sqrt{12}} = 2.70$

Explanation: 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.

Question 4

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?

98% confidence, because $α = 0.02$ corresponds to 98% confidence with two-tailed critical regions (correct answer)
99% confidence, because $t_{0.01,14}$ indicates $α/2 = 0.01$ so $α = 0.02$ and confidence = 98%
99% confidence, because $t_{0.01,14}$ indicates $α = 0.01$ so confidence level = 99%
95% confidence, because this is the standard confidence level used in medical research studies
90% confidence, because $t_{0.01,14}$ represents the one-tailed probability for this confidence level

Explanation: 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.

Question 5

A pharmaceutical company tests a new drug's effectiveness on 18 patients. The mean improvement score is 12.4 points with a standard deviation of 3.1 points. If the 90% confidence interval is (11.0, 13.8), what would happen to the interval if the sample size were increased to 72 patients with the same mean and standard deviation?

The interval would become narrower because the standard error decreases while the critical value remains approximately the same (correct answer)
The interval would become wider because larger samples require higher confidence to maintain statistical validity
The interval would remain the same width because the confidence level and sample statistics are unchanged
The interval would become narrower initially, then wider as the t-distribution approaches the normal distribution
The interval width cannot be predicted without knowing the exact distribution of the improvement scores

Explanation: When you encounter confidence interval questions involving sample size changes, focus on how sample size affects the standard error, which directly impacts interval width. The width of a confidence interval depends on the critical value (determined by confidence level and degrees of freedom) and the standard error. The standard error formula is

SE = \frac{s}{\sqrt{n}}

, where s is the standard deviation and n is the sample size. When sample size increases from 18 to 72 patients, the standard error decreases dramatically:

SE_{18} = \frac{3.1}{\sqrt{18}} = 0.73

versus

SE_{72} = \frac{3.1}{\sqrt{72}} = 0.37

. This halving of the standard error makes the confidence interval much narrower. The critical t-value remains essentially unchanged since both sample sizes use similar confidence levels. Choice A correctly identifies this relationship - larger samples reduce standard error while keeping the critical value approximately constant, resulting in narrower intervals. Choice B incorrectly suggests larger samples require higher confidence; sample size and confidence level are independent choices. Choice C ignores the fundamental relationship between sample size and standard error - even with identical means and standard deviations, the precision (interval width) changes with sample size. Choice D incorrectly suggests the interval would eventually widen; while the t-distribution does approach normal with larger samples, this effect is minimal and doesn't overcome the dramatic standard error reduction. Remember: larger samples always produce narrower confidence intervals (better precision) when other factors remain constant. The relationship

SE = \frac{s}{\sqrt{n}}

is key to confidence interval problems.

Question 6

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?

(1873.1, 2026.9) because removing the outlier decreased both the mean and the variability significantly
(1874.2, 2025.8) because the smaller sample size increases the critical value from 2.064 to 2.069 (correct answer)
(1878.5, 2021.5) because the reduced variability more than compensates for the smaller sample size
(1871.8, 2028.2) because the critical value increases substantially when the sample size decreases
(1880.0, 2020.0) because the outlier removal creates a more symmetric distribution requiring adjustment

Explanation: 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.

Question 7

Two researchers independently study the same population parameter. Researcher A uses n=25 and reports a 95% CI of (14.2, 17.8). Researcher B uses n=100 and reports a 95% CI of (15.1, 16.9). Both intervals are calculated correctly. Which statement best explains the relationship between these results?

Researcher B's interval is narrower because larger samples always produce more precise estimates with smaller standard errors
The intervals overlap substantially, suggesting both studies are measuring the same population parameter consistently
Researcher A's interval is wider due to using a larger t-critical value with fewer degrees of freedom
Researcher B's interval is more reliable because it's based on a sample size that meets the central limit theorem requirements
The difference in interval widths reflects both the sample size effect and potentially different sample variabilities (correct answer)

