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 masses of bags of cement are independent and normal with mean 25 kg and standard deviation 0.5 kg. Find the probability P that the total mass of 16 of them is below 404 kg. Round z to two decimals.
μ = 25 kgσ = 0.5 kgn = 16< 404 kg
P =
2.Normal distributionHard
The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. Drag the cut-off so the shaded bottom 90% is to the left of it. Find the cut-off x.
μ = 500 hσ = 40 h90%
x = h
3.Confidence intervalsMedium
A sample of 12 items has mean 106 and standard deviation s = 18. The population is roughly normal. Find the margin of error E of a 90% confidence interval for the mean.
n = 12x̄ = 106s = 1890%t(11) = 1.796
E =
4.Confidence intervalsHard
The population standard deviation is 36. 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?
σ = 36E = 295%
n =
5.Hypothesis tests & confidence intervalsMedium
A sample of 64 students has a mean score of 501.4 and a sample standard deviation of 4. Find the test statistic t for H₀: μ = 500.
μ₀ = 500x̄ = 501.4s = 4n = 64
t =
6.Hypothesis tests & confidence intervalsHard
A sample of 64 measurements has mean 101.7. The population standard deviation is 8. Find the lower limit L of the 99% confidence interval for the mean.
x̄ = 101.7σ = 8n = 6499%
L =
7.Linear regression & correlationMedium
The table gives 6 paired observations of x = advertising spend in k€ and y = sales in k€. Find Pearson's correlation coefficient r.
n = 6
r =
8.Linear regression & correlationHard
The table gives 6 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 = 11.
n = 6x = 11
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 “urban, low”.
row total = 73column total = 26N = 128
E =
10.Chi-square testsHard
A chi-square test of independence is carried out on a contingency table with 2 rows and 6 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 = 0.9772
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 total
μT = 400 kg, σT = 0.5·√16 = 2 kg
Standardise
z = (404 − 400)/2 = 2
Probability
P = Φ(2) = 0.9772
Answer
P = 0.9772
2.Normal distribution
Answer: x = 551 h
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
Φ(z) = 0.9
z from the table
z ≈ 1.28
Back to the original units
x = 500 + 1.28·40 = 551 h
Answer
x = 551 h
3.Confidence intervals
Answer: E = 9.332
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 = 1.796·18/√12 = 9.332
Answer
E = 9.332
4.Confidence intervals
Answer: n = 1245
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 = (1.96·36/2)² = 1245
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 1245
Answer
n = 1245
5.Hypothesis tests & confidence intervals
Answer: t = 2.8
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 = 4/√64 = 4/8 = 0.5
Test statistic
t = (501.4 − 500)/0.5 = 2.8
Answer
t = 2.8
t0.025; df = 63 = 1.998
|t| > 1.998, so H₀ is rejected at the 5% level (two-sided).
6.Hypothesis tests & confidence intervals
Answer: L = 99.12
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 = 2.576·8/√64 = 2.576
Interval limit
L = 101.7 − 2.576 = 99.12
Answer
L = 99.12
7.Linear regression & correlation
Answer: r = 0.9986
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 = 48/6 = 8
ȳ = 369/6 = 61.5
Sums of squares and products
Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 313
Syy = Σ(y − ȳ)² = 1404
Correlation coefficient
r = 313/√(70·1404) = 0.9986
Answer
r = 0.9986
8.Linear regression & correlation
Answer: ŷ = 42.74
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 = 42/6 = 7
ȳ = 169/6 = 28.17
Sums of squares and products
Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 255
Regression line
b = 255/70 = 3.643
a = 28.17 − 3.643·7 = 2.667
ŷ = 2.667 + 3.643·x
Prediction
ŷ = 2.667 + 3.643·11 = 42.74
Answer
ŷ = 42.74
9.Chi-square tests
Answer: E = 14.83
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 = 73·26/128 = 14.83
Answer
E = 14.83
10.Chi-square tests
Answer: χ²crit = 11.07
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)·(6 − 1) = 5
Table
χ²crit(df = 5, α = 0.05) = 11.070
Answer
χ²crit = 11.070
Reject H₀ if the computed χ² is larger than this value.