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 under the standard normal curve, to the right of z, is 0.0301. Find z to two decimals.
A = 0.0301
z =
2.Normal distributionHard
Hole diameters D are normal with mean 25.04 mm and standard deviation 0.01 mm. Shaft diameters d are independent and normal with mean 25 mm and standard deviation 0.02 mm. A randomly chosen shaft fits when D − d > 0. Find the probability P that it fits. Round z to two decimals.
hole: μ = 25.04 mm, σ = 0.01 mmshaft: μ = 25 mm, σ = 0.02 mm
P =
3.Confidence intervalsMedium
The battery life of two phone models is compared. Model 1: 50 phones, mean 102.5 h, known standard deviation 7 h. Model 2: 30 phones, mean 95.5 h, known standard deviation 12 h. Find the margin of error E of the 95% confidence interval for μ₁ − μ₂.
The population standard deviation is 15. 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 2.5?
σ = 15E = 2.595%
n =
5.Hypothesis tests & confidence intervalsMedium
A sample of 81 students has a mean score of 76.33 and a sample standard deviation of 10. Find the test statistic t for H₀: μ = 75.
μ₀ = 75x̄ = 76.33s = 10n = 81
t =
6.Hypothesis tests & confidence intervalsHard
A one-sided z test at α = 0.01 must detect a shift of the mean of 2 units with a power of 80%. The standard deviation is 11. Find the minimum sample size n.
The table gives 6 paired observations of x = temperature in °C and y = ice creams sold. Find the coefficient of determination r² of the least-squares line.
n = 6
r² =
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 = 5.
n = 5x = 5
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 = 2002 × 2
χ² =
10.Chi-square testsHard
A chi-square goodness-of-fit test compares observed counts in 10 categories with fully specified expected proportions (no parameters are estimated from the data). Using the table, find the critical value χ²crit at significance level α = 0.05.
α = 0.05
χ²crit =
3Solutions
1.Normal distribution
Answer: z = 1.88
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 − 0.0301 = 0.9699
Read the z-table backwards
Find 0.9699 inside the table and read off z.
z = 1.88
Answer
z = 1.88
2.Normal distribution
Answer: P = 0.9633
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 = 25.04 − 25 = 0.04 mm, σ = 0.02236 mm
Standardise
z = (0 − 0.04)/0.02236 = −1.79
Probability
P = 1 − Φ(−1.79) = 0.9633
Answer
P = 0.9633
3.Confidence intervals
Answer: E = 4.712 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 = 102.5 − 95.5 = 7
Standard error
SE = √(7²/50 + 12²/30) = 2.404
Margin of error
E = 1.96·2.404 = 4.712
Answer
E = 4.712 h
The interval does not contain 0, so the means differ significantly.
4.Confidence intervals
Answer: n = 139
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 = (1.96·15/2.5)² = 138.3
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 139
Answer
n = 139
5.Hypothesis tests & confidence intervals
Answer: t = 1.197
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/√81 = 10/9 = 1.111
Test statistic
t = (76.33 − 75)/1.111 = 1.197
Answer
t = 1.197
t0.025; df = 80 = 1.99
|t| < 1.99, so H₀ is not rejected at the 5% level (two-sided).
6.Hypothesis tests & confidence intervals
Answer: n = 304
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 + 0.842)·11/2)² = 303.6
Round up
n = 304
Answer
n = 304
7.Linear regression & correlation
Answer: r² = 0.9845
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 = 141/6 = 23.5
ȳ = 679/6 = 113.2
Sums of squares and products
Sxx = Σ(x − x̄)² = 157.5
Sxy = Σ(x − x̄)(y − ȳ) = 856.5
Syy = Σ(y − ȳ)² = 4731
Coefficient of determination
r² = 856.5²/(157.5·4731) = 0.9845
Answer
r² = 0.9845
About 98.45% of the variation in y is explained by the line.
8.Linear regression & correlation
Answer: ŷ = 11.7
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 = 68/5 = 13.6
Sums of squares and products
Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 76
Regression line
b = 76/40 = 1.9
a = 13.6 − 1.9·6 = 2.2
ŷ = 2.2 + 1.9·x
Prediction
ŷ = 2.2 + 1.9·5 = 11.7
Answer
ŷ = 11.7
9.Chi-square tests
Answer: χ² = 0.7576
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 not rejected.
10.Chi-square tests
Answer: χ²crit = 16.92
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 = 10 − 1 = 9
Table
χ²crit(df = 9, α = 0.05) = 16.919
Answer
χ²crit = 16.919
Reject H₀ if the computed χ² is larger than this value.