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In a quality-control calibration of a 50.00 mL50.00\text{ mL} Class A volumetric burette, five standardized trials yielded delivered volumes with a sample mean of Vˉ=49.962 mL\bar{V} = 49.962\text{ mL} and a sample standard deviation of s=0.018 mLs = 0.018\text{ mL}. Tolerable delivery volumes VV are defined by the experimental acceptance criterion ∣V−Vˉ∣≤2.776sn|V - \bar{V}| \le 2.776\frac{s}{\sqrt{n}} corresponding to a 95%95\% confidence threshold with n=5n = 5 (where t0.025,4=2.776t_{0.025, 4} = 2.776). Calculate the acceptable numerical interval for VV to three decimal places, and describe its precise graphical representation on a volumetric number line calibrated in mL\text{mL}.

Answer and steps

Final answer

Interval notation: [49.940 mL,49.984 mL][49.940\text{ mL}, 49.984\text{ mL}]; on the number line, solid closed circles at V=49.940 mLV = 49.940\text{ mL} and V=49.984 mLV = 49.984\text{ mL} with a shaded segment of total width 0.044 mL0.044\text{ mL} connecting them.

Steps

01

Identify the Given Parameters

From the calibration data:

  • Sample mean: Vˉ=49.962 mL\bar{V} = 49.962\text{ mL}
  • Sample standard deviation: s=0.018 mLs = 0.018\text{ mL}
  • Sample size: n=5n = 5
  • Critical value: t=2.776t = 2.776

The acceptance condition is given by the absolute value inequality: ∣V−Vˉ∣≤2.776sn|V - \bar{V}| \le 2.776\frac{s}{\sqrt{n}}

02

Compute the Acceptance Margin

Calculate the tolerance margin, EE: E=2.776⋅0.0185E = 2.776 \cdot \frac{0.018}{\sqrt{5}}

Evaluating the expression: 5≈2.23607\sqrt{5} \approx 2.23607 E=0.0499682.23607≈0.022346 mL≈0.022 mLE = \frac{0.049968}{2.23607} \approx 0.022346\text{ mL} \approx 0.022\text{ mL}

03

Solve the Absolute Value Inequality

An inequality of the form ∣X∣≤E|X| \le E is equivalent to the compound inequality −E≤X≤E-E \le X \le E: −0.022≤V−49.962≤0.022-0.022 \le V - 49.962 \le 0.022

Add 49.96249.962 across all parts of the inequality: 49.962−0.022≤V≤49.962+0.02249.962 - 0.022 \le V \le 49.962 + 0.022 49.940≤V≤49.98449.940 \le V \le 49.984

In interval notation, the acceptable volume range is: [49.940 mL,49.984 mL][49.940\text{ mL}, 49.984\text{ mL}]

04

Graph the Solution on a Number Line

To represent this continuous solution set geometrically:

  • Endpoints: Because the inequality uses "≤\le" (non-strict inequality), both endpoints are included in the solution set. Plot solid, closed circles at V=49.940 mLV = 49.940\text{ mL} and V=49.984 mLV = 49.984\text{ mL}.
  • Segment: Shade the continuous interval between the two points, which has a total length of: 49.984−49.940=0.044 mL49.984 - 49.940 = 0.044\text{ mL}
  • Center Point: The sample mean Vˉ=49.962 mL\bar{V} = 49.962\text{ mL} lies at the exact midpoint of this shaded segment.
Educational illustration for this question

Typical errors

OverviewACEpath Teacher's Diagnosis

01 The student forgets to divide the sample standard deviation by the square root of n when calculating the acceptance margin.

Sample Student Work

Reviewed by ACEpath Teacher
Step-by-Step Review
Walk Through It with Your Teacher
Incorrect
  1. Step 1

    Tolerance margin\text{Tolerance margin} = 2.776 ×\times s = 2.776 ×\times 0.018 = 0.050 mL\text{ mL}

    Correct

    E = 2.776 ⋅\cdot sn\frac{s}{\sqrt{n}} = 2.776 ⋅\cdot 0.0185\frac{0.018}{\sqrt{5}} ≈\approx 0.022 mL\text{ mL}

    You calculated 2.776×s2.776 \times s directly instead of evaluating 2.776⋅sn2.776 \cdot \frac{s}{\sqrt{n}}. Because the sample size is n=5n = 5, you must divide the sample standard deviation by 5≈2.236\sqrt{5} \approx 2.236.

