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 exam scores are normally distributed with mean 65 points and standard deviation 10 points. What fraction lies between 50.5 and 75 points? Give P as a decimal.
μ = 65σ = 10x1 = 50.5x2 = 75
P =
2.Normal distributionHard
The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. Find the value x below which 10% of them lie (the 10th percentile).
μ = 178σ = 7P = 10%
x = cm
3.Confidence intervalsMedium
A sample of 6 items has mean 40.5 and standard deviation s = 9. The population is roughly normal. Find the upper limit of the 99% confidence interval for the mean.
n = 6x̄ = 40.5s = 999%t(5) = 4.032
upper limit =
4.Confidence intervalsHard
The population standard deviation is 27. 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 = 27E = 2.595%
n =
5.Hypothesis tests & confidence intervalsMedium
Two production lines make the same part. Line 1: a sample of 30 parts has mean length 103 mm, and the line's known standard deviation is 7 mm. Line 2: a sample of 40 parts has mean 106.9 mm, known standard deviation 19 mm. Find the test statistic z for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (line 1 minus line 2).
How large must a sample be so that the 99% confidence interval for a mean has a margin of error of at most 1.5? The standard deviation is 5. Find the smallest sample size n.
σ = 5E = 1.599%
n =
7.Linear regression & correlationMedium
The table gives 6 paired observations of x = car age in years and y = price in k€. Find the coefficient of determination r² of the least-squares line.
n = 6
r² =
8.Linear regression & correlationHard
The table gives 5 paired observations of x = advertising spend in k€ and y = sales in k€. Use the least-squares line ŷ = a + b·x. A new observation (3, 35) is made. Find its residual e = y − ŷ.
n = 5(3, 35)
e =
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 test of independence is carried out on a contingency table with 2 rows and 11 columns (not counting the totals). Using the table, find the critical value χ²crit at significance level α = 0.05.
α = 0.05
χ²crit =
3Solutions
1.Normal distribution
Answer: P = 0.7678
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 = (50.5 − 65)/10 = −1.45
z₂ = (75 − 65)/10 = 1
Area between
P = Φ(1) − Φ(−1.45) = 0.8413 − 0.07353 = 0.7678
Answer
P = 0.7678
2.Normal distribution
Answer: x = 169 cm
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) = 0.1 → z = −1.282
Back to x
x = 178 + −1.282·7 = 169.03 cm
Answer
x = 169.03 cm
3.Confidence intervals
Answer: upper limit = 55.31
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 = 5, t = 4.032
Margin of error
E = 4.032·9/√6 = 14.81
Interval
40.5 ± 14.81 → [25.69, 55.31]
Answer
upper limit = 55.31
4.Confidence intervals
Answer: n = 449
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 = (1.96·27/2.5)² = 448.1
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 449
Answer
n = 449
5.Hypothesis tests & confidence intervals
Answer: z = −1.195
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.
Standard error of the difference
SE = √(7²/30 + 19²/40) = √10.66 = 3.265
Test statistic
z = (103 − 106.9)/3.265 = −1.195
Answer
z = −1.195
|z| < 1.96, so H₀ is not rejected at the 5% level (two-sided).
6.Hypothesis tests & confidence intervals
Answer: n = 74
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).
Solve for n
n = (2.576·5/1.5)² = 73.73
Round up
n = 74
Answer
n = 74
7.Linear regression & correlation
Answer: r² = 0.9479
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 = 98/6 = 16.33
Sums of squares and products
Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −53
Syy = Σ(y − ȳ)² = 169.3
Coefficient of determination
r² = −53²/(17.5·169.3) = 0.9479
Answer
r² = 0.9479
About 94.79% of the variation in y is explained by the line.
8.Linear regression & correlation
Answer: e = 2.45
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 = 198/5 = 39.6
Sums of squares and products
Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 94
Regression line
b = 94/40 = 2.35
a = 39.6 − 2.35·6 = 25.5
ŷ = 25.5 + 2.35·x
Predicted value
ŷ = 25.5 + 2.35·3 = 32.55
Residual
e = 35 − 32.55 = 2.45
Answer
e = 2.45
The point lies above the line.
9.Chi-square tests
Answer: χ² = 1.886
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: χ²crit = 18.31
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.
Degrees of freedom
df = (2 − 1)·(11 − 1) = 10
Table
χ²crit(df = 10, α = 0.05) = 18.307
Answer
χ²crit = 18.307
Reject H₀ if the computed χ² is larger than this value.