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 764.8 g. Round z to two decimals.
μX = 250, σX = 4μY = 500, σY = 9X + Y > 764.8 g
P =
2.Normal distributionHard
Hole diameters D are normal with mean 25.02 mm and standard deviation 0.015 mm. Shaft diameters d are independent and normal with mean 25 mm and standard deviation 0.015 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 = 0.015 mmshaft: μ = 25 mm, σ = 0.015 mm
P =
3.Confidence intervalsMedium
A sample of 15 items has mean 62 and standard deviation s = 11.5. The population is roughly normal. Find the lower limit of the 99% confidence interval for the mean.
n = 15x̄ = 62s = 11.599%t(14) = 2.977
lower limit =
4.Confidence intervalsHard
The population standard deviation is 39. How large a sample n do you need so that a 90% confidence interval for the mean has a margin of error of at most 1?
σ = 39E = 190%
n =
5.Hypothesis tests & confidence intervalsMedium
A sample of 36 students has a mean score of 501.4 and a sample standard deviation of 5. Find the test statistic t for H₀: μ = 500.
μ₀ = 500x̄ = 501.4s = 5n = 36
t =
6.Hypothesis tests & confidence intervalsHard
A sample of 64 measurements has mean 201.2. The population standard deviation is 4. Find the upper limit U of the 99% confidence interval for the mean.
x̄ = 201.2σ = 4n = 6499%
U =
7.Linear regression & correlationMedium
The table gives 5 paired observations of x = car age in years and y = price in k€. 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 = 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 = 103?
n = 6ŷ = 103
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, low”.
row total = 64column total = 78N = 154
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 = 0.01.
α = 0.01
χ²crit =
3Solutions
1.Normal distribution
Answer: P = 0.06681
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 = (764.8 − 750)/9.849 = 1.5
Upper tail
P = 1 − Φ(1.5) = 1 − 0.9332 = 0.06681
Answer
P = 0.06681
2.Normal distribution
Answer: P = 0.8264
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 = 0.02121 mm
Standardise
z = (0 − 0.02)/0.02121 = −0.94
Probability
P = 1 − Φ(−0.94) = 0.8264
Answer
P = 0.8264
3.Confidence intervals
Answer: lower limit = 53.16
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 = 14, t = 2.977
Margin of error
E = 2.977·11.5/√15 = 8.84
Interval
62 ± 8.84 → [53.16, 70.84]
Answer
lower limit = 53.16
4.Confidence intervals
Answer: n = 4116
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 = (1.645·39/1)² = 4116
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 4116
Answer
n = 4116
5.Hypothesis tests & confidence intervals
Answer: t = 1.68
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 = 5/√36 = 5/6 = 0.8333
Test statistic
t = (501.4 − 500)/0.8333 = 1.68
Answer
t = 1.68
t0.025; df = 35 = 2.03
|t| < 2.03, so H₀ is not rejected at the 5% level (two-sided).
6.Hypothesis tests & confidence intervals
Answer: U = 202.5
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 = 2.576·4/√64 = 1.288
Interval limit
U = 201.2 + 1.288 = 202.5
Answer
U = 202.5
7.Linear regression & correlation
Answer: r² = 0.9078
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 = 15/5 = 3
ȳ = 109/5 = 21.8
Sums of squares and products
Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = −32
Syy = Σ(y − ȳ)² = 112.8
Coefficient of determination
r² = −32²/(10·112.8) = 0.9078
Answer
r² = 0.9078
About 90.78% of the variation in y is explained by the line.
8.Linear regression & correlation
Answer: x = 23.01
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 = 126/6 = 21
ȳ = 553/6 = 92.17
Sums of squares and products
Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 377
Regression line
b = 377/70 = 5.386
a = 92.17 − 5.386·21 = −20.93
ŷ = −20.93 + 5.386·x
Solve for x
x = (103 − (−20.93))/5.386 = 23.01
Answer
x = 23.01
9.Chi-square tests
Answer: E = 32.42
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 = 64·78/154 = 32.42
Answer
E = 32.42
10.Chi-square tests
Answer: χ²crit = 9.21
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 = 0.01) = 9.210
Answer
χ²crit = 9.210
Reject H₀ if the computed χ² is larger than this value.