09-01: Exercises — Confidence Interval for a Mean¶
Notes reference: 09-01: Confidence Interval for a Mean
Q1: z-interval (σ known)¶
A sample of n = 64 light bulbs has x̄ = 1,190 hours. The manufacturing process is known to have σ = 120 hours.
Build 90%, 95%, and 99% confidence intervals for μ.
Solution
SE = σ/√n = 120/√64 = 120/8 = 15.0
90%: z = 1.6449 E = 24.67 → 1190 ± 24.67 = (1165.3, 1214.7)
95%: z = 1.9600 E = 29.40 → 1190 ± 29.40 = (1160.6, 1219.4)
99%: z = 2.5758 E = 38.64 → 1190 ± 38.64 = (1151.4, 1228.6)
Higher confidence → WIDER interval.
90% width = 49.3 hours
95% width = 58.8 hours
99% width = 77.3 hours
You cannot buy confidence without paying in precision.
=CONFIDENCE.NORM(0.10, 120, 64) ' 24.673
=CONFIDENCE.NORM(0.05, 120, 64) ' 29.400
=CONFIDENCE.NORM(0.01, 120, 64) ' 38.637
Q2: t-interval (σ unknown)¶
A sample of 18 households has a mean monthly electricity bill of $142.60 with s = $28.40.
Build a 95% confidence interval.
Solution
n = 18, df = 17, t(0.025, 17) = 2.1098
SE = 28.40/√18 = 28.40/4.2426 = 6.6940
E = 2.1098 × 6.6940 = 14.123
CI = 142.60 ± 14.12 = ($128.48, $156.72)
INTERPRETATION
"We are 95% confident that the mean monthly electricity bill for all
households in this population is between $128.48 and $156.72."
The z-interval would have used 1.96 instead of 2.1098, giving ±13.12 —
about a dollar narrower, and WRONG, because σ was estimated from the data.
=CONFIDENCE.T(0.05, 28.40, 18) ' 14.123
=142.60-14.123 ' 128.477
=142.60+14.123 ' 156.723
=T.INV.2T(0.05, 17) ' 2.10982
xbar <- 142.60; s <- 28.40; n <- 18
tc <- qt(0.975, n - 1)
xbar + c(-1, 1) * tc * s/sqrt(n) # 128.477 156.723
xbar, s, n = 142.60, 28.40, 18
tc = stats.t.ppf(0.975, n - 1)
xbar - tc*s/np.sqrt(n), xbar + tc*s/np.sqrt(n)
Q3: z or t?¶
For each, choose the correct interval and give the critical value.
| Situation | z or t? | Critical value (95%) |
|---|---|---|
n = 40, σ = 8 known |
? | ? |
n = 40, s = 8 computed |
? | ? |
n = 12, s = 3.2, population normal |
? | ? |
n = 12, s = 3.2, population strongly skewed |
? | ? |
n = 500, s = 22 |
? | ? |
Solution
| Situation | z or t? | Critical value (95%) |
|---|---|---|
n = 40, σ known |
z | 1.960 |
n = 40, s computed |
t, df = 39 |
2.023 |
n = 12, s, normal population |
t, df = 11 |
2.201 |
n = 12, s, strongly skewed |
Neither is safe | — |
n = 500, s |
t, df = 499 |
1.965 ≈ z |
Row 4: with n = 12 from a strongly skewed population, neither the CLT nor
the normality assumption holds. Options: transform the data (log),
use a nonparametric method (sign test, Wilcoxon), or bootstrap the CI.
Row 5: at df = 499 the t critical value (1.9647) is essentially z (1.9600).
This is why large-sample z and t intervals agree to three decimals.
Q4: Interpret correctly¶
A 95% CI for the mean commute time is (24.1, 31.7) minutes. Mark each statement true or false.
- 95% of commutes are between 24.1 and 31.7 minutes.
- There is a 95% probability that
μlies in(24.1, 31.7). - If we repeated this study many times, about 95% of the intervals produced would contain
μ. - We are 95% confident the true mean commute time is between 24.1 and 31.7 minutes.
- The sample mean is 27.9 minutes.
- A hypothesis test of
H₀: μ = 30would not be rejected atα = 0.05.
Solution
1. FALSE — that describes individual commutes, not the mean. A PREDICTION
interval for one commute would be much wider.
2. FALSE — μ is a fixed constant. It either is or is not in this interval.
The 95% describes the long-run behaviour of the PROCEDURE.
3. TRUE — this is the correct frequentist definition.
4. TRUE — the standard, accepted shorthand for #3.
5. TRUE — the interval is symmetric about x̄, so
x̄ = (24.1 + 31.7)/2 = 27.9 ✓
6. TRUE — 30 lies INSIDE the interval, so H₀: μ = 30 is not rejected
at α = 0.05. (H₀: μ = 22 WOULD be rejected — 22 is outside.)
Q5: Sample size for a mean¶
A researcher wants to estimate mean household income with a margin of error of $500 at 95% confidence. A pilot study gives s ≈ $6,200.
