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 exam scores are normally distributed with mean 65 points and standard deviation 10 points. What fraction lies between 53.5 and 83.5 points? Give P as a decimal.
μ = 65σ = 10x1 = 53.5x2 = 83.5
P =
2.Normal distributionHard
The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. Find the value x below which 80% of them lie (the 80th percentile).
μ = 500σ = 40P = 80%
x = h
3.Confidence intervalsMedium
The blood pressure of 13 patients is measured before and after a treatment. The differences d = before − after (mmHg) have mean 3 and standard deviation 5.2. Find the margin of error E of the 95% confidence interval for the mean difference.
n = 13d̄ = 3sd = 5.295%t(12) = 2.179
E = mmHg
4.Confidence intervalsHard
A pilot study suggests that about 0.1 of products are defective. How many products n must be inspected so that a 99% confidence interval for the proportion has a margin of error of at most 0.02?
p ≈ 0.1E = 0.0299%
n =
5.Hypothesis tests & confidence intervalsMedium
A sample of 16 students has a mean score of 127.3 and a sample standard deviation of 10. Find the test statistic t for H₀: μ = 120.
μ₀ = 120x̄ = 127.3s = 10n = 16
t =
6.Hypothesis tests & confidence intervalsHard
A one-sided z test at α = 0.01 must detect a shift of the mean of 2.5 units with a power of 80%. The standard deviation is 20. Find the minimum sample size n.
The table gives 5 paired observations of x = car age in years and y = price in k€. Find Pearson's correlation coefficient r.
n = 5
r =
8.Linear regression & correlationHard
The table gives 5 paired observations of x = load in kN and y = extension in mm. Use the least-squares line ŷ = a + b·x. A new observation (5, 13) is made. Find its residual e = y − ŷ.
n = 5(5, 13)
e =
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 10 categories with fully specified expected proportions (no parameters are estimated from the data). How many degrees of freedom df does the test have?
k = 10
df =
3Solutions
1.Normal distribution
Answer: P = 0.8428
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 = (53.5 − 65)/10 = −1.15
z₂ = (83.5 − 65)/10 = 1.85
Area between
P = Φ(1.85) − Φ(−1.15) = 0.9678 − 0.1251 = 0.8428
Answer
P = 0.8428
2.Normal distribution
Answer: x = 533.7 h
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) = 0.8 → z = 0.8416
Back to x
x = 500 + 0.8416·40 = 533.66 h
Answer
x = 533.66 h
3.Confidence intervals
Answer: E = 3.143 mmHg
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.
Critical value
df = 12, t = 2.179
Margin of error
E = 2.179·5.2/√13 = 3.143
Answer
E = 3.143 mmHg
4.Confidence intervals
Answer: n = 1494
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 = 2.576²·0.1·0.9/0.02² = 1493
Round up
n = 1494
Answer
n = 1494
5.Hypothesis tests & confidence intervals
Answer: t = 2.92
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 = 10/√16 = 10/4 = 2.5
Test statistic
t = (127.3 − 120)/2.5 = 2.92
Answer
t = 2.92
t0.025; df = 15 = 2.131
|t| > 2.131, so H₀ is rejected at the 5% level (two-sided).
6.Hypothesis tests & confidence intervals
Answer: n = 643
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 = ((2.326 + 0.842)·20/2.5)² = 642.3
Round up
n = 643
Answer
n = 643
7.Linear regression & correlation
Answer: r = −0.9237
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 = 95/5 = 19
Sums of squares and products
Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = −23
Syy = Σ(y − ȳ)² = 62
Correlation coefficient
r = −23/√(10·62) = −0.9237
Answer
r = −0.9237
8.Linear regression & correlation
Answer: e = −4.9
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 = 106/5 = 21.2
Sums of squares and products
Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 132
Regression line
b = 132/40 = 3.3
a = 21.2 − 3.3·6 = 1.4
ŷ = 1.4 + 3.3·x
Predicted value
ŷ = 1.4 + 3.3·5 = 17.9
Residual
e = 13 − 17.9 = −4.9
Answer
e = −4.9
The point lies below the line.
9.Chi-square tests
Answer: χ² = 20.91
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 = 9
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.