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 masses of passengers with luggage are independent and normal with mean 85 kg and standard deviation 12 kg. Find the probability P that the total mass of 8 of them exceeds 663 kg. Round z to two decimals.
μ = 85 kgσ = 12 kgn = 8> 663 kg
P =
2.Normal distributionHard
The exam scores are normally distributed with mean 65 points and standard deviation 10 points. Drag the cut-off so the shaded bottom 25% is to the left of it. Find the cut-off x.
μ = 65 pointsσ = 10 points25%
x = points
3.Confidence intervalsMedium
The battery life of two phone models is compared. Model 1: 80 phones, mean 91.5 h, known standard deviation 20 h. Model 2: 100 phones, mean 94.5 h, known standard deviation 8 h. Find the upper limit of the 90% confidence interval for μ₁ − μ₂.
The population standard deviation is 10. 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 = 10E = 299%
n =
5.Hypothesis tests & confidence intervalsMedium
A sample of 100 students has a mean score of 199.4 and a sample standard deviation of 4. Find the test statistic t for H₀: μ = 200.
μ₀ = 200x̄ = 199.4s = 4n = 100
t =
6.Hypothesis tests & confidence intervalsHard
A sample of 81 measurements has mean 117.1. The population standard deviation is 10. Find the upper limit U of the 95% confidence interval for the mean.
x̄ = 117.1σ = 10n = 8195%
U =
7.Linear regression & correlationMedium
The table gives 6 paired observations of x = load in kN and y = extension in mm. Their sums are given. Find Pearson's correlation coefficient r from these sums.
n = 6Σx = 42Σy = 104Σx² = 364Σy² = 2052Σxy = 856
r =
8.Linear regression & correlationHard
The table gives 5 paired observations of x = car age in years and y = price in k€. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 2.5.
n = 5x = 2.5
ŷ =
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 “day shift, no defect”.
row total = 85column total = 66N = 182
E =
10.Chi-square testsHard
A chi-square test of independence is carried out on a contingency table with 3 rows and 3 columns (not counting the totals). How many degrees of freedom df does the test have?
3 × 3
df =
3Solutions
1.Normal distribution
Answer: P = 0.6915
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 total
μT = 680 kg, σT = 12·√8 = 33.94 kg
Standardise
z = (663 − 680)/33.94 = −0.5
Probability
P = 1 − Φ(−0.5) = 0.6915
Answer
P = 0.6915
2.Normal distribution
Answer: x = 58.3 points
From an area back to z: Φ(z) is the area to the left of z. For a left-tail area A, find z with Φ(z) = A in the table. A right-tail area A means Φ(z) = 1 − A. A central area A between −z and z leaves (1 − A)/2 in each tail, so Φ(z) = (1 + A)/2. For a value x, use x = μ + zσ.
Area to the left
Φ(z) = 0.25
z from the table
z ≈ −0.67
Back to the original units
x = 65 − 0.67·10 = 58.3 points
Answer
x = 58.3 points
3.Confidence intervals
Answer: upper limit = 0.9067 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 = 91.5 − 94.5 = −3
Standard error
SE = √(20²/80 + 8²/100) = 2.375
Margin of error
E = 1.645·2.375 = 3.907
Interval
−3 ± 3.907 → [−6.907, 0.9067]
Answer
upper limit = 0.9067 h
The interval contains 0, so the data do not show a significant difference.
4.Confidence intervals
Answer: n = 166
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 = (2.576·10/2)² = 165.9
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 166
Answer
n = 166
5.Hypothesis tests & confidence intervals
Answer: t = −1.5
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).
Standard error
SE = 4/√100 = 4/10 = 0.4
Test statistic
t = (199.4 − 200)/0.4 = −1.5
Answer
t = −1.5
t0.025; df = 99 = 1.984
|t| < 1.984, so H₀ is not rejected at the 5% level (two-sided).
6.Hypothesis tests & confidence intervals
Answer: U = 119.3
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 = 1.96·10/√81 = 2.178
Interval limit
U = 117.1 + 2.178 = 119.3
Answer
U = 119.3
7.Linear regression & correlation
Answer: r = 0.9689
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 = 856 − 42·104/6 = 128
Sxx = 364 − 42²/6 = 70
Syy = 2052 − 104²/6 = 249.3
Correlation coefficient
r = 128/√(70·249.3) = 0.9689
Answer
r = 0.9689
8.Linear regression & correlation
Answer: ŷ = 20.7
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 = 101/5 = 20.2
Sums of squares and products
Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = −10
Regression line
b = −10/10 = −1
a = 20.2 − (−1)·3 = 23.2
ŷ = 23.2 − 1·x
Prediction
ŷ = 23.2 + (−1)·2.5 = 20.7
Answer
ŷ = 20.7
9.Chi-square tests
Answer: E = 30.82
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 = 85·66/182 = 30.82
Answer
E = 30.82
10.Chi-square tests
Answer: df = 4
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.