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 heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. What fraction lies between 167.15 and 190.6 cm? Give P as a decimal.
μ = 178σ = 7x1 = 167.15x2 = 190.6
P =
2.Normal distributionHard
The mass X of a product from line 1 and Y from line 2 are independent and normal: X has mean 500 g and standard deviation 8 g, Y has mean 490 g and standard deviation 6 g. Find the probability P that X is greater than Y. Round z to two decimals.
μX = 500 g, σX = 8 gμY = 490 g, σY = 6 g
P =
3.Confidence intervalsMedium
The battery life of two phone models is compared. Model 1: 40 phones, mean 123 h, known standard deviation 9 h. Model 2: 100 phones, mean 125.5 h, known standard deviation 10 h. Find the upper limit of the 99% confidence interval for μ₁ − μ₂.
The population standard deviation is 24. How large a sample n do you need so that a 99% confidence interval for the mean has a margin of error of at most 0.5?
σ = 24E = 0.599%
n =
5.Hypothesis tests & confidence intervalsMedium
A sample of 100 students has a mean score of 48.96 and a sample standard deviation of 4. Find the test statistic t for H₀: μ = 50.
μ₀ = 50x̄ = 48.96s = 4n = 100
t =
6.Hypothesis tests & confidence intervalsHard
In a trial, 83 of 150 patients on treatment A recover, and 201 of 300 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.
x₁ = 83n₁ = 150x₂ = 201n₂ = 300
z =
7.Linear regression & correlationMedium
The table gives 5 paired observations of x = load in kN and y = extension in mm. Find the coefficient of determination r² of the least-squares line.
n = 5
r² =
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 (11, 40) is made. Find its residual e = y − ŷ.
n = 6(11, 40)
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, bus”.
row total = 67column total = 32N = 139
E =
10.Chi-square testsHard
A chi-square goodness-of-fit test checks whether counts in 4 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 = 0.9035
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 = (167.15 − 178)/7 = −1.55
z₂ = (190.6 − 178)/7 = 1.8
Area between
P = Φ(1.8) − Φ(−1.55) = 0.9641 − 0.06057 = 0.9035
Answer
P = 0.9035
2.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 difference
μ = 500 − 490 = 10 g, σ = 10 g
Standardise
z = (0 − 10)/10 = −1
Probability
P = 1 − Φ(−1) = 0.8413
Answer
P = 0.8413
3.Confidence intervals
Answer: upper limit = 1.98 h
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.
Point estimate
x̄₁ − x̄₂ = 123 − 125.5 = −2.5
Standard error
SE = √(9²/40 + 10²/100) = 1.739
Margin of error
E = 2.576·1.739 = 4.48
Interval
−2.5 ± 4.48 → [−6.98, 1.98]
Answer
upper limit = 1.98 h
The interval contains 0, so the data do not show a significant difference.
4.Confidence intervals
Answer: n = 15289
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 = (2.576·24/0.5)² = 15289
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 15289
Answer
n = 15289
5.Hypothesis tests & confidence intervals
Answer: t = −2.6
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).
Standard error
SE = 4/√100 = 4/10 = 0.4
Test statistic
t = (48.96 − 50)/0.4 = −2.6
Answer
t = −2.6
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: z = −2.418
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.
Sample proportions
p̂₁ = 83/150 = 0.5533
p̂₂ = 201/300 = 0.67
p̂ = 284/450 = 0.6311
Standard error under H₀
SE = √(0.6311·0.3689·(1/150 + 1/300)) = 0.04825
Test statistic
z = (0.5533 − 0.67)/0.04825 = −2.418
Answer
z = −2.418
|z| > 1.96, so H₀ is rejected at the 5% level (two-sided).
7.Linear regression & correlation
Answer: r² = 0.995
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 = 84/5 = 16.8
Sums of squares and products
Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 80
Syy = Σ(y − ȳ)² = 160.8
Coefficient of determination
r² = 80²/(40·160.8) = 0.995
Answer
r² = 0.995
About 99.5% of the variation in y is explained by the line.
8.Linear regression & correlation
Answer: e = −1.967
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 = 163/6 = 27.17
Sums of squares and products
Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 259
Regression line
b = 259/70 = 3.7
a = 27.17 − 3.7·7 = 1.267
ŷ = 1.267 + 3.7·x
Predicted value
ŷ = 1.267 + 3.7·11 = 41.97
Residual
e = 40 − 41.97 = −1.967
Answer
e = −1.967
The point lies below the line.
9.Chi-square tests
Answer: E = 15.42
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 = 67·32/139 = 15.42
Answer
E = 15.42
10.Chi-square tests
Answer: χ²crit = 5.991
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 = 4 − 1 − 1 = 2
Table
χ²crit(df = 2, α = 0.05) = 5.991
Answer
χ²crit = 5.991
Reject H₀ if the computed χ² is larger than this value.