  2. Step 2

    Lower bound\text{Lower bound} = 49.962 - 0.050 = 49.912 mL\text{ mL}   ⟹  \implies Upper bound\text{Upper bound} = 49.962 + 0.050 = 50.012 mL\text{ mL}   ⟹  \implies [49.912, 50.012] mL\text{ mL}

    Correct

    49.962 - 0.022 ≤\le V ≤\le 49.962 + 0.022   ⟹  \implies [49.940 mL\text{ mL}, 49.984 mL\text{ mL}]

    Your method of subtracting and adding the margin to the sample mean Vˉ=49.962 mL\bar{V} = 49.962\text{ mL} is completely correct, but the resulting boundaries are numerical errors propagated from Step 1.

  3. Step 3

    A closed segment on the number line from \text{A closed segment on the number line from } 49.912 mL to \text{ mL to } 50.012 mL with solid endpoints (and plotted number line)\text{ mL with solid endpoints (and plotted number line)}

    Correct

    A shaded segment from \text{A shaded segment from } 49.940 mL to \text{ mL to } 49.984 mL with solid, closed circles at both endpoints\text{ mL with solid, closed circles at both endpoints}

    You accurately interpreted the non-strict inequality '≤\le' by drawing solid endpoints connected by a continuous segment, but the endpoint coordinates inherit the earlier calculation error.

WHAT YOU DID WELL
Compliments from Your Teacher

You correctly identified the sample mean Vˉ=49.962 mL\bar{V} = 49.962\text{ mL} as the midpoint of the acceptance interval.

You correctly used solid circles and a continuous shaded segment on the number line to represent a non-strict compound inequality.

THE ERROR
Where Your ACEpath Teacher Spotted the Mistake
Omission of sample size factor in standard error

You forgot to divide the sample standard deviation ss by n\sqrt{n} when computing the margin of error.

WHY IT HAPPENED
ACEpath Teacher's Diagnosis

Confusing the margin of error for a sample mean, which requires the standard error sn\frac{s}{\sqrt{n}}, with a tolerance threshold based on raw standard deviation ss.

REASONING PATH
Your ACEpath teacher’s suggested approach
Standard error check

  • Standard error=sn\text{Standard error} = \frac{s}{\sqrt{n}}

  • Always verify whether the given formula contains n\sqrt{n} in the denominator before multiplying by the critical tt-value.

CORRECTED PATH
ACEpath Teacher's Demonstration

  1. Calculate the tolerance margin using the full formula: E=2.776⋅0.0185=0.0499682.23607≈0.022 mLE = 2.776 \cdot \frac{0.018}{\sqrt{5}} = \frac{0.049968}{2.23607} \approx 0.022\text{ mL}.

  2. Unfold the absolute value inequality ∣V−49.962∣≤0.022|V - 49.962| \le 0.022 into −0.022≤V−49.962≤0.022-0.022 \le V - 49.962 \le 0.022.

  3. Add 49.96249.962 to all parts to obtain 49.940 mL≤V≤49.984 mL49.940\text{ mL} \le V \le 49.984\text{ mL}, which is [49.940 mL,49.984 mL][49.940\text{ mL}, 49.984\text{ mL}] in interval notation.

  4. Plot solid closed circles at V=49.940 mLV = 49.940\text{ mL} and V=49.984 mLV = 49.984\text{ mL} on a calibrated number line, then shade the segment connecting them.

Final answer

Interval notation: [49.940 mL,49.984 mL][49.940\text{ mL}, 49.984\text{ mL}]; on the number line, solid closed circles at V=49.940 mLV = 49.940\text{ mL} and V=49.984 mLV = 49.984\text{ mL} with a shaded segment connecting them.

NEXT STEPS
Final Takeaways from Your Teacher

  • Review how standard error sn\frac{s}{\sqrt{n}} differs from standard deviation ss when constructing confidence and acceptance intervals.

