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 bag weights are normally distributed with mean 250 g and standard deviation 5 g. What fraction lies between 244.25 and 259 g? Give P as a decimal.
μ = 250σ = 5x1 = 244.25x2 = 259
P =
2.Normal distributionHard
The mass X of a product from line 1 and Y from line 2 are independent and normal: X has mean 500 g and standard deviation 8 g, Y has mean 490 g and standard deviation 6 g. Find the probability P that X is greater than Y. Round z to two decimals.
μX = 500 g, σX = 8 gμY = 490 g, σY = 6 g
P =
3.Confidence intervalsMedium
A sample of 12 items has mean 91 and standard deviation s = 11. The population is roughly normal. Find the lower limit of the 90% confidence interval for the mean.
n = 12x̄ = 91s = 1190%t(11) = 1.796
lower limit =
4.Confidence intervalsHard
The diameters of 15 machined shafts have a sample standard deviation of 1.2 µm. Assuming the diameters are normally distributed, find the lower limit σ of the 95% confidence interval for the population standard deviation. With df = 14, 2.5% of the χ² distribution lies below 5.629 and 2.5% lies above 26.119.
n = 15s = 1.2 µmχ² = 5.629, 26.119
σ = µm
5.Hypothesis tests & confidence intervalsMedium
Two production lines make the same part. Line 1: a sample of 40 parts has mean length 62 mm, and the line's known standard deviation is 18 mm. Line 2: a sample of 60 parts has mean 69.1 mm, known standard deviation 14 mm. Find the test statistic z for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (line 1 minus line 2).
A one-sided z test at α = 0.01 must detect a shift of the mean of 2 units with a power of 90%. The standard deviation is 15. Find the minimum sample size n.
The table gives 5 paired observations of x = load in kN and y = extension in mm. Their sums are given. Find Pearson's correlation coefficient r from these sums.
n = 5Σx = 30Σy = 81Σx² = 220Σy² = 1531Σxy = 578
r =
8.Linear regression & correlationHard
The table gives 5 paired observations of x = temperature in °C and y = ice creams sold. Use the least-squares line ŷ = a + b·x. A new observation (21.5, 112) is made. Find its residual e = y − ŷ.
n = 5(21.5, 112)
e =
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, bike”.
row total = 55column total = 63N = 141
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). How many degrees of freedom df does the test have?
k = 9
df =
3Solutions
1.Normal distribution
Answer: P = 0.839
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 = (244.25 − 250)/5 = −1.15
z₂ = (259 − 250)/5 = 1.8
Area between
P = Φ(1.8) − Φ(−1.15) = 0.9641 − 0.1251 = 0.839
Answer
P = 0.839
2.Normal distribution
Answer: P = 0.8413
Sums and differences of normal variables: if X and Y are independent and normal, X + Y and X − Y are normal too. Means add or subtract: E(X ± Y) = μX ± μY. Variances always add, also for a difference: Var(X ± Y) = σX² + σY², so σ = √(σX² + σY²). Standard deviations never add directly. The total of n independent items with the same distribution has mean n·μ and standard deviation σ·√n. Then standardise as usual: z = (x − mean)/σ and read Φ(z).
Distribution of the difference
μ = 500 − 490 = 10 g, σ = 10 g
Standardise
z = (0 − 10)/10 = −1
Probability
P = 1 − Φ(−1) = 0.8413
Answer
P = 0.8413
3.Confidence intervals
Answer: lower limit = 85.3
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 = 11, t = 1.796
Margin of error
E = 1.796·11/√12 = 5.703
Interval
91 ± 5.703 → [85.3, 96.7]
Answer
lower limit = 85.3
4.Confidence intervals
Answer: σ = 0.8786 µm
Interval for a variance: (n − 1)s²/σ² has a χ² distribution with n − 1 degrees of freedom. The 95% interval for σ² runs from (n − 1)s² divided by the upper χ² value (2.5% above it) to (n − 1)s² divided by the lower χ² value (2.5% below it); the interval for σ is the square root of both limits. It is not symmetric around s.
Sum of squares
(n − 1)s² = 14·1.2² = 20.16
Limits for σ²
20.16/26.119 = 0.7719
20.16/5.629 = 3.581
Limit for σ
σ = √0.7719 = 0.8786 µm
Answer
σ = 0.8786 µm
95% interval for σ: [0.8786, 1.892] µm.
5.Hypothesis tests & confidence intervals
Answer: z = −2.106
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 = √(18²/40 + 14²/60) = √11.37 = 3.371
Test statistic
z = (62 − 69.1)/3.371 = −2.106
Answer
z = −2.106
|z| > 1.96, so H₀ is rejected at the 5% level (two-sided).
6.Hypothesis tests & confidence intervals
Answer: n = 733
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 = ((2.326 + 1.282)·15/2)² = 732.2
Round up
n = 733
Answer
n = 733
7.Linear regression & correlation
Answer: r = 0.9834
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.
Centred sums
Sxy = 578 − 30·81/5 = 92
Sxx = 220 − 30²/5 = 40
Syy = 1531 − 81²/5 = 218.8
Correlation coefficient
r = 92/√(40·218.8) = 0.9834
Answer
r = 0.9834
8.Linear regression & correlation
Answer: e = −4.45
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 = 115/5 = 23
ȳ = 624/5 = 124.8
Sums of squares and products
Sxx = Σ(x − x̄)² = 90
Sxy = Σ(x − x̄)(y − ȳ) = 501
Regression line
b = 501/90 = 5.567
a = 124.8 − 5.567·23 = −3.233
ŷ = −3.233 + 5.567·x
Predicted value
ŷ = −3.233 + 5.567·21.5 = 116.4
Residual
e = 112 − 116.4 = −4.45
Answer
e = −4.45
The point lies below the line.
9.Chi-square tests
Answer: E = 24.57
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 = 55·63/141 = 24.57
Answer
E = 24.57
10.Chi-square tests
Answer: df = 8
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.