All practice questions

Question 121Question →

An academic research group is designing a prospective cohort study to investigate the association between daily ultra-processed food consumption and the incidence of non-alcoholic fatty liver disease (NAFLD) in young adults. To ensure the study yields reliable conclusions, the principal investigator specifies an acceptable false-positive rate of $5\%$ . Through a pilot analysis and sample size estimation, the research team determines there is an $18\%$ probability that the study will fail to detect a true $1.5$ -fold increased risk of NAFLD, should one actually exist. Based on these specified parameters, what is the statistical power of this cohort study to detect the true association, expressed as a percentage?

Show answer & explanation

Answer: 82

Answer

82

Statistical power is defined as the probability of correctly rejecting the null hypothesis when it is indeed false (i.e., the ability of a study to detect a true effect or association). Mathematically, power is calculated as

1 - \beta

, where

\beta

is the Type II error rate. In this vignette, the researchers note an

18\%

chance of failing to detect a true

1.5

-fold increased risk. This failure to detect a true difference is a Type II error, meaning

\beta = 0.18

(or

18\%

). Therefore, the statistical power of the study is

1 - 0.18 = 0.82

, or

82\%

. The

5\%

false-positive rate mentioned in the stem represents the Type I error rate (

\alpha

), which is used to set the significance level but is not used in the direct arithmetic calculation of power when

\beta

is already known.

Step-by-Step Solution

1

Identify the Type II error rate (

\beta

) from the clinical scenario.

\beta = 18\%

The stem defines an

18\%

probability of failing to detect a true increased risk, which is the definition of a Type II error.

2

Identify the formula for statistical power.

\text{Power} = 1 - \beta

Statistical power represents the probability of correctly rejecting the null hypothesis when it is false, which is the complement of the Type II error rate.

3

Calculate the final percentage.

100\% - 18\% = 82\%

Substituting the

\beta

value into the formula provides the statistical power.

Key Concept

Statistical power is calculated as

1 - \beta

, where

\beta

represents the probability of committing a Type II error (failing to detect a true difference).

Estimated Time:1m 0s

Question 122Question →

A multidisciplinary team at an academic medical center is designing a prospective, randomized, double-blind, placebo-controlled trial to evaluate a new targeted biologic therapy for refractory ulcerative colitis. The primary endpoint is the rate of clinical remission at $52$ weeks. The trial is designed with a two-sided alpha level of $0.05$ . A pre-study statistical analysis determines that enrolling $280$ patients per arm will provide a $90\%$ probability of detecting a true absolute difference in clinical remission rates of $15\%$ or greater between the active drug and placebo.

If the actual difference in remission rates between the biologic and placebo is indeed $15\%$ , what is the probability (expressed as a percentage) that the trial will incorrectly fail to demonstrate a statistically significant difference between the two groups?

Show answer & explanation

Answer: 10

Answer

10

The scenario describes a study with a statistical power of

90\%

. Power is defined as the probability of correctly rejecting the null hypothesis when there is a true difference (i.e., avoiding a Type II error). The question asks for the probability of failing to detect a true difference, which is the exact definition of a Type II error (

\beta

). Since

\text{Power} = 1 - \beta

, the Type II error rate is calculated as

1 - 0.90 = 0.10

, which is

10\%

.

Step-by-Step Solution

1

Define statistical power in the context of the described study.

The statistical power is given as

90\%

.

Power is explicitly defined in the stem as the

90\%

probability of detecting a true absolute difference between the active drug and placebo.

2

Identify what statistical metric the question is asking to find.

The question asks for the probability of incorrectly failing to demonstrate a difference when a true difference exists.

This specific scenario defines a false-negative result, which corresponds to the Type II error rate (

\beta

).

3

Calculate the Type II error (

\beta

) from the given statistical power.

\beta = 100\% - 90\% = 10\%

.

Statistical power and Type II error are mathematically complementary probabilities (

1 - \beta = \text{Power}

).

Key Concept

Statistical power (

1 - \beta

) and Type II error (

\beta

) are complementary probabilities. A Type II error occurs when a study incorrectly fails to reject a false null hypothesis.

Question 123Question →

A $45$ -year-old man is brought to the emergency department after being found unresponsive in a garage. Beside him is a partially empty jug of antifreeze. Upon arrival, he is obtunded. His temperature is $37.0^\circ$ C ( $98.6^\circ$ F), blood pressure is $130/80$ mmHg, heart rate is $112$ /min, and respiratory rate is $24$ /min.

