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 436 and 510 h? Give P as a decimal.
μ = 500σ = 40x1 = 436x2 = 510
P =
2.Normal distributionHard
The delivery times are normally distributed with mean 32 min and standard deviation 6 min. Find the value x below which 25% of them lie (the 25th percentile).
μ = 32σ = 6P = 25%
x = min
3.Confidence intervalsMedium
The blood pressure of 20 patients is measured before and after a treatment. The differences d = before − after (mmHg) have mean 7.5 and standard deviation 0.8. Find the margin of error E of the 99% confidence interval for the mean difference.
n = 20d̄ = 7.5sd = 0.899%t(19) = 2.861
E = mmHg
4.Confidence intervalsHard
The population standard deviation is 22. 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 1.5?
σ = 22E = 1.599%
n =
5.Hypothesis tests & confidence intervalsMedium
Two production lines make the same part. Line 1: a sample of 50 parts has mean length 100 mm, and the line's known standard deviation is 11 mm. Line 2: a sample of 80 parts has mean 92.1 mm, known standard deviation 19 mm. Find the test statistic z for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (line 1 minus line 2).
A one-sided z test at α = 0.01 must detect a shift of the mean of 4 units with a power of 90%. The standard deviation is 9. Find the minimum sample size n.
α = 0.01power = 90%δ = 4σ = 9zα = 2.326zβ = 1.282
n =
7.Linear regression & correlationMedium
The table gives 5 paired observations of x = load in kN and y = extension in mm. Find the coefficient of determination r² of the least-squares line.
n = 5
r² =
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, 15) is made. Find its residual e = y − ŷ.
n = 5(5, 15)
e =
9.Chi-square testsMedium
The contingency table shows observed counts. For a chi-square test of independence, find the expected count E for the cell “day shift, no defect”.
row total = 89column total = 38N = 165
E =
10.Chi-square testsHard
A chi-square goodness-of-fit test checks whether counts in 5 categories follow a Poisson distribution whose mean is estimated from the same data. How many degrees of freedom df does the test have?
k = 5
df =
3Solutions
1.Normal distribution
Answer: P = 0.5439
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 = (436 − 500)/40 = −1.6
z₂ = (510 − 500)/40 = 0.25
Area between
P = Φ(0.25) − Φ(−1.6) = 0.5987 − 0.0548 = 0.5439
Answer
P = 0.5439
2.Normal distribution
Answer: x = 27.95 min
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 = 32 + −0.6745·6 = 27.95 min
Answer
x = 27.95 min
3.Confidence intervals
Answer: E = 0.5118 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 = 19, t = 2.861
Margin of error
E = 2.861·0.8/√20 = 0.5118
Answer
E = 0.5118 mmHg
4.Confidence intervals
Answer: n = 1428
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 = (2.576·22/1.5)² = 1427
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 1428
Answer
n = 1428
5.Hypothesis tests & confidence intervals
Answer: z = 3
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 = √(11²/50 + 19²/80) = √6.933 = 2.633
Test statistic
z = (100 − 92.1)/2.633 = 3
Answer
z = 3
|z| > 1.96, so H₀ is rejected at the 5% level (two-sided).
6.Hypothesis tests & confidence intervals
Answer: n = 66
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 + 1.282)·9/4)² = 65.9
Round up
n = 66
Answer
n = 66
7.Linear regression & correlation
Answer: r² = 0.9882
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 = 114/5 = 22.8
Sums of squares and products
Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 132
Syy = Σ(y − ȳ)² = 440.8
Coefficient of determination
r² = 132²/(40·440.8) = 0.9882
Answer
r² = 0.9882
About 98.82% of the variation in y is explained by the line.
8.Linear regression & correlation
Answer: e = −4.1
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 = 110/5 = 22
Sums of squares and products
Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 116
Regression line
b = 116/40 = 2.9
a = 22 − 2.9·6 = 4.6
ŷ = 4.6 + 2.9·x
Predicted value
ŷ = 4.6 + 2.9·5 = 19.1
Residual
e = 15 − 19.1 = −4.1
Answer
e = −4.1
The point lies below the line.
9.Chi-square tests
Answer: E = 20.5
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.
Expected count
E = 89·38/165 = 20.5
Answer
E = 20.5
10.Chi-square tests
Answer: df = 3
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.