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
X is normal with mean 12 min and standard deviation 1.5 min; Y is independent and normal with mean 20 min and standard deviation 2 min. Find the probability P that X + Y is more than 35.1 min. Round z to two decimals.
μX = 12, σX = 1.5μY = 20, σY = 2X + Y > 35.1 min
P =
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 132.5 h, known standard deviation 10 h. Model 2: 100 phones, mean 135.5 h, known standard deviation 15 h. Find the upper limit of the 95% confidence interval for μ₁ − μ₂.
The population standard deviation is 23. How large a sample n do you need so that a 90% confidence interval for the mean has a margin of error of at most 1?
σ = 23E = 190%
n =
5.Hypothesis tests & confidence intervalsMedium
A sample of 16 students has a mean score of 134.5 and a sample standard deviation of 20. Find the test statistic t for H₀: μ = 120.
μ₀ = 120x̄ = 134.5s = 20n = 16
t =
6.Hypothesis tests & confidence intervalsHard
How large must a sample be so that the 90% confidence interval for a mean has a margin of error of at most 2.5? The standard deviation is 4. Find the smallest sample size n.
σ = 4E = 2.590%
n =
7.Linear regression & correlationMedium
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 = 129Σx² = 220Σy² = 3909Σxy = 926
r =
8.Linear regression & correlationHard
The table gives 5 paired observations of x = load in kN and y = extension in mm. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 7.
n = 5x = 7
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 = 2202 × 2
χ² =
10.Chi-square testsHard
A chi-square goodness-of-fit test checks whether counts in 6 categories follow a Poisson distribution whose mean is estimated from the same data. Using the table, find the critical value χ²crit at significance level α = 0.01.
α = 0.01
χ²crit =
3Solutions
1.Normal distribution
Answer: P = 0.1075
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 sum
μ = 32 min, σ = √(1.5² + 2²) = 2.5 min
Standardise
z = (35.1 − 32)/2.5 = 1.24
Upper tail
P = 1 − Φ(1.24) = 1 − 0.8925 = 0.1075
Answer
P = 0.1075
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: upper limit = 0.5334 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 = 132.5 − 135.5 = −3
Standard error
SE = √(10²/100 + 15²/100) = 1.803
Margin of error
E = 1.96·1.803 = 3.533
Interval
−3 ± 3.533 → [−6.533, 0.5334]
Answer
upper limit = 0.5334 h
The interval contains 0, so the data do not show a significant difference.
4.Confidence intervals
Answer: n = 1432
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 = (1.645·23/1)² = 1431
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 1432
Answer
n = 1432
5.Hypothesis tests & confidence intervals
Answer: t = 2.9
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 = 20/√16 = 20/4 = 5
Test statistic
t = (134.5 − 120)/5 = 2.9
Answer
t = 2.9
t0.025; df = 15 = 2.131
|t| > 2.131, so H₀ is rejected at the 5% level (two-sided).
6.Hypothesis tests & confidence intervals
Answer: n = 7
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 = (1.645·4/2.5)² = 6.927
Round up
n = 7
Answer
n = 7
7.Linear regression & correlation
Answer: r = 0.9972
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 = 926 − 30·129/5 = 152
Sxx = 220 − 30²/5 = 40
Syy = 3909 − 129²/5 = 580.8
Correlation coefficient
r = 152/√(40·580.8) = 0.9972
Answer
r = 0.9972
8.Linear regression & correlation
Answer: ŷ = 30.9
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 = 134/5 = 26.8
Sums of squares and products
Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 164
Regression line
b = 164/40 = 4.1
a = 26.8 − 4.1·6 = 2.2
ŷ = 2.2 + 4.1·x
Prediction
ŷ = 2.2 + 4.1·7 = 30.9
Answer
ŷ = 30.9
9.Chi-square tests
Answer: χ² = 21.41
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 rejected.
10.Chi-square tests
Answer: χ²crit = 13.28
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 = 6 − 1 − 1 = 4
Table
χ²crit(df = 4, α = 0.01) = 13.277
Answer
χ²crit = 13.277
Reject H₀ if the computed χ² is larger than this value.