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 bolts are independent and normal with mean 40 g and standard deviation 1.2 g. Find the probability P that the total mass of 16 of them exceeds 649.6 g. Round z to two decimals.
μ = 40 gσ = 1.2 gn = 16> 649.6 g
P =
2.Normal distributionHard
The delivery times are normally distributed with mean 32 min and standard deviation 6 min. Find the value x below which 5% of them lie (the 5th percentile).
μ = 32σ = 6P = 5%
x = min
3.Confidence intervalsMedium
A 90% confidence interval for a mean, based on 25 observations and a known σ, has a margin of error of 9.2. What is the margin of error E₂ if the same study uses 225 observations instead?
90%n₁ = 25E₁ = 9.2n₂ = 225
E₂ =
4.Confidence intervalsHard
A pilot study suggests that about 0.8 of products are defective. How many products n must be inspected so that a 99% confidence interval for the proportion has a margin of error of at most 0.05?
p ≈ 0.8E = 0.0599%
n =
5.Hypothesis tests & confidence intervalsMedium
A sample of 16 students has a mean score of 503.4 and a sample standard deviation of 8. Find the test statistic t for H₀: μ = 500.
μ₀ = 500x̄ = 503.4s = 8n = 16
t =
6.Hypothesis tests & confidence intervalsHard
A one-sided z test at α = 0.05 must detect a shift of the mean of 3 units with a power of 95%. The standard deviation is 6. Find the minimum sample size n.
α = 0.05power = 95%δ = 3σ = 6zα = 1.645zβ = 1.645
n =
7.Linear regression & correlationMedium
The table gives 5 paired observations of x = hours studied and y = exam score. 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 = hours studied and y = exam score. Use the least-squares line ŷ = a + b·x. For which value x does the line predict ŷ = 79?
n = 5ŷ = 79
x =
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
A chi-square test of independence is carried out on a contingency table with 3 rows and 4 columns (not counting the totals). How many degrees of freedom df does the test have?
3 × 4
df =
3Solutions
1.Normal distribution
Answer: P = 0.02275
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 = 640 g, σT = 1.2·√16 = 4.8 g
Standardise
z = (649.6 − 640)/4.8 = 2
Probability
P = 1 − Φ(2) = 0.02275
Answer
P = 0.02275
2.Normal distribution
Answer: x = 22.13 min
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 for the percentile
Φ(z) = 0.05 → z = −1.645
Back to x
x = 32 + −1.645·6 = 22.13 min
Answer
x = 22.13 min
3.Confidence intervals
Answer: E₂ = 3.067
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 = (1.645/1.645)·√(25/225) = 0.3333
New margin
E₂ = 9.2·0.3333 = 3.067
Answer
E₂ = 3.067
4.Confidence intervals
Answer: n = 425
Confidence interval for a proportion: p̂ ± z·√(p̂(1 − p̂)/n), where p̂ = x/n. To reach a margin E you need n ≥ z²·p(1 − p)/E², rounded up; without a prior estimate use p = 0.5, which gives the largest n.
Solve for n
n = 2.576²·0.8·0.2/0.05² = 424.7
Round up
n = 425
Answer
n = 425
5.Hypothesis tests & confidence intervals
Answer: t = 1.7
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 = 8/√16 = 8/4 = 2
Test statistic
t = (503.4 − 500)/2 = 1.7
Answer
t = 1.7
t0.025; df = 15 = 2.131
|t| < 2.131, so H₀ is not rejected at the 5% level (two-sided).
6.Hypothesis tests & confidence intervals
Answer: n = 44
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 = ((1.645 + 1.645)·6/3)² = 43.3
Round up
n = 44
Answer
n = 44
7.Linear regression & correlation
Answer: r² = 0.98
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 = 394/5 = 78.8
Sums of squares and products
Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 244
Syy = Σ(y − ȳ)² = 1519
Coefficient of determination
r² = 244²/(40·1519) = 0.98
Answer
r² = 0.98
About 98% of the variation in y is explained by the line.
8.Linear regression & correlation
Answer: x = 8.05
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 = 374/5 = 74.8
Sums of squares and products
Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 160
Regression line
b = 160/40 = 4
a = 74.8 − 4·7 = 46.8
ŷ = 46.8 + 4·x
Solve for x
x = (79 − 46.8)/4 = 8.05
Answer
x = 8.05
9.Chi-square tests
Answer: χ² = 3.75
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 = 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.