Formula sheets, the key method steps for every topic and 10 practice problems, with full solutions at the back.
The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. What fraction lies between 440 and 580 h? Give P as a decimal.
The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. Find the value x below which 10% of them lie (the 10th percentile).
A sample of 30 items has mean 90.5 and standard deviation s = 1.5. The population is roughly normal. Find the upper limit of the 95% confidence interval for the mean.
The population standard deviation is 21. How large a sample n do you need so that a 90% confidence interval for the mean has a margin of error of at most 0.5?
Two teaching methods are compared. Group 1: 18 students, mean score 27.5, standard deviation 12. Group 2: 8 students, mean score 20.6, standard deviation 7. 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 16 measurements has mean 505.3. The population standard deviation is 10. Find the lower limit L of the 95% confidence interval for the mean.
The table gives 6 paired observations of x = temperature in °C and y = ice creams sold. Their sums are given. Find Pearson's correlation coefficient r from these sums.
The table gives 5 paired observations of x = advertising spend in k€ and y = sales in k€. Use the least-squares line ŷ = a + b·x. A new observation (2.5, 51) is made. Find its residual e = y − ŷ.
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 checks whether counts in 4 categories follow a Poisson distribution whose mean is estimated from the same data. Using the table, find the critical value χ²crit at significance level α = 0.05.
Answer: P = 0.9104
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 = 448.7 h
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 = 91.06
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: n = 4774
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.
A sample size is a whole number, and rounding down would make the margin too large.
Answer: t = 1.506
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: L = 500.4
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 = 0.993
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 = 5.45
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.
The point lies above the line.
Answer: χ² = 13.2
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.
Answer: χ²crit = 5.991
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.
Reject H₀ if the computed χ² is larger than this value.
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