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 between −z and z under the standard normal curve is 0.9786. Find z to two decimals.
A = 0.9786
z =
2.Normal distributionHard
The bag weights are normally distributed with mean 250 g and standard deviation 5 g. Drag the cut-off so the shaded bottom 90% is to the left of it. Find the cut-off x.
μ = 250 gσ = 5 g90%
x = g
3.Confidence intervalsMedium
A sample of 15 items has mean 15 and standard deviation s = 3.5. The population is roughly normal. Find the margin of error E of a 99% confidence interval for the mean.
n = 15x̄ = 15s = 3.599%t(14) = 2.977
E =
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 90% confidence interval for the proportion has a margin of error of at most 0.05?
p ≈ 0.1E = 0.0590%
n =
5.Hypothesis tests & confidence intervalsMedium
The mean of a population is claimed to be 250; the population standard deviation is known to be 10. A random sample of 100 observations has a mean of 252.04. Test H₀: μ = 250 against H₁: μ > 250. Find the p-value p.
μ₀ = 250σ = 10n = 100x̄ = 252.04H₁: μ > μ₀
p =
6.Hypothesis tests & confidence intervalsHard
A sample of 81 measurements has mean 101. The population standard deviation is 15. Find the upper limit U of the 90% confidence interval for the mean.
x̄ = 101σ = 15n = 8190%
U =
7.Linear regression & correlationMedium
The table gives 6 paired observations of x = hours studied and y = exam score. Find Pearson's correlation coefficient r.
n = 6
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 = 1402 × 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 = 2003 × 2α = 0.05
χ² =
3Solutions
1.Normal distribution
Answer: z = 2.3
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 + 0.9786)/2 = 0.9893
Read the z-table backwards
Find 0.9893 inside the table and read off z.
z = 2.30
Answer
z = 2.30
2.Normal distribution
Answer: x = 256.4 g
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) = 0.9
z from the table
z ≈ 1.28
Back to the original units
x = 250 + 1.28·5 = 256.4 g
Answer
x = 256.4 g
3.Confidence intervals
Answer: E = 2.69
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 = 14, t = 2.977
Margin of error
E = 2.977·3.5/√15 = 2.69
Answer
E = 2.69
4.Confidence intervals
Answer: n = 98
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 = 1.645²·0.1·0.9/0.05² = 97.42
Round up
n = 98
Answer
n = 98
5.Hypothesis tests & confidence intervals
Answer: p = 0.02068
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 = 10/√100 = 1
z = (252.04 − 250)/1 = 2.04
Tail area
Φ(2.04) = 0.9793
1 − Φ(2.04) = 0.0207
p-value
p = 0.02068
Answer
p = 0.02068
p ≤ 0.05, so H₀ is rejected at the 5% level.
6.Hypothesis tests & confidence intervals
Answer: U = 103.7
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 = 1.645·15/√81 = 2.742
Interval limit
U = 101 + 2.742 = 103.7
Answer
U = 103.7
7.Linear regression & correlation
Answer: r = 0.8381
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 = 33/6 = 5.5
ȳ = 371/6 = 61.83
Sums of squares and products
Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = 53.5
Syy = Σ(y − ȳ)² = 232.8
Correlation coefficient
r = 53.5/√(17.5·232.8) = 0.8381
Answer
r = 0.8381
8.Linear regression & correlation
Answer: ŷ = 23.56
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 = 135/6 = 22.5
Sums of squares and products
Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −18.5
Regression line
b = −18.5/17.5 = −1.057
a = 22.5 − (−1.057)·3.5 = 26.2
ŷ = 26.2 − 1.057·x
Prediction
ŷ = 26.2 + (−1.057)·2.5 = 23.56
Answer
ŷ = 23.56
9.Chi-square tests
Answer: χ² = 21.2
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: χ² = 14.7
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.