- Required
n - Required
nforE = $250 - Required
nforE = $500at 99% confidence
Solution
( z σ )²
n = ( ───── ) ALWAYS round UP
( E )
1. n = (1.96 × 6200 / 500)² = (24.304)² = 590.68 → n = 591
2. n = (1.96 × 6200 / 250)² = (48.608)² = 2362.7 → n = 2363
Halving E QUADRUPLES the sample (591 → 2363).
3. n = (2.5758 × 6200 / 500)² = (31.940)² = 1020.2 → n = 1021
Going from 95% to 99% confidence at the same precision nearly
DOUBLES the sample.
=ROUNDUP((NORM.S.INV(0.975)*6200/500)^2, 0) ' 591
=ROUNDUP((NORM.S.INV(0.975)*6200/250)^2, 0) ' 2363
=ROUNDUP((NORM.S.INV(0.995)*6200/500)^2, 0) ' 1021
n_for_mean <- function(sigma, E, conf = 0.95)
ceiling((qnorm(1 - (1-conf)/2) * sigma / E)^2)
n_for_mean(6200, 500); n_for_mean(6200, 250); n_for_mean(6200, 500, 0.99)
Q6: From raw data¶
Fifteen service calls (minutes):
Build a 95% CI for the mean.
Solution
n = 15, Σx = 411, x̄ = 27.4
Σ(x − x̄)² = 295.6 → s² = 295.6/14 = 21.114, s = 4.5951
SE = 4.5951/√15 = 4.5951/3.8730 = 1.18645
df = 14, t(0.025, 14) = 2.14479
E = 2.14479 × 1.18645 = 2.5447
CI = 27.4 ± 2.54 = (24.86, 29.94) minutes
=AVERAGE(A2:A16) ' 27.4
=STDEV.S(A2:A16) ' 4.5951
=CONFIDENCE.T(0.05, STDEV.S(A2:A16), COUNT(A2:A16)) ' 2.5447
' Or: Data ▸ Data Analysis ▸ Descriptive Statistics
' tick "Confidence Level for Mean: 95%" → reports E directly
x <- c(22,31,28,19,35,27,24,33,29,21,30,26,32,25,29)
t.test(x)
# mean of x: 27.4
# 95 percent confidence interval: 24.8553 29.9447
t.test(x, conf.level = 0.99)$conf.int
x = np.array([22,31,28,19,35,27,24,33,29,21,30,26,32,25,29])
stats.t.interval(0.95, len(x)-1, x.mean(), stats.sem(x))
Q7: What makes it narrower?¶
Starting from a 95% CI of (24.86, 29.94) based on n = 15, state the effect of each change.
- Raise confidence to 99%
- Increase
nto 60 (samex̄, sames) - The data turns out to be more variable (
sdoubles) - Drop confidence to 90%
Solution
Current: E = 2.545, width = 5.09
1. 99% confidence: t(0.005, 14) = 2.9768
E = 2.9768 × 1.18645 = 3.532 → width 7.06 WIDER
2. n = 60: SE = 4.5951/√60 = 0.59323, t(0.025, 59) = 2.0010
E = 2.0010 × 0.59323 = 1.187 → width 2.37 NARROWER
Roughly halved — because n quadrupled.
3. s = 9.1902: SE = 2.3729, t unchanged at 2.14479
E = 5.089 → width 10.18 WIDER
Doubling s doubles the width.
4. 90% confidence: t(0.05, 14) = 1.7613
E = 1.7613 × 1.18645 = 2.090 → width 4.18 NARROWER
SUMMARY
↑ confidence → wider
↑ n → narrower (by √n)
↑ s → wider (proportionally)
Q8: Confidence interval for a variance¶
Using the Q6 data (n = 15, s = 4.5951), build a 95% CI for σ² and for σ.
Solution
df = 14, α = 0.05
χ²(0.025, 14) = 26.119 (right-tail area 0.025)
χ²(0.975, 14) = 5.629 (right-tail area 0.975)
(n−1)s² 14 × 21.114 295.6
Lower = ───────────────── = ────────────── = ───────── = 11.318
χ²(α/2) 26.119 26.119
(n−1)s² 295.6
Upper = ───────────────── = ───────────── = 52.514
χ²(1−α/2) 5.629
95% CI for σ²: (11.32, 52.51)
95% CI for σ : (√11.32, √52.51) = (3.36, 7.25) minutes
NOTICE how ASYMMETRIC this is about s² = 21.11 —
21.11 − 11.32 = 9.79 below
52.51 − 21.11 = 31.40 above
The chi-square distribution is right-skewed, so the interval is too.
WARNING: this procedure requires the population to be NORMAL and is
NOT robust to that assumption, unlike the t-interval for the mean.
=(15-1)*STDEV.S(A2:A16)^2/CHISQ.INV.RT(0.025, 14) ' 11.318
=(15-1)*STDEV.S(A2:A16)^2/CHISQ.INV.RT(0.975, 14) ' 52.514
=SQRT(D1) & " to " & SQRT(D2) ' 3.364 to 7.247
n <- 15; s2 <- var(x)
c(lower = (n-1)*s2/qchisq(0.975, n-1),
upper = (n-1)*s2/qchisq(0.025, n-1))
sqrt(c((n-1)*s2/qchisq(0.975, n-1), (n-1)*s2/qchisq(0.025, n-1)))
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