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 battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. What fraction lies between 470 and 540 h? Give P as a decimal.
μ = 500σ = 40x1 = 470x2 = 540
P =
2.Normal distributionHard
The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. Find the value x below which 97.5% of them lie (the 97.5th percentile).
μ = 500σ = 40P = 97.5%
x = h
3.Confidence intervalsMedium
A sample of 16 items has mean 118.5 and standard deviation s = 10.5. The population is roughly normal. Find the lower limit of the 95% confidence interval for the mean.
n = 16x̄ = 118.5s = 10.595%t(15) = 2.131
lower limit =
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 0.5?
σ = 11E = 0.595%
n =
5.Hypothesis tests & confidence intervalsMedium
Two production lines make the same part. Line 1: a sample of 40 parts has mean length 123 mm, and the line's known standard deviation is 20 mm. Line 2: a sample of 40 parts has mean 113.7 mm, known standard deviation 3 mm. Find the test statistic z for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (line 1 minus line 2).
A sample of 100 measurements has mean 75.8. The population standard deviation is 4. Find the upper limit U of the 99% confidence interval for the mean.
x̄ = 75.8σ = 4n = 10099%
U =
7.Linear regression & correlationMedium
The table gives 6 paired observations of x = car age in years and y = price in k€. Find Pearson's correlation coefficient r.
n = 6
r =
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 = 8.
n = 6x = 8
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 “day shift, no defect”.
row total = 81column total = 35N = 148
E =
10.Chi-square testsHard
A chi-square goodness-of-fit test compares observed counts in 9 categories with fully specified expected proportions (no parameters are estimated from the data). Using the table, find the critical value χ²crit at significance level α = 0.05.
α = 0.05
χ²crit =
3Solutions
1.Normal distribution
Answer: P = 0.6147
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 = (470 − 500)/40 = −0.75
z₂ = (540 − 500)/40 = 1
Area between
P = Φ(1) − Φ(−0.75) = 0.8413 − 0.2266 = 0.6147
Answer
P = 0.6147
2.Normal distribution
Answer: x = 578.4 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) = 0.975 → z = 1.96
Back to x
x = 500 + 1.96·40 = 578.4 h
Answer
x = 578.4 h
3.Confidence intervals
Answer: lower limit = 112.9
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 = 15, t = 2.131
Margin of error
E = 2.131·10.5/√16 = 5.594
Interval
118.5 ± 5.594 → [112.9, 124.1]
Answer
lower limit = 112.9
4.Confidence intervals
Answer: n = 1860
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 = (1.96·11/0.5)² = 1859
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 1860
Answer
n = 1860
5.Hypothesis tests & confidence intervals
Answer: z = 2.908
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 = √(20²/40 + 3²/40) = √10.23 = 3.198
Test statistic
z = (123 − 113.7)/3.198 = 2.908
Answer
z = 2.908
|z| > 1.96, so H₀ is rejected at the 5% level (two-sided).
6.Hypothesis tests & confidence intervals
Answer: U = 76.83
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/√100 = 1.03
Interval limit
U = 75.8 + 1.03 = 76.83
Answer
U = 76.83
7.Linear regression & correlation
Answer: r = −0.9011
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 = 100/6 = 16.67
Sums of squares and products
Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −30
Syy = Σ(y − ȳ)² = 63.33
Correlation coefficient
r = −30/√(17.5·63.33) = −0.9011
Answer
r = −0.9011
8.Linear regression & correlation
Answer: ŷ = 72.51
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 = 36/6 = 6
ȳ = 403/6 = 67.17
Sums of squares and products
Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 187
Regression line
b = 187/70 = 2.671
a = 67.17 − 2.671·6 = 51.14
ŷ = 51.14 + 2.671·x
Prediction
ŷ = 51.14 + 2.671·8 = 72.51
Answer
ŷ = 72.51
9.Chi-square tests
Answer: E = 19.16
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 = 81·35/148 = 19.16
Answer
E = 19.16
10.Chi-square tests
Answer: χ²crit = 15.51
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 = 9 − 1 = 8
Table
χ²crit(df = 8, α = 0.05) = 15.507
Answer
χ²crit = 15.507
Reject H₀ if the computed χ² is larger than this value.