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
Drag z so that the shaded area between −z and z under the standard normal curve is 0.8990. Find z to two decimals.
A = 0.8990
z =
2.Normal distributionHard
Hole diameters D are normal with mean 25.04 mm and standard deviation 0.015 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 = 25.04 mm, σ = 0.015 mmshaft: μ = 25 mm, σ = 0.01 mm
P =
3.Confidence intervalsMedium
A 90% confidence interval for a mean, based on 36 observations and a known σ, has a margin of error of 6.3. What is the margin of error E₂ if the same study uses 144 observations instead?
90%n₁ = 36E₁ = 6.3n₂ = 144
E₂ =
4.Confidence intervalsHard
A pilot study suggests that about 0.2 of products are defective. How many products n must be inspected so that a 95% confidence interval for the proportion has a margin of error of at most 0.025?
p ≈ 0.2E = 0.02595%
n =
5.Hypothesis tests & confidence intervalsMedium
Two teaching methods are compared. Group 1: 17 students, mean score 66, standard deviation 9. Group 2: 5 students, mean score 52.7, standard deviation 5.5. Assume equal population variances and use the pooled two-sample t test. Find the test statistic t for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (group 1 minus group 2).
n₁ = 17x̄₁ = 66s₁ = 9n₂ = 5x̄₂ = 52.7s₂ = 5.5
t =
6.Hypothesis tests & confidence intervalsHard
A one-sided z test at α = 0.05 must detect a shift of the mean of 5 units with a power of 80%. The standard deviation is 20. Find the minimum sample size n.
The table gives 6 paired observations of x = temperature in °C and y = ice creams sold. Their sums are given. Find Pearson's correlation coefficient r from these sums.
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. A new observation (5.5, 11) is made. Find its residual e = y − ŷ.
n = 6(5.5, 11)
e =
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 “staff, bike”.
row total = 56column total = 67N = 143
E =
10.Chi-square testsHard
A chi-square test of independence is carried out on a contingency table with 3 rows and 6 columns (not counting the totals). How many degrees of freedom df does the test have?
3 × 6
df =
3Solutions
1.Normal distribution
Answer: z = 1.64
From an area back to z: Φ(z) is the area to the left of z. For a left-tail area A, find z with Φ(z) = A in the table. A right-tail area A means Φ(z) = 1 − A. A central area A between −z and z leaves (1 − A)/2 in each tail, so Φ(z) = (1 + A)/2. For a value x, use x = μ + zσ.
Area to the left of z
Φ(z) = (1 + 0.8990)/2 = 0.9495
Read the z-table backwards
Find 0.9495 inside the table and read off z.
z = 1.64
Answer
z = 1.64
2.Normal distribution
Answer: P = 0.9868
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 = 25.04 − 25 = 0.04 mm, σ = 0.01803 mm
Standardise
z = (0 − 0.04)/0.01803 = −2.22
Probability
P = 1 − Φ(−2.22) = 0.9868
Answer
P = 0.9868
3.Confidence intervals
Answer: E₂ = 3.15
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.
Ratio of margins
E₂/E₁ = (1.645/1.645)·√(36/144) = 0.5
New margin
E₂ = 6.3·0.5 = 3.15
Answer
E₂ = 3.15
4.Confidence intervals
Answer: n = 984
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 = 1.96²·0.2·0.8/0.025² = 983.4
Round up
n = 984
Answer
n = 984
5.Hypothesis tests & confidence intervals
Answer: t = 3.106
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.
Pooled variance
sp² = (16·9² + 4·5.5²)/20 = 70.85
sp = 8.417
Standard error
SE = 8.417·√(1/17 + 1/5) = 4.282
Test statistic
t = (66 − 52.7)/4.282 = 3.106
Answer
t = 3.106
With df = 20, the two-sided 5% critical value is 2.086; |t| is larger, so H₀ is rejected.
6.Hypothesis tests & confidence intervals
Answer: n = 99
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 + 0.842)·20/5)² = 98.96
Round up
n = 99
Answer
n = 99
7.Linear regression & correlation
Answer: r = 0.9751
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 = 7768 − 126·360/6 = 208
Sxx = 2716 − 126²/6 = 70
Syy = 22250 − 360²/6 = 650
Correlation coefficient
r = 208/√(70·650) = 0.9751
Answer
r = 0.9751
8.Linear regression & correlation
Answer: e = −4.419
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 = 122/6 = 20.33
Sums of squares and products
Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −43
Regression line
b = −43/17.5 = −2.457
a = 20.33 − (−2.457)·3.5 = 28.93
ŷ = 28.93 − 2.457·x
Predicted value
ŷ = 28.93 + (−2.457)·5.5 = 15.42
Residual
e = 11 − 15.42 = −4.419
Answer
e = −4.419
The point lies below the line.
9.Chi-square tests
Answer: E = 26.24
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 = 56·67/143 = 26.24
Answer
E = 26.24
10.Chi-square tests
Answer: df = 10
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.