📏 Confidence Intervals
A sample mean is never exactly the true population mean — a confidence interval gives an honest range, with a stated level of confidence, for where that true value actually lies.
2Topics
8Worked Examples
4Eng. Cases
12Practice Qs
📐Known σ (z)Large samples
📉Small Samples (t)Unknown σ, n<30
⚙️ApplicationsReal engineering
🎮Quick-Fire GameTest your speed
📝PracticeGraded questions
🎯What you will learn
Construct confidence intervals for a population mean using the z-distribution when σ is known; use the t-distribution for small samples with unknown σ; and interpret what a confidence level actually means.
💡 A 95% confidence interval does NOT mean "95% probability the true mean is in this range." It means: if you repeated the sampling process many times, about 95% of the intervals constructed this way would contain the true mean.
🧭How to use this resource
1. Read the rules banner at the top of each tab
2. Experiment with the interactive tool
3. Check the key vocabulary panel
4. Study the visual diagram, then the worked examples
📐
Confidence Interval for the Mean — Known σ
When the population standard deviation is known (or the sample is large), the z-distribution gives the interval directly.
🔧Confidence Interval Calculator (z)Enter x̄, σ, n and confidence level
Enter x̄, σ, n and choose a confidence level, then press Calculate for the full interval…
📖 Key Vocabulary
Confidence level — the long-run proportion of intervals (from repeated sampling) that would contain the true mean.
Standard error (SE) — σ/√n, the SD of the sample mean itself, not of individual values.
Margin of error — the ± part of the interval, z×SE.
Point estimate — the single best guess (x̄) before the interval is added around it.
👁️ Visual aid: the interval is centred exactly on the sample mean, stretching out by the same margin on each side.
✏️Worked Example 1 — Constructing a 95% CIType: Standard 95%
Question A sample of 36 shafts has mean diameter 25.40mm; population σ=0.60mm. Find the 95% CI for the true mean.
Standard error
SE = 0.60/√36 = 0.60/6 = 0.10Margin of error
1.96 × 0.10 = 0.196Answer CI = 25.40 ± 0.196 = (25.204, 25.596) mm
📊Worked Example 2 — Using a different confidence levelType: 99% level
Question Using the same data, find the 99% CI instead.
Margin of error
2.576 × 0.10 = 0.2576Answer CI = 25.40 ± 0.2576 = (25.142, 25.658) mm — wider than the 95% interval, since higher confidence requires more room.
📉Worked Example 3 — Effect of increasing sample sizeType: Sample size effect
Question If n increases from 36 to 144 (same x̄=25.40, σ=0.60), find the new 95% margin of error.
New SE
0.60/√144 = 0.60/12 = 0.05New margin
1.96 × 0.05 = 0.098Answer Quadrupling n halves the margin of error (0.196 → 0.098) — precision improves with the square root of sample size, not proportionally.
⚙️Worked Example 4 — Engineering context (process mean estimate)Type: Engineering
Question 50 resistors are sampled; mean resistance x̄=100.4Ω, known σ=2.5Ω. Find the 90% CI and interpret it for a production report.
SE
2.5/√50 = 0.3536Margin
1.645 × 0.3536 = 0.582Answer CI = (99.82, 100.98)Ω — the production team can report with 90% confidence that the true mean resistance of the batch lies in this range.
🧠 Common slip: using n instead of √n in the standard error formula — the denominator is always the square root of the sample size.
📉
Confidence Intervals with Small Samples — the t-distribution
When σ is unknown and n is small, the t-distribution (wider-tailed than normal) accounts for the extra uncertainty.
🔧Confidence Interval Calculator (t)Enter x̄, s, n and t-value
Enter x̄, s, n and the t-value for your confidence level and degrees of freedom, then press Calculate…
📖 Key Vocabulary
t-distribution — similar in shape to normal but with heavier tails, reflecting extra uncertainty from estimating σ.
Degrees of freedom (df) — n−1 for a single-sample confidence interval; determines the exact t-distribution shape.
Sample SD (s) — calculated using n−1 in the denominator (rather than n) to correct for bias.
t-table — a reference table giving critical t-values for different confidence levels and degrees of freedom.
👁️ Visual aid: the t-curve is shorter and wider than the normal curve — that extra spread in the tails is exactly what makes t-intervals wider than z-intervals for the same confidence level.
✏️Worked Example 1 — Constructing a t-based CIType: Standard calculation
Question A sample of 10 castings has mean tensile strength 48.2 MPa, sample SD s=3.1 MPa. Using t=2.262 (95%, df=9), find the CI.
Standard error
SE = 3.1/√10 = 0.9803Margin
2.262 × 0.9803 = 2.217Answer CI = 48.2 ± 2.217 = (45.98, 50.42) MPa
🔢Worked Example 2 — Finding degrees of freedomType: df identification
Question A sample of 15 pressure readings is taken. State the degrees of freedom to use in the t-table.
Apply rule
df = n − 1 = 15 − 1Answer df = 14 — this tells you which row of the t-table to read from.
⚖️Worked Example 3 — Comparing t and z for the same dataType: t vs z comparison
Question For n=10, 95% confidence, t=2.262 but z=1.96. Explain why the t-based interval is wider even with identical x̄ and s.
Reasoning The t-value (2.262) is larger than the z-value (1.96) for the same confidence level
Answer This reflects genuine extra uncertainty: with only 10 data points, the sample SD s is itself an imperfect estimate of the true σ, so the interval must be wider to maintain the stated confidence level.