Explanation: When you encounter confidence interval comparison questions, focus on how sample size affects interval width and what overlapping intervals tell you about the underlying population parameter. Both researchers are studying the same population, and their 95% confidence intervals overlap substantially (14.2-17.8 vs. 15.1-16.9). This overlap strongly suggests they're both successfully capturing the true population parameter, which likely falls somewhere in the overlapping region around 15.1-16.9. The key insight is that different sample sizes from the same population should yield intervals that contain the same true value, even if the intervals have different widths. Option A is partially correct about larger samples producing narrower intervals, but it's incomplete—it doesn't address the crucial point about what these overlapping results tell us about the population parameter. Option B captures this relationship perfectly by noting that the substantial overlap indicates consistent measurement of the same parameter. Option C focuses on a technical detail about t-distributions but misses the bigger picture of what the results mean. Option D incorrectly suggests one study is more "reliable" based on sample size requirements, when both intervals are valid and informative. The most important takeaway is recognizing that overlapping confidence intervals from studies of the same population provide converging evidence about the true parameter value. Don't get distracted by technical details about sample size effects when the question is really asking you to interpret what the combined results tell you about the underlying population.

Question 8

A biostatistician calculates a confidence interval for mean glucose levels using data from 18 diabetic patients. She finds $\bar{x} = 142.5$ mg/dL, $s = 18.2$ mg/dL, and uses $t_{0.05,17} = 1.740$ to get the interval (135.0, 150.0). A colleague questions whether this represents a 90% or 95% confidence interval. Which is correct?

90% confidence, because $t_{0.05,17}$ indicates $α = 0.10$ for a two-tailed confidence interval (correct answer)
95% confidence, because $t_{0.05,17}$ indicates $α = 0.05$ corresponding to 95% confidence level
90% confidence, because the calculated interval width is consistent with this confidence level
95% confidence, because this is the standard confidence level unless otherwise specified in medical research
The confidence level cannot be determined without additional information about the intended Type I error rate

Explanation: When you encounter confidence interval questions involving t-statistics, the key is understanding the relationship between the subscript notation and the confidence level. The subscript in

t_{α/2, df}

represents the tail probability, not the confidence level itself. The notation

t_{0.05,17}

means we're using the t-value that cuts off 0.05 in each tail of the distribution. Since confidence intervals are two-tailed, we have 0.05 in the upper tail and 0.05 in the lower tail, totaling

α = 0.10

. The confidence level is

1 - α = 1 - 0.10 = 0.90

or 90%. Answer A correctly identifies this relationship:

t_{0.05,17}

indicates

α = 0.10

for a two-tailed interval, yielding 90% confidence. Answer B makes the common error of confusing the tail probability with the total alpha level. The subscript 0.05 represents the probability in each tail, not the total

α

. Answer C suggests using interval width to determine confidence level, but width alone cannot determine the confidence level without knowing the underlying distribution and sample size parameters. Answer D incorrectly assumes confidence levels based on convention rather than the mathematical relationship shown in the t-statistic notation. Study tip: Remember that in t-table notation, the first subscript is always the tail probability (

α/2

), so double it to find total

α

, then subtract from 1 to get the confidence level. For 95% confidence, you'd use

t_{0.025, df}

, not

t_{0.05, df}

Question 9

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?

(1.69, 1.91) because the margin of error remains the same since sample size and variability are unchanged (correct answer)
(1.67, 1.93) because younger patients require a larger confidence interval to account for increased variability
(1.71, 1.89) because the confidence interval width should be proportional to the sample mean
(1.68, 1.92) because the critical value changes when studying a different population demographic
(1.70, 1.90) because the standard error needs adjustment for the different age group characteristics

Explanation: 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.

Question 10

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?

The interval will be too narrow, providing less than 98% confidence, because she used $α = 0.04$ instead of $α = 0.02$
The interval will be too wide, providing more than 98% confidence, because she used a larger critical value than necessary
The interval will be correct because $t_{0.02,18}$ is the appropriate critical value for 98% confidence with 18 degrees of freedom
The interval will be too narrow, providing less than 98% confidence, because she should have used $t_{0.01,18} = 2.552$ (correct answer)
The error cannot be evaluated without knowing the exact confidence level she intended to achieve

Explanation: 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}

Question 11

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?

(820, 1180) milliseconds, obtained by multiplying both endpoints by 1000 (correct answer)
(0.00082, 0.00118) milliseconds, obtained by converting from seconds to milliseconds using the standard conversion
(82, 118) milliseconds, obtained by adjusting for the different measurement precision
(8.2, 11.8) milliseconds, obtained by dividing the original interval by 100 to account for unit differences
(820, 1180) milliseconds, but the confidence level decreases because the measurement scale affects interval precision

Explanation: 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.

