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 masses of boxes are independent and normal with mean 12 kg and standard deviation 0.8 kg. Find the probability P that the total mass of 6 of them is below 73 kg. Round z to two decimals.
μ = 12 kgσ = 0.8 kgn = 6< 73 kg
P =
2.Normal distributionHard
The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. Drag the cut-off so the shaded bottom 15% is to the left of it. Find the cut-off x.
μ = 500 hσ = 40 h15%
x = h
3.Confidence intervalsMedium
The battery life of two phone models is compared. Model 1: 50 phones, mean 79 h, known standard deviation 3 h. Model 2: 100 phones, mean 76.5 h, known standard deviation 17 h. Find the lower limit of the 99% confidence interval for μ₁ − μ₂.
Two fertilisers are tested. Plot group 1: 12 plots, mean yield 11.5 kg, standard deviation 2 kg. Plot group 2: 14 plots, mean 6 kg, standard deviation 7 kg. Assume equal variances (pooled t, df = n₁ + n₂ − 2). Find the upper limit of the 90% confidence interval for μ₁ − μ₂.
10 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 9.18 and the standard deviation of the differences is 8.8. Find the paired test statistic t for H₀: μd = 0.
n = 10d̄ = 9.18sd = 8.8
t =
6.Hypothesis tests & confidence intervalsHard
How large must a sample be so that the 95% confidence interval for a mean has a margin of error of at most 2? The standard deviation is 10. Find the smallest sample size n.
σ = 10E = 295%
n =
7.Linear regression & correlationMedium
The table gives 5 paired observations of x = load in kN and y = extension in mm. Find the coefficient of determination r² of the least-squares line.
n = 5
r² =
8.Linear regression & correlationHard
The table gives 6 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 = 21?
n = 6ŷ = 21
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 = 1402 × 2
χ² =
10.Chi-square testsHard
A chi-square goodness-of-fit test compares observed counts in 8 categories with fully specified expected proportions (no parameters are estimated from the data). Using the table, find the critical value χ²crit at significance level α = 0.01.
α = 0.01
χ²crit =
3Solutions
1.Normal distribution
Answer: P = 0.695
Sums and differences of normal variables: if X and Y are independent and normal, X + Y and X − Y are normal too. Means add or subtract: E(X ± Y) = μX ± μY. Variances always add, also for a difference: Var(X ± Y) = σX² + σY², so σ = √(σX² + σY²). Standard deviations never add directly. The total of n independent items with the same distribution has mean n·μ and standard deviation σ·√n. Then standardise as usual: z = (x − mean)/σ and read Φ(z).
Distribution of the total
μT = 72 kg, σT = 0.8·√6 = 1.96 kg
Standardise
z = (73 − 72)/1.96 = 0.51
Probability
P = Φ(0.51) = 0.695
Answer
P = 0.695
2.Normal distribution
Answer: x = 458 h
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) = 0.15
z from the table
z ≈ −1.04
Back to the original units
x = 500 − 1.04·40 = 458 h
Answer
x = 458 h
3.Confidence intervals
Answer: lower limit = −2.014 h
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.
Point estimate
x̄₁ − x̄₂ = 79 − 76.5 = 2.5
Standard error
SE = √(3²/50 + 17²/100) = 1.752
Margin of error
E = 2.576·1.752 = 4.514
Interval
2.5 ± 4.514 → [−2.014, 7.014]
Answer
lower limit = −2.014 h
The interval contains 0, so the data do not show a significant difference.
4.Confidence intervals
Answer: upper limit = 9.086 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 = (11·2² + 13·7²)/24 = 28.38
sp = 5.327
Margin of error
E = 1.711·5.327·√(1/12 + 1/14) = 3.586
Interval
5.5 ± 3.586 → [1.914, 9.086]
Answer
upper limit = 9.086 kg
5.Hypothesis tests & confidence intervals
Answer: t = 3.299
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 = 8.8/√10 = 2.783
Test statistic
t = 9.18/2.783 = 3.299
Answer
t = 3.299
With df = 9, the two-sided 5% critical value is 2.262; |t| is larger, so H₀ is rejected.
6.Hypothesis tests & confidence intervals
Answer: n = 97
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).
Solve for n
n = (1.96·10/2)² = 96.04
Round up
n = 97
Answer
n = 97
7.Linear regression & correlation
Answer: r² = 0.8522
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 = 30/5 = 6
ȳ = 50/5 = 10
Sums of squares and products
Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 56
Syy = Σ(y − ȳ)² = 92
Coefficient of determination
r² = 56²/(40·92) = 0.8522
Answer
r² = 0.8522
About 85.22% of the variation in y is explained by the line.
8.Linear regression & correlation
Answer: x = 1.598
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 = 21/6 = 3.5
ȳ = 96/6 = 16
Sums of squares and products
Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −46
Regression line
b = −46/17.5 = −2.629
a = 16 − (−2.629)·3.5 = 25.2
ŷ = 25.2 − 2.629·x
Solve for x
x = (21 − 25.2)/(−2.629) = 1.598
Answer
x = 1.598
9.Chi-square tests
Answer: χ² = 1.061
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 = 1. At α = 0.05 the critical value is 3.841, so H₀ (independence) is not rejected.
10.Chi-square tests
Answer: χ²crit = 18.48
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 = 8 − 1 = 7
Table
χ²crit(df = 7, α = 0.01) = 18.475
Answer
χ²crit = 18.475
Reject H₀ if the computed χ² is larger than this value.