  • Practice checking each variable in the experimental formula before substituting numerical values.

OverviewACEpath Teacher's Diagnosis

02 The student divides by n instead of the square root of n in the standard error term.

Sample Student Work

Reviewed by ACEpath Teacher
Step-by-Step Review
Walk Through It with Your Teacher
Incorrect
  1. Step 1

    Margin\text{Margin} = 2.776 ×\times (0.018 / 5) = 2.776 ×\times 0.0036 = 0.010 mL\text{ mL}

    Correct

    E = 2.776 ×\times 0.0185\frac{0.018}{\sqrt{5}} ≈\approx 2.776 ×\times 0.00805 ≈\approx 0.022 mL\text{ mL}

    You substituted the sample size n=5n = 5 into the denominator directly as 55 instead of evaluating its square root, 5\sqrt{5}.

  2. Step 2

    Lower limit\text{Lower limit} = 49.962 - 0.010 = 49.952 mL\text{ mL}   ⟹  \implies Upper limit\text{Upper limit} = 49.962 + 0.010 = 49.972 mL\text{ mL}   ⟹  \implies [49.952, 49.972] mL\text{ mL}

    Correct

    Vˉ\bar{V} - E = 49.962 - 0.022 = 49.940 mL\text{ mL}   ⟹  \implies Vˉ\bar{V} + E = 49.962 + 0.022 = 49.984 mL\text{ mL}   ⟹  \implies [49.940 mL\text{ mL}, 49.984 mL\text{ mL}]

    You applied the correct symmetric interval formula Vˉ±E\bar{V} \pm E, but the numerical limits are affected by the standard error miscalculation in Step 1.

  3. Step 3

    A continuous line segment between \text{A continuous line segment between } 49.952 mL and \text{ mL and } 49.972 mL with filled dots at the boundaries.\text{ mL with filled dots at the boundaries.}

    Correct

    Solid closed circles at \text{Solid closed circles at } 49.940 mL and \text{ mL and } 49.984 mL connected by a shaded line segment of width \text{ mL connected by a shaded line segment of width } 0.044 mL.\text{ mL.}

    You correctly recognized that the non-strict inequality ≤\le requires solid filled boundary dots connected by a continuous segment, but the endpoint coordinates were carried over from the incorrect margin.

WHAT YOU DID WELL
Compliments from Your Teacher

You correctly recognized that the absolute value tolerance forms a symmetric interval centered at the sample mean.

You correctly used solid, closed boundary dots connected by a continuous segment to graph a non-strict inequality.

THE ERROR
Where Your ACEpath Teacher Spotted the Mistake
Omitted square root in standard error

You divided by 55 instead of 5\sqrt{5} in the margin of error formula.

WHY IT HAPPENED
ACEpath Teacher's Diagnosis

Confusing sample size nn with the standard error denominator n\sqrt{n}, overlooking the radical in s/ns/\sqrt{n}.

REASONING PATH
Your ACEpath teacher’s suggested approach
Standard Error Formula Verification

  • The standard error of the mean is always sn\frac{s}{\sqrt{n}}, not sn\frac{s}{n}.

  • Always write down n\sqrt{n} explicitly and evaluate n\sqrt{n} before dividing.

CORRECTED PATH
ACEpath Teacher's Demonstration

  1. Recalculate the margin with the square root: E=2.776⋅0.0185≈2.776⋅0.008050≈0.02235 mL≈0.022 mLE = 2.776 \cdot \frac{0.018}{\sqrt{5}} \approx 2.776 \cdot 0.008050 \approx 0.02235\text{ mL} \approx 0.022\text{ mL}.

  2. Find the interval endpoints: Lower=49.962−0.022=49.940 mL\text{Lower} = 49.962 - 0.022 = 49.940\text{ mL} and Upper=49.962+0.022=49.984 mL\text{Upper} = 49.962 + 0.022 = 49.984\text{ mL}.

  3. Plot solid closed circles at V=49.940 mLV = 49.940\text{ mL} and V=49.984 mLV = 49.984\text{ mL} and shade the segment between them.

