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 0.7199. Find z to two decimals.
A = 0.7199
z =
2.Normal distributionHard
The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. Find the value x below which 5% of them lie (the 5th percentile).
μ = 100σ = 15P = 5%
x = points
3.Confidence intervalsMedium
A 90% confidence interval for a mean, based on 64 observations and a known σ, has a margin of error of 11.2. What is the margin of error E₂ of a 99% interval based on 16 observations?
90%n₁ = 64E₁ = 11.299%n₂ = 16
E₂ =
4.Confidence intervalsHard
A pilot study suggests that about 0.6 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 0.03?
p ≈ 0.6E = 0.0390%
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 25 observations has a mean of 246. Test H₀: μ = 250 against H₁: μ ≠ 250. Find the p-value p.
μ₀ = 250σ = 10n = 25x̄ = 246H₁: μ ≠ μ₀
p =
6.Hypothesis tests & confidence intervalsHard
A one-sided z test at α = 0.01 must detect a shift of the mean of 2 units with a power of 90%. The standard deviation is 14. Find the minimum sample size n.
The table gives 6 paired observations of x = load in kN and y = extension in mm. Their sums are given. Find Pearson's correlation coefficient r from these sums.
n = 6Σx = 42Σy = 187Σx² = 364Σy² = 6867Σxy = 1578
r =
8.Linear regression & correlationHard
The table gives 5 paired observations of x = temperature in °C and y = ice creams sold. Use the least-squares line ŷ = a + b·x. For which value x does the line predict ŷ = 113?
n = 5ŷ = 113
x =
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 × 2
χ² =
10.Chi-square testsHard
A chi-square goodness-of-fit test compares observed counts in 8 categories with fully specified expected proportions (no parameters are estimated from the data). How many degrees of freedom df does the test have?
k = 8
df =
3Solutions
1.Normal distribution
Answer: z = 1.08
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 + 0.7199)/2 = 0.8599
Read the z-table backwards
Find 0.8599 inside the table and read off z.
z = 1.08
Answer
z = 1.08
2.Normal distribution
Answer: x = 75.33 points
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) = 0.05 → z = −1.645
Back to x
x = 100 + −1.645·15 = 75.33 points
Answer
x = 75.33 points
3.Confidence intervals
Answer: E₂ = 35.08
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 = (2.576/1.645)·√(64/16) = 3.132
New margin
E₂ = 11.2·3.132 = 35.08
Answer
E₂ = 35.08
4.Confidence intervals
Answer: n = 722
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 = 1.645²·0.6·0.4/0.03² = 721.6
Round up
n = 722
Answer
n = 722
5.Hypothesis tests & confidence intervals
Answer: p = 0.0455
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/√25 = 2
z = (246 − 250)/2 = −2
Tail area
Φ(2) = 0.9772
1 − Φ(2) = 0.0228
p-value
p = 2·0.0228 = 0.0455
Answer
p = 0.0455
p ≤ 0.05, so H₀ is rejected at the 5% level.
6.Hypothesis tests & confidence intervals
Answer: n = 638
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)·14/2)² = 637.9
Round up
n = 638
Answer
n = 638
7.Linear regression & correlation
Answer: r = 0.9975
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 = 1578 − 42·187/6 = 269
Sxx = 364 − 42²/6 = 70
Syy = 6867 − 187²/6 = 1039
Correlation coefficient
r = 269/√(70·1039) = 0.9975
Answer
r = 0.9975
8.Linear regression & correlation
Answer: x = 25.02
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 = 110/5 = 22
ȳ = 479/5 = 95.8
Sums of squares and products
Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 228
Regression line
b = 228/40 = 5.7
a = 95.8 − 5.7·22 = −29.6
ŷ = −29.6 + 5.7·x
Solve for x
x = (113 − (−29.6))/5.7 = 25.02
Answer
x = 25.02
9.Chi-square tests
Answer: χ² = 0.4861
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 = 1. At α = 0.05 the critical value is 3.841, so H₀ (independence) is not rejected.
10.Chi-square tests
Answer: df = 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.