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 bag weights are normally distributed with mean 250 g and standard deviation 5 g. What fraction lies between 244 and 260 g? Give P as a decimal.
μ = 250σ = 5x1 = 244x2 = 260
P =
2.Normal distributionHard
The exam scores are normally distributed with mean 65 points and standard deviation 10 points. Find the value x below which 25% of them lie (the 25th percentile).
μ = 65σ = 10P = 25%
x = points
3.Confidence intervalsMedium
A sample of 7 items has mean 75 and standard deviation s = 7.5. The population is roughly normal. Find the upper limit of the 99% confidence interval for the mean.
n = 7x̄ = 75s = 7.599%t(6) = 3.707
upper limit =
4.Confidence intervalsHard
A pilot study suggests that about 0.15 of products are defective. How many products n must be inspected so that a 90% confidence interval for the proportion has a margin of error of at most 0.025?
p ≈ 0.15E = 0.02590%
n =
5.Hypothesis tests & confidence intervalsMedium
A sample of 81 students has a mean score of 497.2 and a sample standard deviation of 8. Find the test statistic t for H₀: μ = 500.
μ₀ = 500x̄ = 497.2s = 8n = 81
t =
6.Hypothesis tests & confidence intervalsHard
A one-sided z test of H₀: μ = 500 against H₁: μ < 500 is done at α = 0.01 with a sample of 25; the standard deviation is known to be 5. In reality the mean is 497.8. Find the power 1 − β of the test in percent.
μ₀ = 500μ = 497.8σ = 5n = 25α = 0.01
1 − β = %
7.Linear regression & correlationMedium
The table gives 6 paired observations of x = load in kN and y = extension in mm. 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 = car age in years and y = price in k€. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 1.5.
n = 6x = 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 = 1802 × 3
χ² =
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 = 0.8622
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 = (244 − 250)/5 = −1.2
z₂ = (260 − 250)/5 = 2
Area between
P = Φ(2) − Φ(−1.2) = 0.9772 − 0.1151 = 0.8622
Answer
P = 0.8622
2.Normal distribution
Answer: x = 58.26 points
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) = 0.25 → z = −0.6745
Back to x
x = 65 + −0.6745·10 = 58.26 points
Answer
x = 58.26 points
3.Confidence intervals
Answer: upper limit = 85.51
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 = 6, t = 3.707
Margin of error
E = 3.707·7.5/√7 = 10.51
Interval
75 ± 10.51 → [64.49, 85.51]
Answer
upper limit = 85.51
4.Confidence intervals
Answer: n = 553
Confidence interval for a proportion: p̂ ± z·√(p̂(1 − p̂)/n), where p̂ = x/n. To reach a margin E you need n ≥ z²·p(1 − p)/E², rounded up; without a prior estimate use p = 0.5, which gives the largest n.
Solve for n
n = 1.645²·0.15·0.85/0.025² = 552
Round up
n = 553
Answer
n = 553
5.Hypothesis tests & confidence intervals
Answer: t = −3.15
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/√81 = 8/9 = 0.8889
Test statistic
t = (497.2 − 500)/0.8889 = −3.15
Answer
t = −3.15
t0.025; df = 80 = 1.99
|t| > 1.99, so H₀ is rejected at the 5% level (two-sided).
6.Hypothesis tests & confidence intervals
Answer: 1 − β = 44.99 %
Power of a test: power = 1 − β is the probability of rejecting H₀ when it is false. Find the cutoff x̄c from μ₀ and z_α, then compute the probability of landing beyond x̄c using the true mean: z = (x̄c − μ)/(σ/√n).
Critical sample mean
SE = 5/√25 = 1
x̄c = 500 − 2.326·1 = 497.7
z under the true mean
z = (497.7 − 497.8)/1 = −0.126
Power
power = P(Z < −0.126) = 44.99%
1 − β = 44.99%
Answer
1 − β = 44.99%
7.Linear regression & correlation
Answer: r² = 0.9918
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 = 187/6 = 31.17
Sums of squares and products
Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 267
Syy = Σ(y − ȳ)² = 1027
Coefficient of determination
r² = 267²/(70·1027) = 0.9918
Answer
r² = 0.9918
About 99.18% of the variation in y is explained by the line.
8.Linear regression & correlation
Answer: ŷ = 26.42
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 = 139/6 = 23.17
Sums of squares and products
Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −28.5
Regression line
b = −28.5/17.5 = −1.629
a = 23.17 − (−1.629)·3.5 = 28.87
ŷ = 28.87 − 1.629·x
Prediction
ŷ = 28.87 + (−1.629)·1.5 = 26.42
Answer
ŷ = 26.42
9.Chi-square tests
Answer: χ² = 45.6
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 rejected.
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.