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 0.2434. Find z to two decimals.
A = 0.2434
z =
2.Normal distributionHard
The heights X of men and Y of women in a group are independent and normal: X has mean 180 cm and standard deviation 10 cm, Y has mean 170 cm and standard deviation 12 cm. Find the probability P that X is greater than Y. Round z to two decimals.
μX = 180 cm, σX = 10 cmμY = 170 cm, σY = 12 cm
P =
3.Confidence intervalsMedium
The battery life of two phone models is compared. Model 1: 100 phones, mean 67 h, known standard deviation 11 h. Model 2: 50 phones, mean 77.5 h, known standard deviation 10 h. Find the margin of error E of the 99% confidence interval for μ₁ − μ₂.
A pilot study suggests that about 0.7 of products are defective. How many products n must be inspected so that a 95% confidence interval for the proportion has a margin of error of at most 0.04?
p ≈ 0.7E = 0.0495%
n =
5.Hypothesis tests & confidence intervalsMedium
Two production lines make the same part. Line 1: a sample of 30 parts has mean length 122 mm, and the line's known standard deviation is 13 mm. Line 2: a sample of 100 parts has mean 124.8 mm, known standard deviation 9 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 122.1. The population standard deviation is 15. Find the lower limit L of the 99% confidence interval for the mean.
x̄ = 122.1σ = 15n = 10099%
L =
7.Linear regression & correlationMedium
The table gives 5 paired observations of x = hours studied and y = exam score. Their sums are given. Find Pearson's correlation coefficient r from these sums.
The table gives 6 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 (10, 93) is made. Find its residual e = y − ŷ.
n = 6(10, 93)
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 “staff, bus”.
row total = 100column total = 59N = 178
E =
10.Chi-square testsHard
The table shows observed counts. Test at α = 0.05 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 = 2803 × 2α = 0.05
χ² =
3Solutions
1.Normal distribution
Answer: z = 0.31
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 + 0.2434)/2 = 0.6217
Read the z-table backwards
Find 0.6217 inside the table and read off z.
z = 0.31
Answer
z = 0.31
2.Normal distribution
Answer: P = 0.7389
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 = 180 − 170 = 10 cm, σ = 15.62 cm
Standardise
z = (0 − 10)/15.62 = −0.64
Probability
P = 1 − Φ(−0.64) = 0.7389
Answer
P = 0.7389
3.Confidence intervals
Answer: E = 4.615 h
Interval for a difference: estimate ± critical value × standard error. For two independent means with known σ: SE = √(σ₁²/n₁ + σ₂²/n₂) and z. With equal unknown variances: pooled s_p and t with n₁ + n₂ − 2 degrees of freedom. For paired data, work with the differences: d̄ ± t·s_d/√n.
Point estimate
x̄₁ − x̄₂ = 67 − 77.5 = −10.5
Standard error
SE = √(11²/100 + 10²/50) = 1.792
Margin of error
E = 2.576·1.792 = 4.615
Answer
E = 4.615 h
The interval does not contain 0, so the means differ significantly.
4.Confidence intervals
Answer: n = 505
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 = 1.96²·0.7·0.3/0.04² = 504.2
Round up
n = 505
Answer
n = 505
5.Hypothesis tests & confidence intervals
Answer: z = −1.103
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 = √(13²/30 + 9²/100) = √6.443 = 2.538
Test statistic
z = (122 − 124.8)/2.538 = −1.103
Answer
z = −1.103
|z| < 1.96, so H₀ is not rejected at the 5% level (two-sided).
6.Hypothesis tests & confidence intervals
Answer: L = 118.2
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 = 2.576·15/√100 = 3.864
Interval limit
L = 122.1 − 3.864 = 118.2
Answer
L = 118.2
7.Linear regression & correlation
Answer: r = 0.9838
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 = 2404 − 35·322/5 = 150
Sxx = 285 − 35²/5 = 40
Syy = 21318 − 322²/5 = 581.2
Correlation coefficient
r = 150/√(40·581.2) = 0.9838
Answer
r = 0.9838
8.Linear regression & correlation
Answer: e = 4.143
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 = 396/6 = 66
Sums of squares and products
Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 400
Regression line
b = 400/70 = 5.714
a = 66 − 5.714·6 = 31.71
ŷ = 31.71 + 5.714·x
Predicted value
ŷ = 31.71 + 5.714·10 = 88.86
Residual
e = 93 − 88.86 = 4.143
Answer
e = 4.143
The point lies above the line.
9.Chi-square tests
Answer: E = 33.15
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 = 100·59/178 = 33.15
Answer
E = 33.15
10.Chi-square tests
Answer: χ² = 8.762
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.