Formula sheets, the key method steps for every topic and 10 practice problems, with full solutions at the back.
Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.9838. Find z to two decimals.
The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. Find the value x below which 90% of them lie (the 90th percentile).
A sample of 9 items has mean 47.5 and standard deviation s = 6.5. The population is roughly normal. Find the lower limit of the 99% confidence interval for the mean.
Two fertilisers are tested. Plot group 1: 5 plots, mean yield 96.5 kg, standard deviation 6.5 kg. Plot group 2: 7 plots, mean 94.5 kg, standard deviation 2 kg. Assume equal variances (pooled t, df = n₁ + n₂ − 2). Find the lower limit of the 95% confidence interval for μ₁ − μ₂.
Two teaching methods are compared. Group 1: 10 students, mean score 66.5, standard deviation 7. Group 2: 16 students, mean score 71.7, standard deviation 9.5. Assume equal population variances and use the pooled two-sample t test. Find the test statistic t for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (group 1 minus group 2).
A sample of 49 measurements has mean 500.1. The population standard deviation is 8. Find the upper limit U of the 99% confidence interval for the mean.
The table gives 6 paired observations of x = advertising spend in k€ and y = sales in k€. Find the coefficient of determination r² of the least-squares line.
The table gives 6 paired observations of x = advertising spend in k€ and y = sales in k€. Use the least-squares line ŷ = a + b·x. For which value x does the line predict ŷ = 75?
The contingency table shows observed counts. For a chi-square test of independence, find the expected count E for the cell “rural, high”.
A chi-square goodness-of-fit test compares observed counts in 7 categories with fully specified expected proportions (no parameters are estimated from the data). How many degrees of freedom df does the test have?
Answer: z = −2.14
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.
Find 0.0162 inside the table and read off z.
Answer: x = 187 cm
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 = 40.23
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.
Answer: lower limit = −3.731 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: t = −1.492
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.
With df = 24, the two-sided 5% critical value is 2.064; |t| is smaller, so H₀ is not rejected.
Answer: U = 503
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: r² = 0.7935
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.
About 79.35% of the variation in y is explained by the line.
Answer: x = 9.91
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 = 25.24
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 = 6
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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