Formula sheets, the key method steps for every topic and 10 practice problems, with full solutions at the back.
The masses of bolts are independent and normal with mean 40 g and standard deviation 1.2 g. Find the probability P that the total mass of 4 of them is below 164.8 g. Round z to two decimals.
The bag weights are normally distributed with mean 250 g and standard deviation 5 g. Find the value x below which 99% of them lie (the 99th percentile).
The battery life of two phone models is compared. Model 1: 40 phones, mean 113.5 h, known standard deviation 12 h. Model 2: 40 phones, mean 115 h, known standard deviation 12 h. Find the upper limit of the 99% confidence interval for μ₁ − μ₂.
Two fertilisers are tested. Plot group 1: 14 plots, mean yield 11 kg, standard deviation 2.5 kg. Plot group 2: 7 plots, mean 6 kg, standard deviation 8 kg. Assume equal variances (pooled t, df = n₁ + n₂ − 2). Find the upper limit of the 90% confidence interval for μ₁ − μ₂.
The mean of a population is claimed to be 250; the population standard deviation is known to be 15. A random sample of 100 observations has a mean of 248.02. Test H₀: μ = 250 against H₁: μ ≠ 250. Find the p-value p.
A one-sided z test of H₀: μ = 500 against H₁: μ > 500 is done at α = 0.05 with a sample of 36; the standard deviation is known to be 6. In reality the mean is 502.9. Find the probability of a Type II error β in percent.
The table gives 6 paired observations of x = hours studied and y = exam score. Find Pearson's correlation coefficient r.
The table gives 6 paired observations of x = car age in years and y = price in k€. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 2.5.
The contingency table shows observed counts. For a chi-square test of independence, find the expected count E for the cell “rural, low”.
A chi-square goodness-of-fit test compares observed counts in 8 categories with fully specified expected proportions (no parameters are estimated from the data). How many degrees of freedom df does the test have?
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).
Answer: x = 261.6 g
Normal distribution: convert a value x to a z-score, z = (x − μ)/σ, which counts standard deviations from the mean. A z-table (or Φ(z)) gives the area to the left of z, which is P(X < x). The area to the right is 1 − Φ(z), and the area between two values is the difference of their two areas.
Answer: upper limit = 5.412 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.
The interval contains 0, so the data do not show a significant difference.
Answer: upper limit = 8.961 kg
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.
Answer: p = 0.1868
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.
p > 0.05, so H₀ is not rejected at the 5% level.
Answer: β = 10.47 %
Power of a test: power = 1 − β is the probability of rejecting H₀ when it is false. Find the cutoff x̄c from μ₀ and z_α, then compute the probability of landing beyond x̄c using the true mean: z = (x̄c − μ)/(σ/√n).
Answer: r = 0.9968
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.
Answer: ŷ = 23.98
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.
Answer: E = 38.77
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.
Answer: df = 7
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.
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