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 0.9861. Find z to two decimals.
A = 0.9861
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 25% of them lie (the 25th percentile).
μ = 500σ = 40P = 25%
x = h
3.Confidence intervalsMedium
A 99% confidence interval for a mean, based on 25 observations and a known σ, has a margin of error of 2.3. What is the margin of error E₂ if the same study uses 100 observations instead?
99%n₁ = 25E₁ = 2.3n₂ = 100
E₂ =
4.Confidence intervalsHard
A pilot study suggests that about 0.7 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 ≈ 0.7E = 0.0590%
n =
5.Hypothesis tests & confidence intervalsMedium
A sample of 25 students has a mean score of 44.72 and a sample standard deviation of 12. Find the test statistic t for H₀: μ = 50.
μ₀ = 50x̄ = 44.72s = 12n = 25
t =
6.Hypothesis tests & confidence intervalsHard
A sample of 16 measurements has mean 76.5. The population standard deviation is 4. Find the upper limit U of the 99% confidence interval for the mean.
x̄ = 76.5σ = 4n = 1699%
U =
7.Linear regression & correlationMedium
The table gives 5 paired observations of x = car age in years and y = price in k€. Their sums are given. Find Pearson's correlation coefficient r from these sums.
n = 5Σx = 15Σy = 113Σx² = 55Σy² = 2615Σxy = 317
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 = 4.5.
n = 6x = 4.5
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 = 802 × 3
χ² =
10.Chi-square testsHard
The table shows observed counts. Test at α = 0.01 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 = 1803 × 2α = 0.01
χ² =
3Solutions
1.Normal distribution
Answer: z = −2.2
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 − 0.9861 = 0.0139
Read the z-table backwards
Find 0.0139 inside the table and read off z.
z = −2.20
Answer
z = −2.20
2.Normal distribution
Answer: x = 473 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) = 0.25 → z = −0.6745
Back to x
x = 500 + −0.6745·40 = 473.02 h
Answer
x = 473.02 h
3.Confidence intervals
Answer: E₂ = 1.15
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 = (2.576/2.576)·√(25/100) = 0.5
New margin
E₂ = 2.3·0.5 = 1.15
Answer
E₂ = 1.15
4.Confidence intervals
Answer: n = 228
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 = 1.645²·0.7·0.3/0.05² = 227.3
Round up
n = 228
Answer
n = 228
5.Hypothesis tests & confidence intervals
Answer: t = −2.2
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).
Standard error
SE = 12/√25 = 12/5 = 2.4
Test statistic
t = (44.72 − 50)/2.4 = −2.2
Answer
t = −2.2
t0.025; df = 24 = 2.064
|t| > 2.064, so H₀ is rejected at the 5% level (two-sided).
6.Hypothesis tests & confidence intervals
Answer: U = 79.08
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 = 2.576·4/√16 = 2.576
Interval limit
U = 76.5 + 2.576 = 79.08
Answer
U = 79.08
7.Linear regression & correlation
Answer: r = −0.8893
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 = 317 − 15·113/5 = −22
Sxx = 55 − 15²/5 = 10
Syy = 2615 − 113²/5 = 61.2
Correlation coefficient
r = −22/√(10·61.2) = −0.8893
Answer
r = −0.8893
8.Linear regression & correlation
Answer: ŷ = 11.8
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 = 90/6 = 15
Sums of squares and products
Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −56
Regression line
b = −56/17.5 = −3.2
a = 15 − (−3.2)·3.5 = 26.2
ŷ = 26.2 − 3.2·x
Prediction
ŷ = 26.2 + (−3.2)·4.5 = 11.8
Answer
ŷ = 11.8
9.Chi-square tests
Answer: χ² = 1.817
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 = 3.767
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.