Laboratory studies are obtained and reveal the following:

Test	Result	Reference Range
Serum sodium	$138$ mEq/L	$136-145$ mEq/L
Serum potassium	$4.5$ mEq/L	$3.5-5.0$ mEq/L
Serum chloride	$102$ mEq/L	$98-106$ mEq/L
Serum bicarbonate	$12$ mEq/L	$22-28$ mEq/L
Blood urea nitrogen (BUN)	$14$ mg/dL	$8-20$ mg/dL
Serum creatinine	$1.5$ mg/dL	$0.7-1.3$ mg/dL
Serum glucose	$108$ mg/dL	$70-100$ mg/dL
Serum ethanol	Undetectable	Undetectable
Measured serum osmolality	$315$ mOsm/kg H $_2$ O	$275-295$ mOsm/kg H $_2$ O

Based on these findings, what is the patient's calculated osmolar gap?

Show answer & explanation

Answer: 28

Answer

28

The calculated serum osmolality is determined using the formula:

2 \times [\text{Na}^+] + [\text{BUN}]/2.8 + [\text{Glucose}]/18

. Plugging in the patient's values yields

2 \times 138 + 14/2.8 + 108/18 = 276 + 5 + 6 = 287 \text{ mOsm/kg H}_2\text{O}

. The osmolar gap is the measured osmolality minus the calculated osmolality, which is

315 - 287 = 28 \text{ mOsm/kg H}_2\text{O}

. An elevated osmolar gap (typically

>10

) alongside a high anion gap metabolic acidosis is a classic presentation for toxic alcohol poisoning, supported here by the presence of antifreeze (ethylene glycol).

Step-by-Step Solution

1

Calculate the expected serum osmolality using the standard physiological formula.

Calculated osmolality = 287 mOsm/kg H2O

This establishes the baseline osmoles contributed by the major plasma solutes (sodium, urea, and glucose).

2

Subtract the calculated osmolality from the measured osmolality.

315 - 287 = 28 mOsm/kg H2O

The difference between measured and calculated osmolality (osmolar gap) represents unmeasured osmotically active substances, such as toxic alcohols.

Key Concept

Interpreting Laboratory Results: Osmolar Gap Calculation

Estimated Time:1m 30s

Question 124Question →

A hospital quality improvement committee is designing a prospective intervention study to reduce catheter-associated urinary tract infections (CAUTIs) in the intensive care unit. They plan to compare a new silver-alloy catheter to the standard silicone catheter. A priori power analysis determines that 108 patients per intervention group are required to achieve a statistical power of $80\%$ ( $\beta = 0.20$ ) at a significance level of $\alpha = 0.05$ to detect a clinically meaningful reduction in CAUTI incidence. Based on prior hospital initiatives, the committee anticipates a $20\%$ loss of analyzable data due to early patient transfers or incomplete electronic health records.

To ensure the study maintains the desired $80\%$ power, what is the minimum total number of patients the committee must initially enroll across both groups combined?

Show answer & explanation

Answer: 270

Answer

270

Statistical power heavily relies on the final number of subjects analyzed at the conclusion of a study. If an a priori power calculation determines that 108 patients are required per group, and a

20\%

dropout rate is expected, enrolling exactly 108 patients would result in only

\approx 86

analyzable patients per arm. This severely diminishes the study's power and substantially increases the risk of a Type II error (failing to detect a true difference). To ensure the final analyzed sample meets the target for robust statistical power, the initial enrollment per group must account for this loss using the formula:

N_{enrolled} = \frac{N_{needed}}{1 - \text{attrition rate}}

. For this study,

N_{enrolled} = \frac{108}{1 - 0.20} = \frac{108}{0.80} = 135

patients per group. Because the trial compares two groups (a new silver-alloy catheter and a standard silicone catheter), the total minimum enrollment required across both arms combined is

135 \times 2 = 270

patients.

Step-by-Step Solution

1

Identify the target final sample size and anticipated attrition rate.

Target = 108 patients per group; Attrition = 20% (0.20).

These specific variables are required to accurately adjust the initial enrollment target upward.

2

Calculate the required initial enrollment per group.

108 / (1 - 0.20) = 135 patients per group.

Dividing the required final size by the proportion of patients expected to complete the study ensures that, after attrition, the group still contains 108 patients.

3

Calculate the total required enrollment across the entire study.

