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 0.4647. Find z to two decimals.
A = 0.4647
z =
2.Normal distributionHard
The exam scores are normally distributed with mean 65 points and standard deviation 10 points. Drag the cut-off so the shaded top 5% is to the right of it. Find the cut-off x.
μ = 65 pointsσ = 10 points5%
x = points
3.Confidence intervalsMedium
A 99% confidence interval for a mean, based on 36 observations and a known σ, has a margin of error of 4.4. What is the margin of error E₂ of a 90% interval from the same data?
99%n₁ = 36E₁ = 4.490%
E₂ =
4.Confidence intervalsHard
Nothing is known in advance about the proportion of defective products. How many products n must be inspected so that a 90% confidence interval for the proportion has a margin of error of at most 0.04?
p = 0.5E = 0.0490%
n =
5.Hypothesis tests & confidence intervalsMedium
A sample of 100 students has a mean score of 73.92 and a sample standard deviation of 12. Find the test statistic t for H₀: μ = 75.
μ₀ = 75x̄ = 73.92s = 12n = 100
t =
6.Hypothesis tests & confidence intervalsHard
A one-sided z test at α = 0.05 must detect a shift of the mean of 2 units with a power of 90%. The standard deviation is 16. Find the minimum sample size n.
The table gives 5 paired observations of x = hours studied and y = exam score. Find Pearson's correlation coefficient r.
n = 5
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 = 3.5.
n = 6x = 3.5
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 = 1802 × 2
χ² =
10.Chi-square testsHard
A chi-square goodness-of-fit test compares observed counts in 11 categories with fully specified expected proportions (no parameters are estimated from the data). How many degrees of freedom df does the test have?
k = 11
df =
3Solutions
1.Normal distribution
Answer: z = 0.62
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 + 0.4647)/2 = 0.7324
Read the z-table backwards
Find 0.7324 inside the table and read off z.
z = 0.62
Answer
z = 0.62
2.Normal distribution
Answer: x = 81.4 points
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) = 0.95
z from the table
z ≈ 1.64
Back to the original units
x = 65 + 1.64·10 = 81.4 points
Answer
x = 81.4 points
3.Confidence intervals
Answer: E₂ = 2.81
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 = (1.645/2.576)·√(36/36) = 0.6386
New margin
E₂ = 4.4·0.6386 = 2.81
Answer
E₂ = 2.81
4.Confidence intervals
Answer: n = 423
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 = 1.645²·0.5·0.5/0.04² = 422.8
Round up
n = 423
Answer
n = 423
5.Hypothesis tests & confidence intervals
Answer: t = −0.9
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 = 12/√100 = 12/10 = 1.2
Test statistic
t = (73.92 − 75)/1.2 = −0.9
Answer
t = −0.9
t0.025; df = 99 = 1.984
|t| < 1.984, so H₀ is not rejected at the 5% level (two-sided).
6.Hypothesis tests & confidence intervals
Answer: n = 549
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 = ((1.645 + 1.282)·16/2)² = 548.3
Round up
n = 549
Answer
n = 549
7.Linear regression & correlation
Answer: r = 0.9948
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 = 443/5 = 88.6
Sums of squares and products
Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 312
Syy = Σ(y − ȳ)² = 2459
Correlation coefficient
r = 312/√(40·2459) = 0.9948
Answer
r = 0.9948
8.Linear regression & correlation
Answer: ŷ = 19.67
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 = 118/6 = 19.67
Sums of squares and products
Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −47
Regression line
b = −47/17.5 = −2.686
a = 19.67 − (−2.686)·3.5 = 29.07
ŷ = 29.07 − 2.686·x
Prediction
ŷ = 29.07 + (−2.686)·3.5 = 19.67
Answer
ŷ = 19.67
9.Chi-square tests
Answer: χ² = 2.25
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 = 10
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.