Formula sheets, the key method steps for every topic and 10 practice problems, with full solutions at the back.
The bag weights are normally distributed with mean 250 g and standard deviation 5 g. What fraction lies between 248.5 and 257.5 g? Give P as a decimal.
The delivery times are normally distributed with mean 32 min and standard deviation 6 min. Find the value x below which 95% of them lie (the 95th percentile).
The blood pressure of 16 patients is measured before and after a treatment. The differences d = before − after (mmHg) have mean 2.5 and standard deviation 0.6. Find the lower limit of the 95% confidence interval for the mean difference.
Two fertilisers are tested. Plot group 1: 5 plots, mean yield 94 kg, standard deviation 9.5 kg. Plot group 2: 11 plots, mean 101 kg, standard deviation 7 kg. Assume equal variances (pooled t, df = n₁ + n₂ − 2). Find the lower limit of the 95% confidence interval for μ₁ − μ₂.
Two production lines make the same part. Line 1: a sample of 60 parts has mean length 125 mm, and the line's known standard deviation is 18 mm. Line 2: a sample of 40 parts has mean 118.4 mm, known standard deviation 3 mm. Find the test statistic z for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (line 1 minus line 2).
How large must a sample be so that the 90% confidence interval for a mean has a margin of error of at most 2? The standard deviation is 10. Find the smallest sample size n.
The table gives 5 paired observations of x = hours studied and y = exam score. Find Pearson's correlation coefficient r.
The table gives 6 paired observations of x = hours studied and y = exam score. Use the least-squares line ŷ = a + b·x. For which value x does the line predict ŷ = 94?
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.
A chi-square goodness-of-fit test compares observed counts in 9 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 = 0.5511
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: x = 41.87 min
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: lower limit = 2.18 mmHg
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: lower limit = −16.02 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: z = 2.783
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.
|z| > 1.96, so H₀ is rejected at the 5% level (two-sided).
Answer: n = 68
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).
Answer: r = 0.9943
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: x = 8.109
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: χ² = 4.286
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.
Answer: df = 8
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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