Final answer

[49.940 mL,49.984 mL][49.940\text{ mL}, 49.984\text{ mL}]; solid closed circles at 49.940 mL49.940\text{ mL} and 49.984 mL49.984\text{ mL} connected by a shaded segment.

NEXT STEPS
Final Takeaways from Your Teacher

  • Review the formula for standard error SE=s/n\text{SE} = s/\sqrt{n} whenever calculating confidence or tolerance margins.

  • Practice computing radical denominators on your calculator before multiplying by the critical tt-value.

OverviewACEpath Teacher's Diagnosis

03 The student misinterprets the weak inequality as a strict inequality, representing the interval as open with hollow endpoints.

Sample Student Work

Reviewed by ACEpath Teacher
Step-by-Step Review
Walk Through It with Your Teacher
Partially correct
  1. Step 1

    Margin of error\text{Margin of error} = 2.776 ×\times (0.0185)\left(\frac{0.018}{\sqrt{5}}\right) = 0.022 mL\text{ mL}   ⟹  \implies Lower limit\text{Lower limit} = 49.962 - 0.022 = 49.940 mL\text{ mL}   ⟹  \implies Upper limit\text{Upper limit} = 49.962 + 0.022 = 49.984 mL\text{ mL}

  2. Step 2

    (49.940, 49.984) mL\text{ mL}   ⟹  \implies An open segment between \text{An open segment between } 49.940 mL\text{ mL}  and \text{ and } 49.984 mL\text{ mL}  with open hollow circles at both ends.\text{ with open hollow circles at both ends.}

    Correct

    [49.940 mL\text{ mL}, 49.984 mL\text{ mL}]  with solid, closed circles at both endpoints connected by a shaded line segment\text{ with solid, closed circles at both endpoints connected by a shaded line segment}

    The criterion is given as ∣V−Vˉ∣≤2.776sn|V - \bar{V}| \le 2.776\frac{s}{\sqrt{n}}, where the symbol '≤\le' means 'less than or equal to'. Because equality is allowed, the boundary values are part of the solution set, which requires square brackets [… ][\dots] and filled/solid circles on the number line.

WHAT YOU DID WELL
Compliments from Your Teacher

You correctly substituted the sample standard deviation and sample size into the confidence threshold formula.

You calculated the margin of error accurately as 0.022 mL0.022\text{ mL}.

You correctly subtracted and added the margin of error to the sample mean to obtain the boundary values 49.940 mL49.940\text{ mL} and 49.984 mL49.984\text{ mL}.

THE ERROR
Where Your ACEpath Teacher Spotted the Mistake
Strict vs. Non-Strict Inequality Boundary Misinterpretation

You initially sketched closed endpoints, but then crossed them out and replaced them with open interval notation and hollow circles.

WHY IT HAPPENED
ACEpath Teacher's Diagnosis

You treated a non-strict inequality (≤\le) as a strict inequality (<<), associating the acceptance boundary with an open interval rather than an inclusive closed interval.

REASONING PATH
Your ACEpath teacher’s suggested approach
Check the inequality symbol for equality

  • The symbols ≤\le and ≥\ge include boundary equality, whereas << and >> exclude the boundary values.

  • Inspect whether the inequality sign has an underline/bar: if it does, use square brackets [… ][\dots] and solid filled circles ∙\bullet.

Translate to interval notation and graphing symbols

  • Included endpoints (≤,≥\le, \ge) correspond to [… ][\dots] and solid dots; excluded endpoints (<,><, >) correspond to (… )(\dots) and open/hollow circles ∘\circ.

  • Always align square brackets with solid dots and round parentheses with open circles.

CORRECTED PATH
ACEpath Teacher's Demonstration

  1. Identify the original inequality symbol: ∣V−49.962∣≤0.022|V - 49.962| \le 0.022.

  2. Expand the absolute value inequality: −0.022≤V−49.962≤0.022-0.022 \le V - 49.962 \le 0.022.

  3. Add 49.96249.962 to all parts: 49.940≤V≤49.98449.940 \le V \le 49.984.

  4. Convert the inclusive compound inequality to interval notation using brackets: [49.940 mL,49.984 mL][49.940\text{ mL}, 49.984\text{ mL}].