⚙️Worked Example 4 — Engineering context (limited prototype testing)Type: Engineering
Question Only 6 prototype units could be tested for fatigue life (hours): mean=1250, s=85. Using t=2.571 (95%, df=5), find the CI for the true mean fatigue life.
SE
85/√6 = 34.70Margin
2.571 × 34.70 = 89.22Answer CI = 1250 ± 89.22 = (1160.8, 1339.2) hours — a wide interval, honestly reflecting how little data was available for this expensive prototype test.
🧠 Common slip: using z-values from a standard normal table when the sample is small and σ is unknown — always switch to the t-distribution with df=n−1 in that situation.
⚙️
Engineering Applications
Four live scenarios — each with its own calculator. Enter your own numbers.
🔩 Batch Quality Reporting
Reporting a confidence interval (not just a single mean) gives customers an honest picture of measurement uncertainty.
CI = x̄ ± z(σ/√n)
📟 Try it — Batch CI Calculator
Enter x̄, σ and n, then press Calculate (95% CI)…
🔬 Material Testing — Small Sample CI
Destructive or expensive tests often only allow small samples, making the t-distribution essential.
CI = x̄ ± t(s/√n)
📟 Try it — Material Test CI Calculator
Enter x̄, s, n and t, then press Calculate…
📊 Required Sample Size Planning
Working backwards from a desired margin of error tells engineers how large a sample they need to collect.
n = (zσ/E)² (E = desired margin of error)
📟 Try it — Sample Size Calculator
Enter σ, desired margin and z, then press Calculate…
🌡️ Sensor Calibration Confidence
A confidence interval around a calibration measurement quantifies how tightly the true calibration value is known.
CI = x̄ ± z(σ/√n)
📟 Try it — Calibration CI Calculator
Enter mean reading, σ and n, then press Calculate (99% CI)…
🎮
Quick-Fire Confidence Interval Game
12 questions on z-based and t-based intervals. How many can you get?
⚡ Confidence Interval Challenge
Score: 0Question: 1/12Streak: 0🔥
Press Start to begin!
📝
Practice Questions
Graded from warm-up to challenge. Attempt fully before revealing the working.
Question 1 🥉2 marks
State the z-value used for a 95% confidence interval.
z = 1.96
Question 2 🥉2 marks
Find the standard error for σ=4, n=64.
4/√64 = 4/8 = 0.5
Question 3 🥉2 marks
For n=20, find the degrees of freedom for a t-based interval.
df = n−1 = 20−1 = 19
Question 4 🥉2 marks
State whether increasing the confidence level makes the interval wider or narrower.
Wider — higher confidence requires a larger margin of error
Question 5 🥈4 marks
A sample of 49 items has x̄=60.0, σ=7.0. Find the 95% CI.
SE=7/√49=1.0
Margin=1.96×1.0=1.96
CI=(58.04, 61.96)
Margin=1.96×1.0=1.96
CI=(58.04, 61.96)
Question 6 🥈4 marks
Using the same data as Q5, find the 99% CI (z=2.576).
Margin=2.576×1.0=2.576
CI=(57.42, 62.58)
CI=(57.42, 62.58)
Question 7 🥈3 marks
A sample of 9 has mean=15.0, s=2.4. Using t=2.306 (95%, df=8), find the margin of error.
SE=2.4/√9=0.8
Margin=2.306×0.8
=1.845
Margin=2.306×0.8
=1.845
Question 8 🥈4 marks
Using the previous question's data, state the full 95% CI.
15.0 ± 1.845
CI=(13.155, 16.845)
CI=(13.155, 16.845)
Question 9 🥇5 marks
A quality engineer wants a margin of error no more than 0.2 with σ=1.2 and 95% confidence. Find the required sample size.
n=(zσ/E)² = (1.96×1.2/0.2)²
= (11.76)²
= 138.3, round up to n=139
= (11.76)²
= 138.3, round up to n=139
Question 10 🥇5 marks
A sample of 16 pressure vessel tests gives mean burst pressure 340 bar, s=18 bar. Using t=2.131 (95%, df=15), find the CI.
SE=18/√16=4.5
Margin=2.131×4.5=9.590
CI=(330.4, 349.6) bar
Margin=2.131×4.5=9.590
CI=(330.4, 349.6) bar
Question 11 🥇5 marks
Explain why a 99% CI is always wider than a 90% CI for the same data, referencing the z-values (1.645 vs 2.576).
Higher confidence (99% vs 90%) requires a larger z-value (2.576 vs 1.645), which directly increases the margin of error z×SE, widening the interval — this is the necessary trade-off for greater certainty.
Question 12 🥇5 marks
A calibration lab tests 8 reference weights: mean=100.02g, s=0.05g. Using t=2.365 (95%, df=7), find the CI and state whether it supports the weight being genuinely 100.00g.
SE=0.05/√8=0.01768
Margin=2.365×0.01768=0.04182
CI=(100.02−0.0418, 100.02+0.0418)
CI≈(99.978, 100.062)g — since 100.00g falls within this interval, the data is consistent with the true weight being exactly 100.00g
Margin=2.365×0.01768=0.04182
CI=(100.02−0.0418, 100.02+0.0418)
CI≈(99.978, 100.062)g — since 100.00g falls within this interval, the data is consistent with the true weight being exactly 100.00g