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 107.5 points? Give P as a decimal.
μ = 100σ = 15x1 = 76x2 = 107.5
P =
2.Normal distributionHard
Hole diameters D are normal with mean 25.03 mm and standard deviation 0.02 mm. Shaft diameters d are independent and normal with mean 25 mm and standard deviation 0.02 mm. A randomly chosen shaft fits when D − d > 0. Find the probability P that it fits. Round z to two decimals.
hole: μ = 25.03 mm, σ = 0.02 mmshaft: μ = 25 mm, σ = 0.02 mm
P =
3.Confidence intervalsMedium
A sample of 10 items has mean 112 and standard deviation s = 12.5. The population is roughly normal. Find the lower limit of the 99% confidence interval for the mean.
n = 10x̄ = 112s = 12.599%t(9) = 3.25
lower limit =
4.Confidence intervalsHard
The population standard deviation is 9. 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 1?
σ = 9E = 199%
n =
5.Hypothesis tests & confidence intervalsMedium
A sample of 100 students has a mean score of 199 and a sample standard deviation of 4. Find the test statistic t for H₀: μ = 200.
μ₀ = 200x̄ = 199s = 4n = 100
t =
6.Hypothesis tests & confidence intervalsHard
A one-sided z test at α = 0.01 must detect a shift of the mean of 2.5 units with a power of 80%. The standard deviation is 8. Find the minimum sample size n.
The table gives 6 paired observations of x = car age in years and y = price in k€. Find the coefficient of determination r² of the least-squares line.
n = 6
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. Predict ŷ for x = 5.5.
n = 6x = 5.5
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 = 1002 × 2
χ² =
10.Chi-square testsHard
A chi-square test of independence is carried out on a contingency table with 3 rows and 3 columns (not counting the totals). How many degrees of freedom df does the test have?
3 × 3
df =
3Solutions
1.Normal distribution
Answer: P = 0.6367
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 = (107.5 − 100)/15 = 0.5
Area between
P = Φ(0.5) − Φ(−1.6) = 0.6915 − 0.0548 = 0.6367
Answer
P = 0.6367
2.Normal distribution
Answer: P = 0.8554
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 = 25.03 − 25 = 0.03 mm, σ = 0.02828 mm
Standardise
z = (0 − 0.03)/0.02828 = −1.06
Probability
P = 1 − Φ(−1.06) = 0.8554
Answer
P = 0.8554
3.Confidence intervals
Answer: lower limit = 99.15
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 = 9, t = 3.25
Margin of error
E = 3.25·12.5/√10 = 12.85
Interval
112 ± 12.85 → [99.15, 124.8]
Answer
lower limit = 99.15
4.Confidence intervals
Answer: n = 538
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 = (2.576·9/1)² = 537.5
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 538
Answer
n = 538
5.Hypothesis tests & confidence intervals
Answer: t = −2.5
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
SE = 4/√100 = 4/10 = 0.4
Test statistic
t = (199 − 200)/0.4 = −2.5
Answer
t = −2.5
t0.025; df = 99 = 1.984
|t| > 1.984, so H₀ is rejected at the 5% level (two-sided).
6.Hypothesis tests & confidence intervals
Answer: n = 103
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 + 0.842)·8/2.5)² = 102.8
Round up
n = 103
Answer
n = 103
7.Linear regression & correlation
Answer: r² = 0.7741
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 = 140/6 = 23.33
Sums of squares and products
Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −34
Syy = Σ(y − ȳ)² = 85.33
Coefficient of determination
r² = −34²/(17.5·85.33) = 0.7741
Answer
r² = 0.7741
About 77.41% of the variation in y is explained by the line.
8.Linear regression & correlation
Answer: ŷ = 19.03
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 = 131/6 = 21.83
Sums of squares and products
Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −24.5
Regression line
b = −24.5/17.5 = −1.4
a = 21.83 − (−1.4)·3.5 = 26.73
ŷ = 26.73 − 1.4·x
Prediction
ŷ = 26.73 + (−1.4)·5.5 = 19.03
Answer
ŷ = 19.03
9.Chi-square tests
Answer: χ² = 1.5
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: df = 4
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.