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 250 g and standard deviation 4 g; Y is independent and normal with mean 500 g and standard deviation 9 g. Find the probability P that X + Y is more than 757.9 g. Round z to two decimals.
μX = 250, σX = 4μY = 500, σY = 9X + Y > 757.9 g
P =
2.Normal distributionHard
Runner A's lap time X and runner B's lap time Y are independent and normal: X has mean 52 s and standard deviation 4 s, Y has mean 48 s and standard deviation 3 s. Find the probability P that X is greater than Y. Round z to two decimals.
μX = 52 s, σX = 4 sμY = 48 s, σY = 3 s
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 3.3. What is the margin of error E₂ if the same study uses 324 observations instead?
90%n₁ = 36E₁ = 3.3n₂ = 324
E₂ =
4.Confidence intervalsHard
Two fertilisers are tested. Plot group 1: 9 plots, mean yield 16.5 kg, standard deviation 4.5 kg. Plot group 2: 5 plots, mean 4.5 kg, standard deviation 3.5 kg. Assume equal variances (pooled t, df = n₁ + n₂ − 2). Find the lower limit of the 95% confidence interval for μ₁ − μ₂.
Two teaching methods are compared. Group 1: 8 students, mean score 73.5, standard deviation 5. Group 2: 8 students, mean score 81.9, 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 = 8x̄₁ = 73.5s₁ = 5n₂ = 8x̄₂ = 81.9s₂ = 9
t =
6.Hypothesis tests & confidence intervalsHard
A sample of 25 measurements has mean 104.8. The population standard deviation is 20. Find the upper limit U of the 95% confidence interval for the mean.
x̄ = 104.8σ = 20n = 2595%
U =
7.Linear regression & correlationMedium
The table gives 6 paired observations of x = temperature in °C and y = ice creams sold. Find Pearson's correlation coefficient r.
n = 6
r =
8.Linear regression & correlationHard
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. For which value x does the line predict ŷ = 14?
n = 5ŷ = 14
x =
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 “urban, high”.
row total = 103column total = 64N = 207
E =
10.Chi-square testsHard
A chi-square goodness-of-fit test checks whether counts in 7 categories follow a Poisson distribution whose mean is estimated from the same data. Using the table, find the critical value χ²crit at significance level α = 0.05.
α = 0.05
χ²crit =
3Solutions
1.Normal distribution
Answer: P = 0.2119
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 = 750 g, σ = √(4² + 9²) = 9.849 g
Standardise
z = (757.9 − 750)/9.849 = 0.8
Upper tail
P = 1 − Φ(0.8) = 1 − 0.7881 = 0.2119
Answer
P = 0.2119
2.Normal distribution
Answer: P = 0.7881
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 = 52 − 48 = 4 s, σ = 5 s
Standardise
z = (0 − 4)/5 = −0.8
Probability
P = 1 − Φ(−0.8) = 0.7881
Answer
P = 0.7881
3.Confidence intervals
Answer: E₂ = 1.1
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)·√(36/324) = 0.3333
New margin
E₂ = 3.3·0.3333 = 1.1
Answer
E₂ = 1.1
4.Confidence intervals
Answer: lower limit = 6.904 kg
Interval for a difference: estimate ± critical value × standard error. For two independent means with known σ: SE = √(σ₁²/n₁ + σ₂²/n₂) and z. With equal unknown variances: pooled s_p and t with n₁ + n₂ − 2 degrees of freedom. For paired data, work with the differences: d̄ ± t·s_d/√n.
Pooled standard deviation
sp² = (8·4.5² + 4·3.5²)/12 = 17.58
sp = 4.193
Margin of error
E = 2.179·4.193·√(1/9 + 1/5) = 5.096
Interval
12 ± 5.096 → [6.904, 17.1]
Answer
lower limit = 6.904 kg
5.Hypothesis tests & confidence intervals
Answer: t = −2.308
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 = (7·5² + 7·9²)/14 = 53
sp = 7.28
Standard error
SE = 7.28·√(1/8 + 1/8) = 3.64
Test statistic
t = (73.5 − 81.9)/3.64 = −2.308
Answer
t = −2.308
With df = 14, the two-sided 5% critical value is 2.145; |t| is larger, so H₀ is rejected.
6.Hypothesis tests & confidence intervals
Answer: U = 112.6
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 = 1.96·20/√25 = 7.84
Interval limit
U = 104.8 + 7.84 = 112.6
Answer
U = 112.6
7.Linear regression & correlation
Answer: r = 0.9661
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 = 114/6 = 19
ȳ = 301/6 = 50.17
Sums of squares and products
Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 273
Syy = Σ(y − ȳ)² = 1141
Correlation coefficient
r = 273/√(70·1141) = 0.9661
Answer
r = 0.9661
8.Linear regression & correlation
Answer: x = 7.2
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 = 58/5 = 11.6
Sums of squares and products
Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 80
Regression line
b = 80/40 = 2
a = 11.6 − 2·6 = −0.4
ŷ = −0.4 + 2·x
Solve for x
x = (14 − (−0.4))/2 = 7.2
Answer
x = 7.2
9.Chi-square tests
Answer: E = 31.85
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 = 103·64/207 = 31.85
Answer
E = 31.85
10.Chi-square tests
Answer: χ²crit = 11.07
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.
Degrees of freedom
df = 7 − 1 − 1 = 5
Table
χ²crit(df = 5, α = 0.05) = 11.070
Answer
χ²crit = 11.070
Reject H₀ if the computed χ² is larger than this value.