Descriptive statistics
- Mean
- x̄ = Σx/n
- Sample standard deviation
- s = √(Σ(x − x̄)²/(n − 1))
- Median
- middle value of the sorted data
- z-score
- z = (x − μ)/σ
- Normal distribution
- 68% within ±1σ, 95% within ±2σ
Formula sheets, the key method steps for every topic and 16 practice problems, with full solutions at the back.
A researcher wants to test whether there is an association between gender (male/female) and preferred film genre (four categories). Which test is appropriate?
The "replication crisis" in psychology refers to:
Find the sample standard deviation s of the data: 52, 54, 11, 53, 29, 33.
A quantity is normally distributed with mean 50 and standard deviation 12. What proportion P of values is below 57.2?
For the distribution shown, find the standard deviation σ of X.
X has the distribution shown, and Y = 10X + 4. Find the standard deviation σY of Y.
Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.0202. Find z to two decimals.
The mass X of a product from line 1 and Y from line 2 are independent and normal: X has mean 500 g and standard deviation 8 g, Y has mean 490 g and standard deviation 6 g. Find the probability P that X is greater than Y. Round z to two decimals.
A sample of 21 items has mean 98.5 and standard deviation s = 14. The population is roughly normal. Find the lower limit of the 99% confidence interval for the mean.
Two fertilisers are tested. Plot group 1: 12 plots, mean yield 41 kg, standard deviation 1.5 kg. Plot group 2: 19 plots, mean 40.5 kg, standard deviation 6 kg. Assume equal variances (pooled t, df = n₁ + n₂ − 2). Find the margin of error E of the 95% confidence interval for μ₁ − μ₂.
A sample of 49 students has a mean score of 76.71 and a sample standard deviation of 5. Find the test statistic t for H₀: μ = 75.
How large must a sample be so that the 99% confidence interval for a mean has a margin of error of at most 2.5? The standard deviation is 8. Find the smallest sample size n.
The table gives 5 paired observations of x = car age in years and y = price in k€. Their sums are given. Find Pearson's correlation coefficient r from these sums.
The table gives 6 paired observations of x = temperature in °C and y = ice creams sold. Use the least-squares line ŷ = a + b·x. A new observation (22.5, 112) 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.
The table shows observed counts. Test at α = 0.05 whether the two variables are independent (no continuity correction). Compute the test statistic χ², then compare it with the critical value from the table.
Answer: D · Chi-square test of independence
The chi-square test compares observed with expected frequencies in a contingency table to test whether two categorical variables are associated.
Answer: B · Many published findings failing to reproduce in new studies
Large replication projects found many effects were smaller or absent, linked to publication bias, small samples and flexible analysis. Pre-registration is one response.
Answer: s = 17.37
Descriptive statistics: the mean is the sum divided by n; the median is the middle value of the sorted data (the average of the two middle values when n is even). The sample standard deviation is s = √(Σ(x − x̄)²/(n − 1)). A z-score z = (x − μ)/σ says how many standard deviations a value lies from the mean.
Answer: P = 0.7257
Descriptive statistics: the mean is the sum divided by n; the median is the middle value of the sorted data (the average of the two middle values when n is even). The sample standard deviation is s = √(Σ(x − x̄)²/(n − 1)). A z-score z = (x − μ)/σ says how many standard deviations a value lies from the mean.
Answer: σ = 2.408
Discrete random variables: a probability distribution lists every value x a variable can take with its probability p(x); the probabilities are between 0 and 1 and add up to 1. The expected value (mean) is E(X) = Σx·p(x), the long-run average. The variance is Var(X) = Σ(x − μ)²·p(x) = E(X²) − μ², and the standard deviation σ = √Var(X). For a linear change Y = aX + b: E(Y) = a·E(X) + b, Var(Y) = a²·Var(X) and σY = |a|·σX; adding a constant does not change the spread.
Answer: σY = 10.62
Discrete random variables: a probability distribution lists every value x a variable can take with its probability p(x); the probabilities are between 0 and 1 and add up to 1. The expected value (mean) is E(X) = Σx·p(x), the long-run average. The variance is Var(X) = Σ(x − μ)²·p(x) = E(X²) − μ², and the standard deviation σ = √Var(X). For a linear change Y = aX + b: E(Y) = a·E(X) + b, Var(Y) = a²·Var(X) and σY = |a|·σX; adding a constant does not change the spread.
The mean does move: E(Y) = 10·1.65 + 4 = 20.5.
Answer: z = 2.05
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 0.9798 inside the table and read off z.
Answer: P = 0.8413
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: lower limit = 89.81
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: E = 3.632 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 = 2.394
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).
|t| > 2.011, 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 = −0.9061
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 = −2.667
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 below the line.
Answer: χ² = 5.533
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 not rejected.
Answer: χ² = 7.282
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.
χ² is above the critical value, so H₀ is rejected: the variables appear to be related.
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