E = row total × column total / grand total, df = (r − 1)(c − 1)
Critical values (α = 0.05)
df 1: 3.841, 2: 5.991, 3: 7.815, 4: 9.488
Decision
reject H0 if χ² > χ²crit
2Practice problems
1.Normal distributionMedium
The delivery times are normally distributed with mean 32 min and standard deviation 6 min. What fraction lies between 22.7 and 38 min? Give P as a decimal.
μ = 32σ = 6x1 = 22.7x2 = 38
P =
2.Normal distributionHard
The bag weights are normally distributed with mean 250 g and standard deviation 5 g. Drag the cut-off so the shaded top 80% is to the right of it. Find the cut-off x.
μ = 250 gσ = 5 g80%
x = g
3.Confidence intervalsMedium
A sample of 20 items has mean 39 and standard deviation s = 11.5. The population is roughly normal. Find the upper limit of the 90% confidence interval for the mean.
n = 20x̄ = 39s = 11.590%t(19) = 1.729
upper limit =
4.Confidence intervalsHard
Two fertilisers are tested. Plot group 1: 16 plots, mean yield 28.5 kg, standard deviation 2.5 kg. Plot group 2: 5 plots, mean 34.5 kg, standard deviation 10 kg. Assume equal variances (pooled t, df = n₁ + n₂ − 2). Find the upper limit of the 90% confidence interval for μ₁ − μ₂.
12 employees take a typing test before and after a training course. For each person the difference d = after − before (words per minute) is recorded. The mean difference is 8.57 and the standard deviation of the differences is 9.9. Find the paired test statistic t for H₀: μd = 0.
n = 12d̄ = 8.57sd = 9.9
t =
6.Hypothesis tests & confidence intervalsHard
A sample of 36 measurements has mean 101.5. The population standard deviation is 4. Find the lower limit L of the 95% confidence interval for the mean.
x̄ = 101.5σ = 4n = 3695%
L =
7.Linear regression & correlationMedium
The table gives 6 paired observations of x = car age in years and y = price in k€. Their sums are given. Find Pearson's correlation coefficient r from these sums.
n = 6Σx = 21Σy = 138Σx² = 91Σy² = 3202Σxy = 464
r =
8.Linear regression & correlationHard
The table gives 5 paired observations of x = car age in years and y = price in k€. Use the least-squares line ŷ = a + b·x. For which value x does the line predict ŷ = 24?
n = 5ŷ = 24
x =
9.Chi-square testsMedium
The contingency table shows observed counts. Compute the chi-square statistic χ² for the test of independence between the row and column variables. Do not use a continuity correction.
N = 2002 × 3
χ² =
10.Chi-square testsHard
A chi-square test of independence is carried out on a contingency table with 3 rows and 4 columns (not counting the totals). Using the table, find the critical value χ²crit at significance level α = 0.01.
α = 0.01
χ²crit =
3Solutions
1.Normal distribution
Answer: P = 0.7808
Normal distribution: convert a value x to a z-score, z = (x − μ)/σ, which counts standard deviations from the mean. A z-table (or Φ(z)) gives the area to the left of z, which is P(X < x). The area to the right is 1 − Φ(z), and the area between two values is the difference of their two areas.
z-scores
z₁ = (22.7 − 32)/6 = −1.55
z₂ = (38 − 32)/6 = 1
Area between
P = Φ(1) − Φ(−1.55) = 0.8413 − 0.06057 = 0.7808
Answer
P = 0.7808
2.Normal distribution
Answer: x = 245.8 g
From an area back to z: Φ(z) is the area to the left of z. For a left-tail area A, find z with Φ(z) = A in the table. A right-tail area A means Φ(z) = 1 − A. A central area A between −z and z leaves (1 − A)/2 in each tail, so Φ(z) = (1 + A)/2. For a value x, use x = μ + zσ.
Area to the left
Φ(z) = 0.2
z from the table
z ≈ −0.84
Back to the original units
x = 250 − 0.84·5 = 245.8 g
Answer
x = 245.8 g
3.Confidence intervals
Answer: upper limit = 43.45
Confidence interval for a mean: x̄ ± margin of error. With a known population σ the margin is z·σ/√n; with only the sample standard deviation s, use t with n − 1 degrees of freedom: t·s/√n. Critical values: z = 1.645 (90%), 1.96 (95%), 2.576 (99%). To get a margin E or smaller, you need n ≥ (z·σ/E)², rounded up.
