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 heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. What fraction lies between 178 and 183.6 cm? Give P as a decimal.
μ = 178σ = 7x1 = 178x2 = 183.6
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 5% of them lie (the 5th percentile).
μ = 250σ = 5P = 5%
x = g
3.Confidence intervalsMedium
A sample of 21 items has mean 32 and standard deviation s = 1.5. The population is roughly normal. Find the lower limit of the 90% confidence interval for the mean.
n = 21x̄ = 32s = 1.590%t(20) = 1.725
lower limit =
4.Confidence intervalsHard
Nothing is known in advance about the proportion of defective products. How many products n must be inspected so that a 95% confidence interval for the proportion has a margin of error of at most 0.04?
p = 0.5E = 0.0495%
n =
5.Hypothesis tests & confidence intervalsMedium
A sample of 64 students has a mean score of 502.9 and a sample standard deviation of 8. Find the test statistic t for H₀: μ = 500.
μ₀ = 500x̄ = 502.9s = 8n = 64
t =
6.Hypothesis tests & confidence intervalsHard
A sample of 25 measurements has mean 101.2. The population standard deviation is 20. Find the lower limit L of the 95% confidence interval for the mean.
x̄ = 101.2σ = 20n = 2595%
L =
7.Linear regression & correlationMedium
The table gives 5 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 = 5
r² =
8.Linear regression & correlationHard
The table gives 6 paired observations of x = hours studied and y = exam score. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 5.5.
n = 6x = 5.5
ŷ =
9.Chi-square testsMedium
The contingency table shows observed counts. For a chi-square test of independence, find the expected count E for the cell “students, car”.
row total = 94column total = 81N = 196
E =
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: P = 0.2881
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 = (178 − 178)/7 = 0
z₂ = (183.6 − 178)/7 = 0.8
Area between
P = Φ(0.8) − Φ(0) = 0.7881 − 0.5 = 0.2881
Answer
P = 0.2881
2.Normal distribution
Answer: x = 241.8 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) = 0.05 → z = −1.645
Back to x
x = 250 + −1.645·5 = 241.78 g
Answer
x = 241.78 g
3.Confidence intervals
Answer: lower limit = 31.44
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 = 20, t = 1.725
Margin of error
E = 1.725·1.5/√21 = 0.5646
Interval
32 ± 0.5646 → [31.44, 32.56]
Answer
lower limit = 31.44
4.Confidence intervals
Answer: n = 601
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 = 1.96²·0.5·0.5/0.04² = 600.2
Round up
n = 601
Answer
n = 601
5.Hypothesis tests & confidence intervals
Answer: t = 2.9
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 = 8/√64 = 8/8 = 1
Test statistic
t = (502.9 − 500)/1 = 2.9
Answer
t = 2.9
t0.025; df = 63 = 1.998
|t| > 1.998, so H₀ is rejected at the 5% level (two-sided).
6.Hypothesis tests & confidence intervals
Answer: L = 93.36
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 = 1.96·20/√25 = 7.84
Interval limit
L = 101.2 − 7.84 = 93.36
Answer
L = 93.36
7.Linear regression & correlation
Answer: r² = 0.8977
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 = 20/5 = 4
ȳ = 272/5 = 54.4
Sums of squares and products
Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = 52
Syy = Σ(y − ȳ)² = 301.2
Coefficient of determination
r² = 52²/(10·301.2) = 0.8977
Answer
r² = 0.8977
About 89.77% of the variation in y is explained by the line.
8.Linear regression & correlation
Answer: ŷ = 61.96
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 = 332/6 = 55.33
Sums of squares and products
Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = 58
Regression line
b = 58/17.5 = 3.314
a = 55.33 − 3.314·3.5 = 43.73
ŷ = 43.73 + 3.314·x
Prediction
ŷ = 43.73 + 3.314·5.5 = 61.96
Answer
ŷ = 61.96
9.Chi-square tests
Answer: E = 38.85
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.
Expected count
E = 94·81/196 = 38.85
Answer
E = 38.85
10.Chi-square tests
Answer: χ² = 14.87
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.