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 battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. What fraction lies between 462 and 582 h? Give P as a decimal.
μ = 500σ = 40x1 = 462x2 = 582
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 0.015 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 = 0.015 mm
P =
3.Confidence intervalsMedium
A sample of 13 items has mean 65.5 and standard deviation s = 10.5. The population is roughly normal. Find the lower limit of the 99% confidence interval for the mean.
n = 13x̄ = 65.5s = 10.599%t(12) = 3.055
lower limit =
4.Confidence intervalsHard
The population standard deviation is 16. How large a sample n do you need so that a 95% confidence interval for the mean has a margin of error of at most 5?
σ = 16E = 595%
n =
5.Hypothesis tests & confidence intervalsMedium
The mean of a population is claimed to be 250; the population standard deviation is known to be 12. A random sample of 36 observations has a mean of 247.56. Test H₀: μ = 250 against H₁: μ < 250. Find the p-value p.
μ₀ = 250σ = 12n = 36x̄ = 247.56H₁: μ < μ₀
p =
6.Hypothesis tests & confidence intervalsHard
A one-sided z test at α = 0.01 must detect a shift of the mean of 5 units with a power of 90%. The standard deviation is 24. Find the minimum sample size n.
The table gives 6 paired observations of x = load in kN and y = extension in mm. Find Pearson's correlation coefficient r.
n = 6
r =
8.Linear regression & correlationHard
The table gives 5 paired observations of x = hours studied and y = exam score. Use the least-squares line ŷ = a + b·x. For which value x does the line predict ŷ = 69?
n = 5ŷ = 69
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 “students, bike”.
row total = 77column total = 37N = 152
E =
10.Chi-square testsHard
The table shows observed counts. Test at α = 0.05 whether the two variables are independent (no continuity correction). Compute the test statistic χ², then compare it with the critical value from the table.
N = 2203 × 2α = 0.05
χ² =
3Solutions
1.Normal distribution
Answer: P = 0.8088
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 = (462 − 500)/40 = −0.95
z₂ = (582 − 500)/40 = 2.05
Area between
P = Φ(2.05) − Φ(−0.95) = 0.9798 − 0.1711 = 0.8088
Answer
P = 0.8088
2.Normal distribution
Answer: P = 0.8665
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 = 0.01803 mm
Standardise
z = (0 − 0.02)/0.01803 = −1.11
Probability
P = 1 − Φ(−1.11) = 0.8665
Answer
P = 0.8665
3.Confidence intervals
Answer: lower limit = 56.6
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 = 12, t = 3.055
Margin of error
E = 3.055·10.5/√13 = 8.897
Interval
65.5 ± 8.897 → [56.6, 74.4]
Answer
lower limit = 56.6
4.Confidence intervals
Answer: n = 40
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 = (1.96·16/5)² = 39.34
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 40
Answer
n = 40
5.Hypothesis tests & confidence intervals
Answer: p = 0.1112
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.
Test statistic
SE = σ/√n = 12/√36 = 2
z = (247.56 − 250)/2 = −1.22
Tail area
Φ(1.22) = 0.8888
1 − Φ(1.22) = 0.1112
p-value
p = 0.1112
Answer
p = 0.1112
p > 0.05, so H₀ is not rejected at the 5% level.
6.Hypothesis tests & confidence intervals
Answer: n = 300
Sample size for a test: to detect a true shift δ with power 1 − β in a one-sided z test at level α, the standard error must satisfy δ = (z_α + z_β)·σ/√n, so n = ((z_α + z_β)·σ/δ)², rounded up.
Solve for n
n = ((2.326 + 1.282)·24/5)² = 299.9
Round up
n = 300
Answer
n = 300
7.Linear regression & correlation
Answer: r = 0.9916
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 = 42/6 = 7
ȳ = 145/6 = 24.17
Sums of squares and products
Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 211
Syy = Σ(y − ȳ)² = 646.8
Correlation coefficient
r = 211/√(70·646.8) = 0.9916
Answer
r = 0.9916
8.Linear regression & correlation
Answer: x = 8.089
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 = 25/5 = 5
ȳ = 284/5 = 56.8
Sums of squares and products
Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 158
Regression line
b = 158/40 = 3.95
a = 56.8 − 3.95·5 = 37.05
ŷ = 37.05 + 3.95·x
Solve for x
x = (69 − 37.05)/3.95 = 8.089
Answer
x = 8.089
9.Chi-square tests
Answer: E = 18.74
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 = 77·37/152 = 18.74
Answer
E = 18.74
10.Chi-square tests
Answer: χ² = 30.03
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.