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 bag weights are normally distributed with mean 250 g and standard deviation 5 g. What fraction lies between 246 and 256.25 g? Give P as a decimal.
μ = 250σ = 5x1 = 246x2 = 256.25
P =
2.Normal distributionHard
Hole diameters D are normal with mean 25.03 mm and standard deviation 0.015 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 = 0.015 mmshaft: μ = 25 mm, σ = 0.02 mm
P =
3.Confidence intervalsMedium
A 99% confidence interval for a mean, based on 100 observations and a known σ, has a margin of error of 10.9. What is the margin of error E₂ of a 95% interval based on 25 observations?
99%n₁ = 100E₁ = 10.995%n₂ = 25
E₂ =
4.Confidence intervalsHard
The population standard deviation is 14. 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 = 14E = 595%
n =
5.Hypothesis tests & confidence intervalsMedium
A sample of 100 students has a mean score of 203.6 and a sample standard deviation of 15. Find the test statistic t for H₀: μ = 200.
μ₀ = 200x̄ = 203.6s = 15n = 100
t =
6.Hypothesis tests & confidence intervalsHard
A one-sided z test at α = 0.01 must detect a shift of the mean of 4 units with a power of 95%. The standard deviation is 17. Find the minimum sample size n.
The table gives 6 paired observations of x = advertising spend in k€ and y = sales in k€. Their sums are given. Find Pearson's correlation coefficient r from these sums.
The table gives 5 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 = 5x = 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 = 1602 × 3
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 = 1603 × 2α = 0.05
χ² =
3Solutions
1.Normal distribution
Answer: P = 0.6825
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 = (246 − 250)/5 = −0.8
z₂ = (256.25 − 250)/5 = 1.25
Area between
P = Φ(1.25) − Φ(−0.8) = 0.8944 − 0.2119 = 0.6825
Answer
P = 0.6825
2.Normal distribution
Answer: P = 0.8849
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 = 0.025 mm
Standardise
z = (0 − 0.03)/0.025 = −1.2
Probability
P = 1 − Φ(−1.2) = 0.8849
Answer
P = 0.8849
3.Confidence intervals
Answer: E₂ = 16.59
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.
Ratio of margins
E₂/E₁ = (1.96/2.576)·√(100/25) = 1.522
New margin
E₂ = 10.9·1.522 = 16.59
Answer
E₂ = 16.59
4.Confidence intervals
Answer: n = 31
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 = (1.96·14/5)² = 30.12
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 31
Answer
n = 31
5.Hypothesis tests & confidence intervals
Answer: t = 2.4
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).
Standard error
SE = 15/√100 = 15/10 = 1.5
Test statistic
t = (203.6 − 200)/1.5 = 2.4
Answer
t = 2.4
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 = 285
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 + 1.645)·17/4)² = 284.8
Round up
n = 285
Answer
n = 285
7.Linear regression & correlation
Answer: r = 0.991
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.
Centred sums
Sxy = 3796 − 48·426/6 = 388
Sxx = 454 − 48²/6 = 70
Syy = 32436 − 426²/6 = 2190
Correlation coefficient
r = 388/√(70·2190) = 0.991
Answer
r = 0.991
8.Linear regression & correlation
Answer: ŷ = 15.15
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 = 63/5 = 12.6
Sums of squares and products
Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 102
Regression line
b = 102/40 = 2.55
a = 12.6 − 2.55·6 = −2.7
ŷ = −2.7 + 2.55·x
Prediction
ŷ = −2.7 + 2.55·7 = 15.15
Answer
ŷ = 15.15
9.Chi-square tests
Answer: χ² = 4.857
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 = 2. At α = 0.05 the critical value is 5.991, so H₀ (independence) is not rejected.
10.Chi-square tests
Answer: χ² = 12.86
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.