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 passengers with luggage are independent and normal with mean 85 kg and standard deviation 12 kg. Find the probability P that the total mass of 5 of them exceeds 411.6 kg. Round z to two decimals.
μ = 85 kgσ = 12 kgn = 5> 411.6 kg
P =
2.Normal distributionHard
Hole diameters D are normal with mean 25.02 mm and standard deviation 0.01 mm. Shaft diameters d are independent and normal with mean 25 mm and standard deviation 0.01 mm. A randomly chosen shaft fits when D − d > 0. Find the probability P that it fits. Round z to two decimals.
hole: μ = 25.02 mm, σ = 0.01 mmshaft: μ = 25 mm, σ = 0.01 mm
P =
3.Confidence intervalsMedium
A sample of 8 items has mean 78.5 and standard deviation s = 9.5. The population is roughly normal. Find the lower limit of the 95% confidence interval for the mean.
n = 8x̄ = 78.5s = 9.595%t(7) = 2.365
lower limit =
4.Confidence intervalsHard
The population standard deviation is 35. How large a sample n do you need so that a 99% confidence interval for the mean has a margin of error of at most 5?
σ = 35E = 599%
n =
5.Hypothesis tests & confidence intervalsMedium
Two production lines make the same part. Line 1: a sample of 100 parts has mean length 109 mm, and the line's known standard deviation is 20 mm. Line 2: a sample of 50 parts has mean 101.2 mm, known standard deviation 17 mm. Find the test statistic z for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (line 1 minus line 2).
A sample of 100 measurements has mean 498.5. The population standard deviation is 15. Find the upper limit U of the 99% confidence interval for the mean.
x̄ = 498.5σ = 15n = 10099%
U =
7.Linear regression & correlationMedium
The table gives 5 paired observations of x = advertising spend in k€ and y = sales in k€. Find the coefficient of determination r² of the least-squares line.
n = 5
r² =
8.Linear regression & correlationHard
The table gives 5 paired observations of x = advertising spend in k€ and y = sales in k€. Use the least-squares line ŷ = a + b·x. For which value x does the line predict ŷ = 77?
n = 5ŷ = 77
x =
9.Chi-square testsMedium
The contingency table shows observed counts. For a chi-square test of independence, find the expected count E for the cell “staff, bike”.
row total = 85column total = 76N = 186
E =
10.Chi-square testsHard
A chi-square goodness-of-fit test compares observed counts in 3 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 = 0.05.
α = 0.05
χ²crit =
3Solutions
1.Normal distribution
Answer: P = 0.6915
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 = 425 kg, σT = 12·√5 = 26.83 kg
Standardise
z = (411.6 − 425)/26.83 = −0.5
Probability
P = 1 − Φ(−0.5) = 0.6915
Answer
P = 0.6915
2.Normal distribution
Answer: P = 0.9207
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 difference
μ = 25.02 − 25 = 0.02 mm, σ = 0.01414 mm
Standardise
z = (0 − 0.02)/0.01414 = −1.41
Probability
P = 1 − Φ(−1.41) = 0.9207
Answer
P = 0.9207
3.Confidence intervals
Answer: lower limit = 70.56
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 = 7, t = 2.365
Margin of error
E = 2.365·9.5/√8 = 7.943
Interval
78.5 ± 7.943 → [70.56, 86.44]
Answer
lower limit = 70.56
4.Confidence intervals
Answer: n = 326
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.
Solve for n
n = (2.576·35/5)² = 325.2
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 326
Answer
n = 326
5.Hypothesis tests & confidence intervals
Answer: z = 2.494
Hypothesis test: the test statistic = (estimate − value under H₀)/(standard error under H₀). The p-value is the probability, assuming H₀ is true, of a statistic at least as extreme as the one observed: the area in one tail for a one-sided H₁, twice the tail area for a two-sided H₁. Reject H₀ when p ≤ α.
Standard error of the difference
SE = √(20²/100 + 17²/50) = √9.78 = 3.127
Test statistic
z = (109 − 101.2)/3.127 = 2.494
Answer
z = 2.494
|z| > 1.96, so H₀ is rejected at the 5% level (two-sided).
6.Hypothesis tests & confidence intervals
Answer: U = 502.4
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 = 2.576·15/√100 = 3.864
Interval limit
U = 498.5 + 3.864 = 502.4
Answer
U = 502.4
7.Linear regression & correlation
Answer: r² = 0.9452
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 = 35/5 = 7
ȳ = 238/5 = 47.6
Sums of squares and products
Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 136
Syy = Σ(y − ȳ)² = 489.2
Coefficient of determination
r² = 136²/(40·489.2) = 0.9452
Answer
r² = 0.9452
About 94.52% of the variation in y is explained by the line.
8.Linear regression & correlation
Answer: x = 8.103
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 = 35/5 = 7
ȳ = 361/5 = 72.2
Sums of squares and products
Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 174
Regression line
b = 174/40 = 4.35
a = 72.2 − 4.35·7 = 41.75
ŷ = 41.75 + 4.35·x
Solve for x
x = (77 − 41.75)/4.35 = 8.103
Answer
x = 8.103
9.Chi-square tests
Answer: E = 34.73
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.
Expected count
E = 85·76/186 = 34.73
Answer
E = 34.73
10.Chi-square tests
Answer: χ²crit = 5.991
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 = 2
Table
χ²crit(df = 2, α = 0.05) = 5.991
Answer
χ²crit = 5.991
Reject H₀ if the computed χ² is larger than this value.