135 patients/group * 2 groups = 270 patients.

The study compares two separate groups (silver-alloy vs. standard catheter). Therefore, the calculated per-group enrollment must be multiplied by 2.

Key Concept

To maintain statistical power despite anticipated patient dropout or data loss, the initial enrollment target per study arm must be calculated as: Required Sample Size / (1 - Attrition Rate).

Question 125Question →

An attending physician is evaluating a hospitalized patient who has developed worsening lower extremity edema and frothy urine. Initial laboratory studies reveal a serum albumin level of $1.5$ g/dL. A basic metabolic panel is obtained concurrently, which demonstrates a measured total serum calcium level of $6.8$ mg/dL. Assuming a standard reference albumin value of $4.0$ g/dL for calculation purposes, what is this patient's corrected total serum calcium concentration?

Show answer & explanation

Answer: 8.8

Answer

The corrected total serum calcium concentration is 8.8 mg/dL.

The correct calculation yields 8.8 mg/dL. Approximately 40% to 45% of total serum calcium is bound to plasma proteins, primarily albumin. When albumin levels are significantly reduced (such as in nephrotic syndrome), the measured total calcium drops accordingly. However, the physiologically active ionized calcium generally remains normal. To avoid erroneously diagnosing true hypocalcemia (pseudohypocalcemia), the measured total calcium must be corrected by adding 0.8 mg/dL for every 1.0 g/dL decrease in serum albumin below the standard reference of 4.0 g/dL. In this patient, the albumin deficit is 2.5 g/dL, resulting in an upward correction of 2.0 mg/dL to the measured 6.8 mg/dL.

Step-by-Step Solution

1

Extract the given laboratory values from the clinical vignette.

Measured calcium =

6.8

mg/dL, Albumin =

1.5

g/dL.

These core variables are required to adjust total calcium for hypoalbuminemia.

2

Determine the difference between the standard reference albumin and the patient's albumin.

4.0 - 1.5 = 2.5

g/dL.

To calculate the correction factor, we must first quantify how much the albumin has decreased from the normal baseline.

3

Calculate the calcium correction factor.

2.5 \times 0.8 = 2.0

mg/dL.

For every

1.0

g/dL decrease in serum albumin below

4.0

, total serum calcium decreases by approximately

0.8

mg/dL due to reduced protein binding.

4

Add the correction factor to the measured total calcium.

6.8 + 2.0 = 8.8

mg/dL.

This yields the corrected total calcium, which accurately reflects the patient's physiologically active ionized calcium status.

Key Concept

Correcting total serum calcium in the setting of hypoalbuminemia to distinguish true hypocalcemia from pseudohypocalcemia.

Estimated Time:1m 0s

Question 126Question →

A 45-year-old man is brought to the emergency department after being found obtunded in his garage. His family reports that he has a history of severe alcohol use disorder and recent major depressive disorder. An empty container of commercial antifreeze was found nearby. Vital signs are temperature $36.5^\circ\text{C}$ ( $97.7^\circ\text{F}$ ), blood pressure $135/85\text{ mmHg}$ , pulse $98/\text{min}$ , and respiratory rate $26/\text{min}$ with deep inspirations (Kussmaul respirations). Physical examination reveals a comatose patient with no focal neurologic deficits.

Initial laboratory results are obtained:

Test	Result
Sodium	$140\text{ mEq/L}$
Potassium	$4.5\text{ mEq/L}$
Chloride	$105\text{ mEq/L}$
Bicarbonate	$10\text{ mEq/L}$
Blood urea nitrogen	$28\text{ mg/dL}$
Creatinine	$1.2\text{ mg/dL}$
Glucose	$108\text{ mg/dL}$
Serum osmolality (measured)	$326\text{ mOsm/kg}$
Arterial blood gas, pH	$7.15$

Based on the provided laboratory data, calculate the patient's serum osmolar gap. What is the value in $\text{mOsm/kg}$ ?

Show answer & explanation

Answer: 30

Answer

The correct serum osmolar gap is

30\text{ mOsm/kg}

.

The patient's osmolar gap is determined by subtracting the calculated serum osmolarity from the measured serum osmolality. The calculated osmolarity is

2 \times 140 + \frac{108}{18} + \frac{28}{2.8} = 280 + 6 + 10 = 296\text{ mOsm/L}

. The osmolar gap is

326 - 296 = 30\text{ mOsm/kg}

. An elevated osmolar gap (normal

\leq 10

) combined with a high anion gap metabolic acidosis in a patient found near antifreeze confirms ethylene glycol toxicity.

