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
X is normal with mean 60 s and standard deviation 5 s; Y is independent and normal with mean 45 s and standard deviation 4 s. Find the probability P that X + Y is more than 98.6 s. Round z to two decimals.
μX = 60, σX = 5μY = 45, σY = 4X + Y > 98.6 s
P =
2.Normal distributionHard
The exam scores are normally distributed with mean 65 points and standard deviation 10 points. Find the value x below which 75% of them lie (the 75th percentile).
μ = 65σ = 10P = 75%
x = points
3.Confidence intervalsMedium
A 95% confidence interval for a mean, based on 25 observations and a known σ, has a margin of error of 12. What is the margin of error E₂ of a 99% interval from the same data?
95%n₁ = 25E₁ = 1299%
E₂ =
4.Confidence intervalsHard
The population standard deviation is 33. 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 0.5?
σ = 33E = 0.595%
n =
5.Hypothesis tests & confidence intervalsMedium
Two teaching methods are compared. Group 1: 10 students, mean score 31, standard deviation 10. Group 2: 7 students, mean score 38.1, standard deviation 9. 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 = 10x̄₁ = 31s₁ = 10n₂ = 7x̄₂ = 38.1s₂ = 9
t =
6.Hypothesis tests & confidence intervalsHard
A sample of 81 measurements has mean 499.5. The population standard deviation is 4. Find the upper limit U of the 95% confidence interval for the mean.
x̄ = 499.5σ = 4n = 8195%
U =
7.Linear regression & correlationMedium
The table gives 5 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.
n = 5Σx = 15Σy = 231Σx² = 55Σy² = 10867Σxy = 736
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. A new observation (3, 6) is made. Find its residual e = y − ŷ.
n = 6(3, 6)
e =
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 = 1402 × 2
χ² =
10.Chi-square testsHard
The table shows observed counts. Test at α = 0.01 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 = 0.01
χ² =
3Solutions
1.Normal distribution
Answer: P = 0.8413
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 sum
μ = 105 s, σ = √(5² + 4²) = 6.403 s
Standardise
z = (98.6 − 105)/6.403 = −1
Upper tail
P = 1 − Φ(−1) = 1 − 0.1587 = 0.8413
Answer
P = 0.8413
2.Normal distribution
Answer: x = 71.75 points
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) = 0.75 → z = 0.6745
Back to x
x = 65 + 0.6745·10 = 71.75 points
Answer
x = 71.75 points
3.Confidence intervals
Answer: E₂ = 15.77
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 = (2.576/1.96)·√(25/25) = 1.314
New margin
E₂ = 12·1.314 = 15.77
Answer
E₂ = 15.77
4.Confidence intervals
Answer: n = 16735
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 = (1.96·33/0.5)² = 16734
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 16735
Answer
n = 16735
5.Hypothesis tests & confidence intervals
Answer: t = −1.499
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 = (9·10² + 6·9²)/15 = 92.4
sp = 9.612
Standard error
SE = 9.612·√(1/10 + 1/7) = 4.737
Test statistic
t = (31 − 38.1)/4.737 = −1.499
Answer
t = −1.499
With df = 15, the two-sided 5% critical value is 2.131; |t| is smaller, so H₀ is not rejected.
6.Hypothesis tests & confidence intervals
Answer: U = 500.4
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).
Margin of error
E = 1.96·4/√81 = 0.8711
Interval limit
U = 499.5 + 0.8711 = 500.4
Answer
U = 500.4
7.Linear regression & correlation
Answer: r = 0.9743
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 = 736 − 15·231/5 = 43
Sxx = 55 − 15²/5 = 10
Syy = 10867 − 231²/5 = 194.8
Correlation coefficient
r = 43/√(10·194.8) = 0.9743
Answer
r = 0.9743
8.Linear regression & correlation
Answer: e = 1.924
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 = 56/6 = 9.333
Sums of squares and products
Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 92
Regression line
b = 92/70 = 1.314
a = 9.333 − 1.314·7 = 0.1333
ŷ = 0.1333 + 1.314·x
Predicted value
ŷ = 0.1333 + 1.314·3 = 4.076
Residual
e = 6 − 4.076 = 1.924
Answer
e = 1.924
The point lies above the line.
9.Chi-square tests
Answer: χ² = 16.97
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 rejected.
10.Chi-square tests
Answer: χ² = 8.717
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.