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 battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. What fraction lies between 452 and 542 h? Give P as a decimal.
μ = 500σ = 40x1 = 452x2 = 542
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 95% of them lie (the 95th percentile).
μ = 250σ = 5P = 95%
x = g
3.Confidence intervalsMedium
A sample of 10 items has mean 92 and standard deviation s = 3. The population is roughly normal. Find the margin of error E of a 99% confidence interval for the mean.
n = 10x̄ = 92s = 399%t(9) = 3.25
E =
4.Confidence intervalsHard
The diameters of 10 machined shafts have a sample standard deviation of 3.7 µm. Assuming the diameters are normally distributed, find the upper limit σ of the 95% confidence interval for the population standard deviation. With df = 9, 2.5% of the χ² distribution lies below 2.7 and 2.5% lies above 19.023.
n = 10s = 3.7 µmχ² = 2.7, 19.023
σ = µm
5.Hypothesis tests & confidence intervalsMedium
The mean of a population is claimed to be 100; the population standard deviation is known to be 20. A random sample of 100 observations has a mean of 96.12. Test H₀: μ = 100 against H₁: μ < 100. Find the p-value p.
μ₀ = 100σ = 20n = 100x̄ = 96.12H₁: μ < μ₀
p =
6.Hypothesis tests & confidence intervalsHard
A sample of 25 measurements has mean 98.8. The population standard deviation is 15. Find the upper limit U of the 90% confidence interval for the mean.
x̄ = 98.8σ = 15n = 2590%
U =
7.Linear regression & correlationMedium
The table gives 5 paired observations of x = car age in years and y = price in k€. Their sums are given. Find Pearson's correlation coefficient r from these sums.
n = 5Σx = 15Σy = 89Σx² = 55Σy² = 1615Σxy = 250
r =
8.Linear regression & correlationHard
The table gives 5 paired observations of x = advertising spend in k€ and y = sales in k€. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 7.
n = 5x = 7
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 = 1602 × 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 = 0.01.
α = 0.01
χ²crit =
3Solutions
1.Normal distribution
Answer: P = 0.7381
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 = (452 − 500)/40 = −1.2
z₂ = (542 − 500)/40 = 1.05
Area between
P = Φ(1.05) − Φ(−1.2) = 0.8531 − 0.1151 = 0.7381
Answer
P = 0.7381
2.Normal distribution
Answer: x = 258.2 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) = 0.95 → z = 1.645
Back to x
x = 250 + 1.645·5 = 258.22 g
Answer
x = 258.22 g
3.Confidence intervals
Answer: E = 3.083
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 = 3.25
Margin of error
E = 3.25·3/√10 = 3.083
Answer
E = 3.083
4.Confidence intervals
Answer: σ = 6.755 µm
Interval for a variance: (n − 1)s²/σ² has a χ² distribution with n − 1 degrees of freedom. The 95% interval for σ² runs from (n − 1)s² divided by the upper χ² value (2.5% above it) to (n − 1)s² divided by the lower χ² value (2.5% below it); the interval for σ is the square root of both limits. It is not symmetric around s.
Sum of squares
(n − 1)s² = 9·3.7² = 123.2
Limits for σ²
123.2/19.023 = 6.477
123.2/2.7 = 45.63
Limit for σ
σ = √45.63 = 6.755 µm
Answer
σ = 6.755 µm
95% interval for σ: [2.545, 6.755] µm.
5.Hypothesis tests & confidence intervals
Answer: p = 0.02619
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 = 20/√100 = 2
z = (96.12 − 100)/2 = −1.94
Tail area
Φ(1.94) = 0.9738
1 − Φ(1.94) = 0.0262
p-value
p = 0.02619
Answer
p = 0.02619
p ≤ 0.05, so H₀ is rejected at the 5% level.
6.Hypothesis tests & confidence intervals
Answer: U = 103.7
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 = 1.645·15/√25 = 4.935
Interval limit
U = 98.8 + 4.935 = 103.7
Answer
U = 103.7
7.Linear regression & correlation
Answer: r = −0.9687
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 = 250 − 15·89/5 = −17
Sxx = 55 − 15²/5 = 10
Syy = 1615 − 89²/5 = 30.8
Correlation coefficient
r = −17/√(10·30.8) = −0.9687
Answer
r = −0.9687
8.Linear regression & correlation
Answer: ŷ = 45
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 = 30/5 = 6
ȳ = 210/5 = 42
Sums of squares and products
Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 120
Regression line
b = 120/40 = 3
a = 42 − 3·6 = 24
ŷ = 24 + 3·x
Prediction
ŷ = 24 + 3·7 = 45
Answer
ŷ = 45
9.Chi-square tests
Answer: χ² = 3.75
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 = 18.48
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 = 0.01) = 18.475
Answer
χ²crit = 18.475
Reject H₀ if the computed χ² is larger than this value.