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 delivery times are normally distributed with mean 32 min and standard deviation 6 min. What fraction lies between 29 and 43.1 min? Give P as a decimal.
μ = 32σ = 6x1 = 29x2 = 43.1
P =
2.Normal distributionHard
Hole diameters D are normal with mean 25.02 mm and standard deviation 0.02 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 = 25.02 mm, σ = 0.02 mmshaft: μ = 25 mm, σ = 0.01 mm
P =
3.Confidence intervalsMedium
A sample of 10 items has mean 78.5 and standard deviation s = 2. The population is roughly normal. Find the upper limit of the 95% confidence interval for the mean.
n = 10x̄ = 78.5s = 295%t(9) = 2.262
upper limit =
4.Confidence intervalsHard
The population standard deviation is 12. 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 2?
σ = 12E = 299%
n =
5.Hypothesis tests & confidence intervalsMedium
The mean of a population is claimed to be 250; the population standard deviation is known to be 40. A random sample of 100 observations has a mean of 242. Test H₀: μ = 250 against H₁: μ ≠ 250. Find the p-value p.
μ₀ = 250σ = 40n = 100x̄ = 242H₁: μ ≠ μ₀
p =
6.Hypothesis tests & confidence intervalsHard
In a trial, 175 of 250 patients on treatment A recover, and 160 of 250 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.
x₁ = 175n₁ = 250x₂ = 160n₂ = 250
z =
7.Linear regression & correlationMedium
The table gives 6 paired observations of x = advertising spend in k€ and y = sales in k€. Find the coefficient of determination r² of the least-squares line.
n = 6
r² =
8.Linear regression & correlationHard
The table gives 6 paired observations of x = temperature in °C and y = ice creams sold. Use the least-squares line ŷ = a + b·x. For which value x does the line predict ŷ = 84?
n = 6ŷ = 84
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 “staff, car”.
row total = 81column total = 55N = 145
E =
10.Chi-square testsHard
A chi-square goodness-of-fit test checks whether counts in 9 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 = 0.6593
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 = (29 − 32)/6 = −0.5
z₂ = (43.1 − 32)/6 = 1.85
Area between
P = Φ(1.85) − Φ(−0.5) = 0.9678 − 0.3085 = 0.6593
Answer
P = 0.6593
2.Normal distribution
Answer: P = 0.8133
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 = 25.02 − 25 = 0.02 mm, σ = 0.02236 mm
Standardise
z = (0 − 0.02)/0.02236 = −0.89
Probability
P = 1 − Φ(−0.89) = 0.8133
Answer
P = 0.8133
3.Confidence intervals
Answer: upper limit = 79.93
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.
Critical value
df = n − 1 = 9, t = 2.262
Margin of error
E = 2.262·2/√10 = 1.431
Interval
78.5 ± 1.431 → [77.07, 79.93]
Answer
upper limit = 79.93
4.Confidence intervals
Answer: n = 239
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 = (2.576·12/2)² = 238.9
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 239
Answer
n = 239
5.Hypothesis tests & confidence intervals
Answer: p = 0.0455
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.
Test statistic
SE = σ/√n = 40/√100 = 4
z = (242 − 250)/4 = −2
Tail area
Φ(2) = 0.9772
1 − Φ(2) = 0.0228
p-value
p = 2·0.0228 = 0.0455
Answer
p = 0.0455
p ≤ 0.05, so H₀ is rejected at the 5% level.
6.Hypothesis tests & confidence intervals
Answer: z = 1.427
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 = 175/250 = 0.7
p̂₂ = 160/250 = 0.64
p̂ = 335/500 = 0.67
Standard error under H₀
SE = √(0.67·0.33·(1/250 + 1/250)) = 0.04206
Test statistic
z = (0.7 − 0.64)/0.04206 = 1.427
Answer
z = 1.427
|z| < 1.96, so H₀ is not rejected at the 5% level (two-sided).
7.Linear regression & correlation
Answer: r² = 0.9828
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 = 284/6 = 47.33
Sums of squares and products
Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 292
Syy = Σ(y − ȳ)² = 1239
Coefficient of determination
r² = 292²/(70·1239) = 0.9828
Answer
r² = 0.9828
About 98.28% of the variation in y is explained by the line.
8.Linear regression & correlation
Answer: x = 17.04
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 = 568/6 = 94.67
Sums of squares and products
Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 380
Regression line
b = 380/70 = 5.429
a = 94.67 − 5.429·19 = −8.476
ŷ = −8.476 + 5.429·x
Solve for x
x = (84 − (−8.476))/5.429 = 17.04
Answer
x = 17.04
9.Chi-square tests
Answer: E = 30.72
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 = 81·55/145 = 30.72
Answer
E = 30.72
10.Chi-square tests
Answer: χ²crit = 14.07
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 = 9 − 1 − 1 = 7
Table
χ²crit(df = 7, α = 0.05) = 14.067
Answer
χ²crit = 14.067
Reject H₀ if the computed χ² is larger than this value.