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 bags of cement are independent and normal with mean 25 kg and standard deviation 0.5 kg. Find the probability P that the total mass of 16 of them exceeds 404 kg. Round z to two decimals.
μ = 25 kgσ = 0.5 kgn = 16> 404 kg
P =
2.Normal distributionHard
The bag weights are normally distributed with mean 250 g and standard deviation 5 g. Find the value x below which 20% of them lie (the 20th percentile).
μ = 250σ = 5P = 20%
x = g
3.Confidence intervalsMedium
The battery life of two phone models is compared. Model 1: 50 phones, mean 149.5 h, known standard deviation 16 h. Model 2: 60 phones, mean 154 h, known standard deviation 3 h. Find the lower limit of the 99% confidence interval for μ₁ − μ₂.
The population standard deviation is 23. 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 2.5?
σ = 23E = 2.599%
n =
5.Hypothesis tests & confidence intervalsMedium
A sample of 16 students has a mean score of 98.2 and a sample standard deviation of 8. Find the test statistic t for H₀: μ = 100.
μ₀ = 100x̄ = 98.2s = 8n = 16
t =
6.Hypothesis tests & confidence intervalsHard
A one-sided z test at α = 0.05 must detect a shift of the mean of 2 units with a power of 80%. The standard deviation is 13. Find the minimum sample size n.
The table gives 6 paired observations of x = car age in years and y = price in k€. Find Pearson's correlation coefficient r.
n = 6
r =
8.Linear regression & correlationHard
The table gives 5 paired observations of x = hours studied and y = exam score. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 1.5.
n = 5x = 1.5
ŷ =
9.Chi-square testsMedium
The contingency table shows observed counts. Compute the chi-square statistic χ² for the test of independence between the row and column variables. Do not use a continuity correction.
N = 2002 × 3
χ² =
10.Chi-square testsHard
A chi-square goodness-of-fit test compares observed counts in 7 categories with fully specified expected proportions (no parameters are estimated from the 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 = 0.02275
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 = 400 kg, σT = 0.5·√16 = 2 kg
Standardise
z = (404 − 400)/2 = 2
Probability
P = 1 − Φ(2) = 0.02275
Answer
P = 0.02275
2.Normal distribution
Answer: x = 245.8 g
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 for the percentile
Φ(z) = 0.2 → z = −0.8416
Back to x
x = 250 + −0.8416·5 = 245.79 g
Answer
x = 245.79 g
3.Confidence intervals
Answer: lower limit = −10.41 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 = 149.5 − 154 = −4.5
Standard error
SE = √(16²/50 + 3²/60) = 2.296
Margin of error
E = 2.576·2.296 = 5.914
Interval
−4.5 ± 5.914 → [−10.41, 1.414]
Answer
lower limit = −10.41 h
The interval contains 0, so the data do not show a significant difference.
4.Confidence intervals
Answer: n = 562
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 = (2.576·23/2.5)² = 561.7
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 562
Answer
n = 562
5.Hypothesis tests & confidence intervals
Answer: t = −0.9
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 = 8/√16 = 8/4 = 2
Test statistic
t = (98.2 − 100)/2 = −0.9
Answer
t = −0.9
t0.025; df = 15 = 2.131
|t| < 2.131, so H₀ is not rejected at the 5% level (two-sided).
6.Hypothesis tests & confidence intervals
Answer: n = 262
Sample size for a test: to detect a true shift δ with power 1 − β in a one-sided z test at level α, the standard error must satisfy δ = (z_α + z_β)·σ/√n, so n = ((z_α + z_β)·σ/δ)², rounded up.
Solve for n
n = ((1.645 + 0.842)·13/2)² = 261.3
Round up
n = 262
Answer
n = 262
7.Linear regression & correlation
Answer: r = −0.991
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 = 21/6 = 3.5
ȳ = 76/6 = 12.67
Sums of squares and products
Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −52
Syy = Σ(y − ȳ)² = 157.3
Correlation coefficient
r = −52/√(17.5·157.3) = −0.991
Answer
r = −0.991
8.Linear regression & correlation
Answer: ŷ = 54.65
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 = 304/5 = 60.8
Sums of squares and products
Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = 41
Regression line
b = 41/10 = 4.1
a = 60.8 − 4.1·3 = 48.5
ŷ = 48.5 + 4.1·x
Prediction
ŷ = 48.5 + 4.1·1.5 = 54.65
Answer
ŷ = 54.65
9.Chi-square tests
Answer: χ² = 4.516
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.
df = 2. At α = 0.05 the critical value is 5.991, so H₀ (independence) is not rejected.
10.Chi-square tests
Answer: χ²crit = 16.81
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 = 6
Table
χ²crit(df = 6, α = 0.01) = 16.812
Answer
χ²crit = 16.812
Reject H₀ if the computed χ² is larger than this value.