Step-by-Step Solution

1

Extract relevant laboratory values required for the osmolarity equation.

Sodium =

140\text{ mEq/L}

, Glucose =

108\text{ mg/dL}

, BUN =

28\text{ mg/dL}

.

These three solutes are the primary physiologic determinants of expected serum osmolarity.

2

Calculate the expected serum osmolarity using the standard formula.

(2 \times 140) + \frac{108}{18} + \frac{28}{2.8} = 280 + 6 + 10 = 296\text{ mOsm/L}

.

The formula standardizes glucose and BUN from mg/dL to mmol/L, and multiplying sodium by 2 accounts for its accompanying anions (chloride and bicarbonate).

3

Calculate the final osmolar gap.

326 \text{ (measured osmolality)} - 296 \text{ (calculated osmolarity)} = 30\text{ mOsm/kg}

.

The gap represents the presence of unmeasured, osmotically active substances in the blood, which in this clinical scenario strongly points to ethylene glycol toxicity.

Key Concept

The serum osmolar gap is calculated as: Measured Osmolality - Calculated Osmolarity, where Calculated Osmolarity =

2 \times [Na^+] + \frac{\text{Glucose}}{18} + \frac{\text{BUN}}{2.8}

. A normal gap is

\leq 10\text{ mOsm/kg}

; an elevated gap indicates the presence of unmeasured exogenous osmoles (e.g., toxic alcohols).

Alternative Method

In patients with suspected mixed ingestions, if an ethanol level is obtained, the calculated osmolarity formula should be expanded by adding

\frac{\text{Ethanol (mg/dL)}}{4.6}

to account for ethanol's osmotic contribution before calculating the final gap.

Estimated Time:1m 15s

Question 127Question →

A 45-year-old woman presents to the clinic with a 3-month history of chronic, watery diarrhea. She reports having 4 to 6 large-volume, loose bowel movements per day and notes that the diarrhea significantly improves when she fasts or avoids meals. Her medical history is notable for type 2 diabetes mellitus, which is managed with metformin and a recently initiated alpha-glucosidase inhibitor (acarbose). Her vital signs are within normal limits, and physical examination is unremarkable.

Laboratory results are as follows:

Serum chemistry:
- Sodium: $140$ mEq/L
- Potassium: $3.8$ mEq/L
- Chloride: $102$ mEq/L
- Bicarbonate: $24$ mEq/L

Stool studies:
- Stool osmolality: $290$ mOsm/kg
- Stool sodium: $40$ mEq/L
- Stool potassium: $35$ mEq/L
- Fecal leukocytes: Negative
- Fecal occult blood: Negative

Based on the provided laboratory values, calculate the patient's stool osmotic gap in mOsm/kg.

Show answer & explanation

Answer: 140

Answer

140

The correct calculation yields 140 mOsm/kg. An elevated stool osmotic gap (typically defined as > 100 mOsm/kg) strongly suggests the presence of unmeasured osmotically active substances in the bowel lumen. In this patient, the recent initiation of acarbose, which inhibits alpha-glucosidase and prevents carbohydrate absorption, leads to unabsorbed carbohydrates acting as osmotic agents, thus causing osmotic diarrhea.

Step-by-Step Solution

1

Identify the formula for stool osmotic gap.

Stool Osmotic Gap = Measured Stool Osmolality - Calculated Stool Osmolality.

This formula differentiates between osmotic and secretory diarrhea by estimating unmeasured osmotically active substances in the stool.

2

Determine the calculated stool osmolality.

Calculated osmolality = 2 \times (Stool Na + Stool K) = 2 \times (40 + 35) = 150 mOsm/kg.

Sodium and potassium, along with their accompanying anions, account for the majority of fecal osmolality in normal conditions.

3

Calculate the final gap.

Stool Osmotic Gap = 290 - 150 = 140 mOsm/kg.

Subtracting the calculated osmolality from the measured osmolality yields the gap, which is elevated (> 50-100) indicating osmotic diarrhea.

Key Concept

The stool osmotic gap is calculated as

290 - 2 \times (Stool Na + Stool K)

and is used to differentiate osmotic diarrhea (gap > 100 mOsm/kg) from secretory diarrhea (gap < 50 mOsm/kg).

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Alternative Method

Answer

Step-by-Step Solution

Key Concept

Common Mistakes