Question 12

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?

The critical value should be $t_{0.05,23} = 1.714$ instead of using this value for the wrong confidence level
The degrees of freedom in the standard error should be 24, not 23, to match the sample size
The critical value should be $t_{0.10,23} = 1.319$ for a 90% confidence interval with 23 degrees of freedom
The standard error calculation should use $\sqrt{24}$ in the denominator, not $\sqrt{23}$ (correct answer)
The setup is correct; the critical value 1.714 is appropriate for 90% confidence with 23 degrees of freedom

Explanation: 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.

Question 13

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?

The z-interval will be slightly narrower because $z_{0.005} = 2.576 < t_{0.005,34} = 2.728$ , improving precision
The z-interval will be wider because the normal distribution has greater variability than the t-distribution
The intervals will be nearly identical because t-values approximate z-values when n > 30
The z-interval will provide more accurate results because it uses the correct theoretical distribution for large samples
The change cannot be determined without knowing whether the population standard deviation is known or estimated (correct answer)

Explanation: 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.

Question 14

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?

The 90% interval (244.7, 256.1) because it provides adequate confidence with better precision for quality control
The 95% interval (243.4, 257.4) because it is wider and meets the specification requirement (correct answer)
Either interval is acceptable because the difference in width is negligible for practical quality control purposes
The 90% interval (244.7, 256.1) because higher confidence levels are not necessary for pharmaceutical applications
The 95% interval (243.4, 257.4) because regulatory standards typically require this minimum confidence level

Explanation: 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.

Question 15

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?

The confidence level should be 95% rather than 90% to meet standard statistical reporting requirements
The sample size of 14 may be too small to justify using the t-distribution approximation effectively
The reported interval assumes the population is normally distributed, which may not be valid for blood pressure (correct answer)
The margin of error of 4.6 mmHg seems too large for a clinical study of this type
The calculation should account for potential correlation between multiple measurements on the same participants

Explanation: 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.

Question 16

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?

The meter is significantly biased because the sample mean difference is positive at +2.1 mg/dL
The meter accuracy is acceptable because the confidence interval is relatively narrow compared to the standard deviation
No significant bias can be concluded because the confidence interval includes zero, indicating the true bias could be zero (correct answer)
The meter is significantly accurate because 95% of individual measurements fall within the confidence interval range
The bias is inconclusive because the sample size of 28 is insufficient for reliable confidence interval estimation

Explanation: 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.

Question 17

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?

The degrees of freedom should be 22 instead of 21 for this sample size calculation
The standard error formula should use $\sqrt{n}$ in the denominator, not $\sqrt{n-1}$ (correct answer)
The confidence level requires $t_{0.05,21}$ instead of $t_{0.025,21}$ for a two-tailed test
The formula should use the population standard deviation σ rather than the sample standard deviation s
The critical value should be $t_{0.025,20}$ because the sample standard deviation has n-1 degrees of freedom

Explanation: 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.

Question 18

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?

(48.4, 52.0) because the z-critical value is smaller than the t-critical value, creating a narrower interval (correct answer)
(48.3, 52.1) because using the normal distribution provides a more accurate estimate for this sample size
(48.0, 52.4) because the z-interval calculation uses the same standard error but different critical value
(47.9, 52.5) because the z-interval assumes the population standard deviation is known, requiring adjustment
(48.2, 52.2) because the difference between t and z intervals is negligible at this sample size

Explanation: 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.

Question 19

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?

40 patients, because halving the margin of error requires doubling the sample size
60 patients, because the relationship involves both sample size and critical value changes
80 patients, because the margin of error is inversely proportional to the square root of sample size (correct answer)
100 patients, because reducing margin of error by half requires increasing sample size by a factor of four
120 patients, because the t-distribution critical values change significantly as sample size increases

Explanation: 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.

Question 20

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?

The 95% interval will be approximately $(97.85, 98.55)°F$ , moderately narrower while maintaining statistical validity
The 95% interval will be approximately $(97.95, 98.45)°F$ , significantly narrower due to the lower confidence requirement
The 95% interval will be $(98.0, 98.4)°F$ , much narrower because 95% confidence requires less precision than 99%
The 95% interval will be $(97.9, 98.5)°F$ , narrower than the original but with the same center point of 98.2°F (correct answer)

Explanation: 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.