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 60 s and standard deviation 5 s; Y is independent and normal with mean 45 s and standard deviation 4 s. Find the probability P that X + Y is more than 114.6 s. Round z to two decimals.
μX = 60, σX = 5μY = 45, σY = 4X + Y > 114.6 s
P =
2.Normal distributionHard
The bag weights are normally distributed with mean 250 g and standard deviation 5 g. Drag the cut-off so the shaded bottom 30% is to the left of it. Find the cut-off x.
μ = 250 gσ = 5 g30%
x = g
3.Confidence intervalsMedium
A 95% confidence interval for a mean, based on 36 observations and a known σ, has a margin of error of 10.8. What is the margin of error E₂ if the same study uses 9 observations instead?
95%n₁ = 36E₁ = 10.8n₂ = 9
E₂ =
4.Confidence intervalsHard
The diameters of 11 machined shafts have a sample standard deviation of 7.6 µm. Assuming the diameters are normally distributed, find the upper limit σ of the 95% confidence interval for the population standard deviation. With df = 10, 2.5% of the χ² distribution lies below 3.247 and 2.5% lies above 20.483.
n = 11s = 7.6 µmχ² = 3.247, 20.483
σ = µm
5.Hypothesis tests & confidence intervalsMedium
A sample of 16 students has a mean score of 94.5 and a sample standard deviation of 10. Find the test statistic t for H₀: μ = 100.
μ₀ = 100x̄ = 94.5s = 10n = 16
t =
6.Hypothesis tests & confidence intervalsHard
A one-sided z test at α = 0.01 must detect a shift of the mean of 3 units with a power of 90%. The standard deviation is 12. Find the minimum sample size n.
The table gives 5 paired observations of x = advertising spend in k€ and y = sales in k€. Find Pearson's correlation coefficient r.
n = 5
r =
8.Linear regression & correlationHard
The table gives 5 paired observations of x = hours studied and y = exam score. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 9.
n = 5x = 9
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 = 2402 × 3
χ² =
10.Chi-square testsHard
A chi-square test of independence is carried out on a contingency table with 2 rows and 4 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 = 0.06681
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 = 105 s, σ = √(5² + 4²) = 6.403 s
Standardise
z = (114.6 − 105)/6.403 = 1.5
Upper tail
P = 1 − Φ(1.5) = 1 − 0.9332 = 0.06681
Answer
P = 0.06681
2.Normal distribution
Answer: x = 247.4 g
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) = 0.3
z from the table
z ≈ −0.52
Back to the original units
x = 250 − 0.52·5 = 247.4 g
Answer
x = 247.4 g
3.Confidence intervals
Answer: E₂ = 21.6
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.
Ratio of margins
E₂/E₁ = (1.96/1.96)·√(36/9) = 2
New margin
E₂ = 10.8·2 = 21.6
Answer
E₂ = 21.6
4.Confidence intervals
Answer: σ = 13.34 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 = 10·7.6² = 577.6
Limits for σ²
577.6/20.483 = 28.2
577.6/3.247 = 177.9
Limit for σ
σ = √177.9 = 13.34 µm
Answer
σ = 13.34 µm
95% interval for σ: [5.31, 13.34] µm.
5.Hypothesis tests & confidence intervals
Answer: t = −2.2
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).
Standard error
SE = 10/√16 = 10/4 = 2.5
Test statistic
t = (94.5 − 100)/2.5 = −2.2
Answer
t = −2.2
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 = 209
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)·12/3)² = 208.3
Round up
n = 209
Answer
n = 209
7.Linear regression & correlation
Answer: r = 0.9502
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 = 15/5 = 3
ȳ = 237/5 = 47.4
Sums of squares and products
Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = 40
Syy = Σ(y − ȳ)² = 177.2
Correlation coefficient
r = 40/√(10·177.2) = 0.9502
Answer
r = 0.9502
8.Linear regression & correlation
Answer: ŷ = 80.5
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 = 344/5 = 68.8
Sums of squares and products
Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 156
Regression line
b = 156/40 = 3.9
a = 68.8 − 3.9·6 = 45.4
ŷ = 45.4 + 3.9·x
Prediction
ŷ = 45.4 + 3.9·9 = 80.5
Answer
ŷ = 80.5
9.Chi-square tests
Answer: χ² = 7.3
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 = 2. At α = 0.05 the critical value is 5.991, so H₀ (independence) is rejected.
10.Chi-square tests
Answer: χ²crit = 7.815
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)·(4 − 1) = 3
Table
χ²crit(df = 3, α = 0.05) = 7.815
Answer
χ²crit = 7.815
Reject H₀ if the computed χ² is larger than this value.