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 240 and 260 g? Give P as a decimal.
μ = 250σ = 5x1 = 240x2 = 260
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 80% of them lie (the 80th percentile).
μ = 250σ = 5P = 80%
x = g
3.Confidence intervalsMedium
The battery life of two phone models is compared. Model 1: 60 phones, mean 125.5 h, known standard deviation 5 h. Model 2: 60 phones, mean 137.5 h, known standard deviation 13 h. Find the margin of error E of the 90% confidence interval for μ₁ − μ₂.
The population standard deviation is 12. 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 = 12E = 1.599%
n =
5.Hypothesis tests & confidence intervalsMedium
The mean of a population is claimed to be 250; the population standard deviation is known to be 6. A random sample of 36 observations has a mean of 250.92. Test H₀: μ = 250 against H₁: μ > 250. Find the p-value p.
μ₀ = 250σ = 6n = 36x̄ = 250.92H₁: μ > μ₀
p =
6.Hypothesis tests & confidence intervalsHard
A one-sided z test at α = 0.05 must detect a shift of the mean of 3 units with a power of 90%. The standard deviation is 7. Find the minimum sample size n.
α = 0.05power = 90%δ = 3σ = 7zα = 1.645zβ = 1.282
n =
7.Linear regression & correlationMedium
The table gives 6 paired observations of x = advertising spend in k€ and y = sales in k€. Find the coefficient of determination r² of the least-squares line.
n = 6
r² =
8.Linear regression & correlationHard
The table gives 5 paired observations of x = hours studied and y = exam score. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 6.
n = 5x = 6
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 = 2202 × 3
χ² =
10.Chi-square testsHard
The table shows observed counts. Test at α = 0.05 whether the two variables are independent (no continuity correction). Compute the test statistic χ², then compare it with the critical value from the table.
N = 2603 × 2α = 0.05
χ² =
3Solutions
1.Normal distribution
Answer: P = 0.9545
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 = (240 − 250)/5 = −2
z₂ = (260 − 250)/5 = 2
Area between
P = Φ(2) − Φ(−2) = 0.9772 − 0.02275 = 0.9545
Answer
P = 0.9545
2.Normal distribution
Answer: x = 254.2 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) = 0.8 → z = 0.8416
Back to x
x = 250 + 0.8416·5 = 254.21 g
Answer
x = 254.21 g
3.Confidence intervals
Answer: E = 2.958 h
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.
Point estimate
x̄₁ − x̄₂ = 125.5 − 137.5 = −12
Standard error
SE = √(5²/60 + 13²/60) = 1.798
Margin of error
E = 1.645·1.798 = 2.958
Answer
E = 2.958 h
The interval does not contain 0, so the means differ significantly.
4.Confidence intervals
Answer: n = 425
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 = (2.576·12/1.5)² = 424.7
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 425
Answer
n = 425
5.Hypothesis tests & confidence intervals
Answer: p = 0.1788
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 = 6/√36 = 1
z = (250.92 − 250)/1 = 0.92
Tail area
Φ(0.92) = 0.8212
1 − Φ(0.92) = 0.1788
p-value
p = 0.1788
Answer
p = 0.1788
p > 0.05, so H₀ is not rejected at the 5% level.
6.Hypothesis tests & confidence intervals
Answer: n = 47
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 = ((1.645 + 1.282)·7/3)² = 46.64
Round up
n = 47
Answer
n = 47
7.Linear regression & correlation
Answer: r² = 0.9796
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 = 290/6 = 48.33
Sums of squares and products
Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 280
Syy = Σ(y − ȳ)² = 1143
Coefficient of determination
r² = 280²/(70·1143) = 0.9796
Answer
r² = 0.9796
About 97.96% of the variation in y is explained by the line.
8.Linear regression & correlation
Answer: ŷ = 74.1
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 = 35/5 = 7
ȳ = 401/5 = 80.2
Sums of squares and products
Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 244
Regression line
b = 244/40 = 6.1
a = 80.2 − 6.1·7 = 37.5
ŷ = 37.5 + 6.1·x
Prediction
ŷ = 37.5 + 6.1·6 = 74.1
Answer
ŷ = 74.1
9.Chi-square tests
Answer: χ² = 4.371
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: χ² = 28.84
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.