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 IQ scores are normally distributed with mean 100 points and standard deviation 15 points. What fraction lies between 76 and 112.75 points? Give P as a decimal.
μ = 100σ = 15x1 = 76x2 = 112.75
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
The blood pressure of 7 patients is measured before and after a treatment. The differences d = before − after (mmHg) have mean −4.3 and standard deviation 5.2. Find the upper limit of the 99% confidence interval for the mean difference.
n = 7d̄ = −4.3sd = 5.299%t(6) = 3.707
upper limit = mmHg
4.Confidence intervalsHard
The population standard deviation is 17. 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 4?
σ = 17E = 495%
n =
5.Hypothesis tests & confidence intervalsMedium
Two production lines make the same part. Line 1: a sample of 40 parts has mean length 80.5 mm, and the line's known standard deviation is 3 mm. Line 2: a sample of 30 parts has mean 82.8 mm, known standard deviation 10 mm. Find the test statistic z for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (line 1 minus line 2).
A sample of 49 measurements has mean 50.8. The population standard deviation is 5. Find the upper limit U of the 99% confidence interval for the mean.
x̄ = 50.8σ = 5n = 4999%
U =
7.Linear regression & correlationMedium
The table gives 5 paired observations of x = load in kN and y = extension in mm. Find Pearson's correlation coefficient r.
n = 5
r =
8.Linear regression & correlationHard
The table gives 6 paired observations of x = load in kN and y = extension in mm. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 7.
n = 6x = 7
ŷ =
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 = 1802 × 2
χ² =
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 = 0.7475
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 = (76 − 100)/15 = −1.6
z₂ = (112.75 − 100)/15 = 0.85
Area between
P = Φ(0.85) − Φ(−1.6) = 0.8023 − 0.0548 = 0.7475
Answer
P = 0.7475
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: upper limit = 2.986 mmHg
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.
Critical value
df = 6, t = 3.707
Margin of error
E = 3.707·5.2/√7 = 7.286
Interval
−4.3 ± 7.286 → [−11.59, 2.986]
Answer
upper limit = 2.986 mmHg
4.Confidence intervals
Answer: n = 70
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 = (1.96·17/4)² = 69.39
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 70
Answer
n = 70
5.Hypothesis tests & confidence intervals
Answer: z = −1.219
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 = √(3²/40 + 10²/30) = √3.558 = 1.886
Test statistic
z = (80.5 − 82.8)/1.886 = −1.219
Answer
z = −1.219
|z| < 1.96, so H₀ is not rejected at the 5% level (two-sided).
6.Hypothesis tests & confidence intervals
Answer: U = 52.64
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 = 2.576·5/√49 = 1.84
Interval limit
U = 50.8 + 1.84 = 52.64
Answer
U = 52.64
7.Linear regression & correlation
Answer: r = 0.9979
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 = 96/5 = 19.2
Sums of squares and products
Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 134
Syy = Σ(y − ȳ)² = 450.8
Correlation coefficient
r = 134/√(40·450.8) = 0.9979
Answer
r = 0.9979
8.Linear regression & correlation
Answer: ŷ = 23
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 = 138/6 = 23
Sums of squares and products
Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 224
Regression line
b = 224/70 = 3.2
a = 23 − 3.2·7 = 0.6
ŷ = 0.6 + 3.2·x
Prediction
ŷ = 0.6 + 3.2·7 = 23
Answer
ŷ = 23
9.Chi-square tests
Answer: χ² = 1.6
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: χ² = 2.032
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.