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 245.25 and 257.75 g? Give P as a decimal.
μ = 250σ = 5x1 = 245.25x2 = 257.75
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 top 90% is to the right of it. Find the cut-off x.
μ = 65 pointsσ = 10 points90%
x = points
3.Confidence intervalsMedium
A sample of 10 items has mean 14.5 and standard deviation s = 17. The population is roughly normal. Find the margin of error E of a 90% confidence interval for the mean.
n = 10x̄ = 14.5s = 1790%t(9) = 1.833
E =
4.Confidence intervalsHard
Nothing is known in advance about the proportion of defective products. How many products n must be inspected so that a 95% confidence interval for the proportion has a margin of error of at most 0.03?
p = 0.5E = 0.0395%
n =
5.Hypothesis tests & confidence intervalsMedium
Two production lines make the same part. Line 1: a sample of 50 parts has mean length 55 mm, and the line's known standard deviation is 17 mm. Line 2: a sample of 80 parts has mean 60.4 mm, known standard deviation 5 mm. Find the test statistic z for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (line 1 minus line 2).
A sample of 36 measurements has mean 71.2. The population standard deviation is 10. Find the upper limit U of the 95% confidence interval for the mean.
x̄ = 71.2σ = 10n = 3695%
U =
7.Linear regression & correlationMedium
The table gives 5 paired observations of x = temperature in °C and y = ice creams sold. 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 = car age in years and y = price in k€. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 5.5.
n = 6x = 5.5
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 = 802 × 3
χ² =
10.Chi-square testsHard
A chi-square goodness-of-fit test compares observed counts in 11 categories with fully specified expected proportions (no parameters are estimated from the data). How many degrees of freedom df does the test have?
k = 11
df =
3Solutions
1.Normal distribution
Answer: P = 0.7684
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 = (245.25 − 250)/5 = −0.95
z₂ = (257.75 − 250)/5 = 1.55
Area between
P = Φ(1.55) − Φ(−0.95) = 0.9394 − 0.1711 = 0.7684
Answer
P = 0.7684
2.Normal distribution
Answer: x = 52.2 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) = 0.1
z from the table
z ≈ −1.28
Back to the original units
x = 65 − 1.28·10 = 52.2 points
Answer
x = 52.2 points
3.Confidence intervals
Answer: E = 9.854
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 = 1.833
Margin of error
E = 1.833·17/√10 = 9.854
Answer
E = 9.854
4.Confidence intervals
Answer: n = 1068
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 = 1.96²·0.5·0.5/0.03² = 1067
Round up
n = 1068
Answer
n = 1068
5.Hypothesis tests & confidence intervals
Answer: z = −2.188
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.
Standard error of the difference
SE = √(17²/50 + 5²/80) = √6.093 = 2.468
Test statistic
z = (55 − 60.4)/2.468 = −2.188
Answer
z = −2.188
|z| > 1.96, so H₀ is rejected at the 5% level (two-sided).
6.Hypothesis tests & confidence intervals
Answer: U = 74.47
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/√36 = 3.267
Interval limit
U = 71.2 + 3.267 = 74.47
Answer
U = 74.47
7.Linear regression & correlation
Answer: r² = 0.9925
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 = 120/5 = 24
ȳ = 451/5 = 90.2
Sums of squares and products
Sxx = Σ(x − x̄)² = 90
Sxy = Σ(x − x̄)(y − ȳ) = 420
Syy = Σ(y − ȳ)² = 1975
Coefficient of determination
r² = 420²/(90·1975) = 0.9925
Answer
r² = 0.9925
About 99.25% of the variation in y is explained by the line.
8.Linear regression & correlation
Answer: ŷ = 17.04
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 = 118/6 = 19.67
Sums of squares and products
Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −23
Regression line
b = −23/17.5 = −1.314
a = 19.67 − (−1.314)·3.5 = 24.27
ŷ = 24.27 − 1.314·x
Prediction
ŷ = 24.27 + (−1.314)·5.5 = 17.04
Answer
ŷ = 17.04
9.Chi-square tests
Answer: χ² = 9.143
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: df = 10
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.