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
Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.8315. Find z to two decimals.
A = 0.8315
z =
2.Normal distributionHard
The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. Find the value x below which 10% of them lie (the 10th percentile).
μ = 500σ = 40P = 10%
x = h
3.Confidence intervalsMedium
A 99% confidence interval for a mean, based on 50 observations and a known σ, has a margin of error of 9.9. What is the margin of error E₂ of a 90% interval from the same data?
99%n₁ = 50E₁ = 9.990%
E₂ =
4.Confidence intervalsHard
The population standard deviation is 39. 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 2?
σ = 39E = 299%
n =
5.Hypothesis tests & confidence intervalsMedium
Two teaching methods are compared. Group 1: 14 students, mean score 28.5, standard deviation 2. Group 2: 17 students, mean score 27.8, standard deviation 2. Assume equal population variances and use the pooled two-sample t test. Find the test statistic t for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (group 1 minus group 2).
n₁ = 14x̄₁ = 28.5s₁ = 2n₂ = 17x̄₂ = 27.8s₂ = 2
t =
6.Hypothesis tests & confidence intervalsHard
How large must a sample be so that the 95% confidence interval for a mean has a margin of error of at most 2.5? The standard deviation is 10. Find the smallest sample size n.
σ = 10E = 2.595%
n =
7.Linear regression & correlationMedium
The table gives 5 paired observations of x = advertising spend in k€ and y = sales in k€. Their sums are given. Find Pearson's correlation coefficient r from these sums.
n = 5Σx = 15Σy = 205Σx² = 55Σy² = 8645Σxy = 662
r =
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 = 2.5.
n = 6x = 2.5
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 = 1602 × 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 = 1603 × 2α = 0.05
χ² =
3Solutions
1.Normal distribution
Answer: z = −0.96
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 of z
Φ(z) = 1 − 0.8315 = 0.1685
Read the z-table backwards
Find 0.1685 inside the table and read off z.
z = −0.96
Answer
z = −0.96
2.Normal distribution
Answer: x = 448.7 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) = 0.1 → z = −1.282
Back to x
x = 500 + −1.282·40 = 448.74 h
Answer
x = 448.74 h
3.Confidence intervals
Answer: E₂ = 6.322
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.
Ratio of margins
E₂/E₁ = (1.645/2.576)·√(50/50) = 0.6386
New margin
E₂ = 9.9·0.6386 = 6.322
Answer
E₂ = 6.322
4.Confidence intervals
Answer: n = 2524
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 = (2.576·39/2)² = 2523
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 2524
Answer
n = 2524
5.Hypothesis tests & confidence intervals
Answer: t = 0.9698
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.
Pooled variance
sp² = (13·2² + 16·2²)/29 = 4
sp = 2
Standard error
SE = 2·√(1/14 + 1/17) = 0.7218
Test statistic
t = (28.5 − 27.8)/0.7218 = 0.9698
Answer
t = 0.9698
With df = 29, the two-sided 5% critical value is 2.045; |t| is smaller, so H₀ is not rejected.
6.Hypothesis tests & confidence intervals
Answer: n = 62
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).
Solve for n
n = (1.96·10/2.5)² = 61.47
Round up
n = 62
Answer
n = 62
7.Linear regression & correlation
Answer: r = 0.9594
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 = 662 − 15·205/5 = 47
Sxx = 55 − 15²/5 = 10
Syy = 8645 − 205²/5 = 240
Correlation coefficient
r = 47/√(10·240) = 0.9594
Answer
r = 0.9594
8.Linear regression & correlation
Answer: ŷ = 25.47
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 = 148/6 = 24.67
Sums of squares and products
Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −14
Regression line
b = −14/17.5 = −0.8
a = 24.67 − (−0.8)·3.5 = 27.47
ŷ = 27.47 − 0.8·x
Prediction
ŷ = 27.47 + (−0.8)·2.5 = 25.47
Answer
ŷ = 25.47
9.Chi-square tests
Answer: χ² = 5.222
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 = 30.63
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.