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 IQ scores are normally distributed with mean 100 points and standard deviation 15 points. What fraction lies between 70.75 and 115.75 points? Give P as a decimal.
μ = 100σ = 15x1 = 70.75x2 = 115.75
P =
2.Normal distributionHard
The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. Drag the cut-off so the shaded top 20% is to the right of it. Find the cut-off x.
μ = 100 pointsσ = 15 points20%
x = points
3.Confidence intervalsMedium
A 90% confidence interval for a mean, based on 25 observations and a known σ, has a margin of error of 5.6. What is the margin of error E₂ of a 99% interval based on 100 observations?
90%n₁ = 25E₁ = 5.699%n₂ = 100
E₂ =
4.Confidence intervalsHard
The population standard deviation is 11. How large a sample n do you need so that a 95% confidence interval for the mean has a margin of error of at most 2.5?
σ = 11E = 2.595%
n =
5.Hypothesis tests & confidence intervalsMedium
A sample of 64 students has a mean score of 98.45 and a sample standard deviation of 4. Find the test statistic t for H₀: μ = 100.
μ₀ = 100x̄ = 98.45s = 4n = 64
t =
6.Hypothesis tests & confidence intervalsHard
A sample of 100 measurements has mean 48.2. The population standard deviation is 15. Find the lower limit L of the 90% confidence interval for the mean.
x̄ = 48.2σ = 15n = 10090%
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 5 paired observations of x = load in kN and y = extension in mm. Use the least-squares line ŷ = a + b·x. For which value x does the line predict ŷ = 36?
n = 5ŷ = 36
x =
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 = 89column total = 58N = 160
E =
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 = 2003 × 2α = 0.01
χ² =
3Solutions
1.Normal distribution
Answer: P = 0.8276
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.
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) = 0.8
z from the table
z ≈ 0.84
Back to the original units
x = 100 + 0.84·15 = 112.6 points
Answer
x = 112.6 points
3.Confidence intervals
Answer: E₂ = 4.385
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)·√(25/100) = 0.783
New margin
E₂ = 5.6·0.783 = 4.385
Answer
E₂ = 4.385
4.Confidence intervals
Answer: n = 75
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 = (1.96·11/2.5)² = 74.37
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 75
Answer
n = 75
5.Hypothesis tests & confidence intervals
Answer: t = −3.1
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 = 4/√64 = 4/8 = 0.5
Test statistic
t = (98.45 − 100)/0.5 = −3.1
Answer
t = −3.1
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 = 45.73
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/√100 = 2.468
Interval limit
L = 48.2 − 2.468 = 45.73
Answer
L = 45.73
7.Linear regression & correlation
Answer: r² = 0.9951
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 = 25/5 = 5
ȳ = 230/5 = 46
Sums of squares and products
Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 180
Syy = Σ(y − ȳ)² = 814
Coefficient of determination
r² = 180²/(40·814) = 0.9951
Answer
r² = 0.9951
About 99.51% of the variation in y is explained by the line.
8.Linear regression & correlation
Answer: x = 9.05
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 = 30/5 = 6
ȳ = 119/5 = 23.8
Sums of squares and products
Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 160
Regression line
b = 160/40 = 4
a = 23.8 − 4·6 = −0.2
ŷ = −0.2 + 4·x
Solve for x
x = (36 − (−0.2))/4 = 9.05
Answer
x = 9.05
9.Chi-square tests
Answer: E = 32.26
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 = 89·58/160 = 32.26
Answer
E = 32.26
10.Chi-square tests
Answer: χ² = 28.78
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.