Critical value
df = n − 1 = 19, t = 1.729
Margin of error
E = 1.729·11.5/√20 = 4.446
Interval
39 ± 4.446 → [34.55, 43.45]
Answer
upper limit = 43.45
4.Confidence intervals
Answer: upper limit = −1.484 kg
Interval for a difference: estimate ± critical value × standard error. For two independent means with known σ: SE = √(σ₁²/n₁ + σ₂²/n₂) and z. With equal unknown variances: pooled s_p and t with n₁ + n₂ − 2 degrees of freedom. For paired data, work with the differences: d̄ ± t·s_d/√n.
Pooled standard deviation
sp² = (15·2.5² + 4·10²)/19 = 25.99
sp = 5.098
Margin of error
E = 1.729·5.098·√(1/16 + 1/5) = 4.516
Interval
−6 ± 4.516 → [−10.52, −1.484]
Answer
upper limit = −1.484 kg
5.Hypothesis tests & confidence intervals
Answer: t = 2.999
Testing a mean: the test statistic measures how many standard errors the sample mean lies from the value claimed in H₀: z = (x̄ − μ₀)/(σ/√n) when σ is known, t = (x̄ − μ₀)/(s/√n) when it is estimated from the sample. A confidence interval is x̄ ± (critical value)·(standard error).
Standard error of the mean difference
SE = 9.9/√12 = 2.858
Test statistic
t = 8.57/2.858 = 2.999
Answer
t = 2.999
With df = 11, the two-sided 5% critical value is 2.201; |t| is larger, so H₀ is rejected.
6.Hypothesis tests & confidence intervals
Answer: L = 100.2
Testing a mean: the test statistic measures how many standard errors the sample mean lies from the value claimed in H₀: z = (x̄ − μ₀)/(σ/√n) when σ is known, t = (x̄ − μ₀)/(s/√n) when it is estimated from the sample. A confidence interval is x̄ ± (critical value)·(standard error).
Margin of error
E = 1.96·4/√36 = 1.307
Interval limit
L = 101.5 − 1.307 = 100.2
Answer
L = 100.2
7.Linear regression & correlation
Answer: r = −0.8583
Least-squares regression: with x̄ and ȳ the means, Sxx = Σ(x − x̄)², Syy = Σ(y − ȳ)² and Sxy = Σ(x − x̄)(y − ȳ), the line ŷ = a + b·x has slope b = Sxy/Sxx and intercept a = ȳ − b·x̄, so it always passes through (x̄, ȳ). Pearson's correlation coefficient r = Sxy/√(Sxx·Syy) lies between −1 and 1; r² is the fraction of the variation in y explained by the line.
Centred sums
Sxy = 464 − 21·138/6 = −19
Sxx = 91 − 21²/6 = 17.5
Syy = 3202 − 138²/6 = 28
Correlation coefficient
r = −19/√(17.5·28) = −0.8583
Answer
r = −0.8583
8.Linear regression & correlation
Answer: x = 2.4
Least-squares regression: with x̄ and ȳ the means, Sxx = Σ(x − x̄)², Syy = Σ(y − ȳ)² and Sxy = Σ(x − x̄)(y − ȳ), the line ŷ = a + b·x has slope b = Sxy/Sxx and intercept a = ȳ − b·x̄, so it always passes through (x̄, ȳ). Pearson's correlation coefficient r = Sxy/√(Sxx·Syy) lies between −1 and 1; r² is the fraction of the variation in y explained by the line.
Means
x̄ = 15/5 = 3
ȳ = 111/5 = 22.2
Sums of squares and products
Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = −30
Regression line
b = −30/10 = −3
a = 22.2 − (−3)·3 = 31.2
ŷ = 31.2 − 3·x
Solve for x
x = (24 − 31.2)/(−3) = 2.4
Answer
x = 2.4
9.Chi-square tests
Answer: χ² = 12.36
Chi-square tests: compare observed counts O with the counts E expected under H₀ using χ² = Σ(O − E)²/E. Goodness of fit: E = n·p for each category, df = k − 1 (minus one more for each parameter estimated from the data). Independence in an r × c table: E = row total × column total / grand total, df = (r − 1)(c − 1). Reject H₀ when χ² exceeds the critical value for the chosen α.
df = 2. At α = 0.05 the critical value is 5.991, so H₀ (independence) is rejected.
10.Chi-square tests
Answer: χ²crit = 16.81
Chi-square tests: compare observed counts O with the counts E expected under H₀ using χ² = Σ(O − E)²/E. Goodness of fit: E = n·p for each category, df = k − 1 (minus one more for each parameter estimated from the data). Independence in an r × c table: E = row total × column total / grand total, df = (r − 1)(c − 1). Reject H₀ when χ² exceeds the critical value for the chosen α.
Degrees of freedom
df = (3 − 1)·(4 − 1) = 6
Table
χ²crit(df = 6, α = 0.01) = 16.812
Answer
χ²crit = 16.812
Reject H₀ if the computed χ² is larger than this value.