  5. Draw a horizontal number line with solid, filled circles at 49.94049.940 and 49.98449.984, and shade the continuous line segment between them.

Final answer

[49.940 mL,49.984 mL][49.940\text{ mL}, 49.984\text{ mL}]; solid closed circles at V=49.940 mLV = 49.940\text{ mL} and V=49.984 mLV = 49.984\text{ mL} connected by a shaded line segment.

NEXT STEPS
Final Takeaways from Your Teacher

  • Practice matching inequality symbols directly with their visual representations: ≤,≥⟷[ ],∙\le, \ge \longleftrightarrow [\ ], \bullet versus <,>⟷( ),∘<, > \longleftrightarrow (\ ), \circ.

  • When double-checking your work, reread the original inequality statement to ensure you did not accidentally change '≤\le' to '<<'.

Lock it in

Question

In confidence intervals and tolerance bounds involving sample means, what is the standard error formula for a sample of size nn?

Answer

SE=snSE = \frac{s}{\sqrt{n}}. Always divide the sample standard deviation ss by the square root of the sample size nn.

Similar question

01

A quality analyst tests n=4n = 4 identical pipettes. The delivery volume mean is Vˉ=10.005 mL\bar{V} = 10.005\text{ mL} with s=0.008 mLs = 0.008\text{ mL}. The acceptance criterion is ∣V−Vˉ∣≤3.182sn|V - \bar{V}| \le 3.182\frac{s}{\sqrt{n}}. Find the acceptable interval for VV to three decimal places.

Answer

[9.992 mL,10.018 mL][9.992\text{ mL}, 10.018\text{ mL}]. Key steps: sn=0.0084=0.004 mL\frac{s}{\sqrt{n}} = \frac{0.008}{\sqrt{4}} = 0.004\text{ mL}; Margin E=3.182×0.004=0.012728≈0.013 mLE = 3.182 \times 0.004 = 0.012728 \approx 0.013\text{ mL}; Interval =10.005±0.013=[9.992,10.018]= 10.005 \pm 0.013 = [9.992, 10.018].

02

An industrial scale calibration uses n=16n = 16 test mass readings with mˉ=250.00 g\bar{m} = 250.00\text{ g} and s=0.40 gs = 0.40\text{ g}. Tolerable readings satisfy ∣m−mˉ∣≤2.131sn|m - \bar{m}| \le 2.131\frac{s}{\sqrt{n}}. Determine the tolerance interval and describe its representation on a number line.

Answer

[249.79 g,250.21 g][249.79\text{ g}, 250.21\text{ g}], represented by solid closed dots at 249.79 g249.79\text{ g} and 250.21 g250.21\text{ g} with a shaded segment of length 0.42 g0.42\text{ g} between them. Key steps: E=2.131×0.4016=2.131×0.10=0.2131≈0.21 gE = 2.131 \times \frac{0.40}{\sqrt{16}} = 2.131 \times 0.10 = 0.2131 \approx 0.21\text{ g}; Bounds: 250.00±0.21250.00 \pm 0.21.

03

An automated micro-pipette has mean dispensing volume Vˉ=1.000 mL\bar{V} = 1.000\text{ mL}, s=0.012 mLs = 0.012\text{ mL}, and n=6n = 6. The manufacturer requires ∣V−1.000∣≤tsn|V - 1.000| \le t \frac{s}{\sqrt{n}} with t=2.571t = 2.571. Compute the maximum allowable deviation EE to three decimal places, state the interval, and determine whether a trial delivery of 1.015 mL1.015\text{ mL} passes inspection.

Answer

E=0.013 mLE = 0.013\text{ mL}, acceptable interval [0.987 mL,1.013 mL][0.987\text{ mL}, 1.013\text{ mL}], and the delivery of 1.015 mL1.015\text{ mL} fails (is rejected). Key steps: E=2.571×0.0126≈2.571×0.004899≈0.0126≈0.013 mLE = 2.571 \times \frac{0.012}{\sqrt{6}} \approx 2.571 \times 0.004899 \approx 0.0126 \approx 0.013\text{ mL}; Interval: [0.987,1.013][0.987, 1.013]; Since 1.015>1.0131.015 > 1.013, it falls outside the acceptance region.

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