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
Introductory Statistics: cram pack
Formula sheets, the key method steps for every topic and 12 practice problems, with full solutions at the back.
1Formulas and methods
Probability
- Complement
- P(not A) = 1 − P(A)
- Addition rule
- P(A ∪ B) = P(A) + P(B) − P(A ∩ B)
- Independent events
- P(A ∩ B) = P(A)·P(B)
- Conditional probability
- P(A|B) = P(A ∩ B)/P(B)
- Binomial
- P(X = k) = C(n,k)·pk(1 − p)n−k
Normal distribution
- z-score
- z = (x − μ)/σ
- Below x
- P(X < x) = Φ(z)
- Above x
- P(X > x) = 1 − Φ(z)
- Between
- Φ(z2) − Φ(z1)
- Percentile
- x = μ + z·σ
Method
- Convert the value to a z-score: z = (x − μ)/σ.
- Look up Φ(z), the area to the left of z.
- For the area to the right use 1 − Φ(z); for an interval subtract the two left areas.
- For a percentile, work backwards: find z for the area, then x = μ + z·σ.
Confidence intervals
- σ known
- x̄ ± z·σ/√n
- σ unknown
- x̄ ± t·s/√n, df = n − 1
- Critical z
- 90%: 1.645, 95%: 1.96, 99%: 2.576
- Sample size
- n = (z·σ/E)², rounded up
Method
- Decide between z (population σ known) and t (only the sample s, with n − 1 degrees of freedom).
- Find the critical value for the confidence level.
- Margin of error: E = critical value × standard deviation/√n.
- Interval: x̄ ± E.
Hypothesis tests & confidence intervals
- z-test for a mean
- z = (x̄ − μ0)/(σ/√n)
- t-test for a mean
- t = (x̄ − μ0)/(s/√n), df = n − 1
- Two-sided at 5%
- reject H0 if |z| > 1.96
- Confidence interval
- x̄ ± z·σ/√n
Method
- State H₀ and H₁ and decide whether the test is one- or two-sided.
- Calculate the test statistic: (x̄ − μ₀)/(σ/√n), or with s and t.
- Compare with the critical value, or find the p-value.
- Reject H₀ if the statistic is beyond the critical value (p below α), and state the conclusion in words.
Linear regression & correlation
- Sums of squares
- Sxx = Σ(x − x̄)² = Σx² − (Σx)²/n, Sxy = Σxy − ΣxΣy/n
- Slope
- b = Sxy/Sxx
- Intercept
- a = ȳ − b·x̄
- Correlation
- r = Sxy/√(SxxSyy)
- Residual
- e = y − ŷ
2Practice problems
1.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 14, 42, 17, 20, 33, 33.
n = 6
s =
2.Descriptive statisticsHard
Exam scores are normally distributed with mean 70 and standard deviation 5. Find the z-score z of a score of 72.
μ = 70σ = 5x = 72
z =
3.ProbabilityMedium
For the distribution shown, find the variance Var(X) of X.
x: 0, 2, 4, 6p(x): 0.25, 0.35, 0.15, 0.25
Var(X) =
4.ProbabilityHard
A game costs €3 to play. You win €6 with probability 0.2 and €30 with probability 0.02; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.
stake = €3€6: p = 0.2€30: p = 0.02
E(G) = €
5.Normal distributionMedium
X is normal with mean 120 mm and standard deviation 0.3 mm; Y is independent and normal with mean 80 mm and standard deviation 0.4 mm. Find the probability P that X + Y is more than 199.8 mm. Round z to two decimals.
μX = 120, σX = 0.3μY = 80, σY = 0.4X + Y > 199.8 mm
P =
6.Normal distributionHard
The exam scores are normally distributed with mean 65 points and standard deviation 10 points. Find the value x below which 95% of them lie (the 95th percentile).
μ = 65σ = 10P = 95%
x = points
7.Confidence intervalsMedium
A sample of 20 items has mean 112.5 and standard deviation s = 4. The population is roughly normal. Find the margin of error E of a 99% confidence interval for the mean.
n = 20x̄ = 112.5s = 499%t(19) = 2.861
E =
8.Confidence intervalsHard
The population standard deviation is 25. 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 5?
σ = 25E = 590%
n =
9.Hypothesis tests & confidence intervalsMedium
Two teaching methods are compared. Group 1: 13 students, mean score 58, standard deviation 2. Group 2: 8 students, mean score 65.9, standard deviation 9. 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).
n₁ = 13x̄₁ = 58s₁ = 2n₂ = 8x̄₂ = 65.9s₂ = 9
t =
10.Hypothesis tests & confidence intervalsHard
In a trial, 60 of 300 patients on treatment A recover, and 6 of 100 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.
x₁ = 60n₁ = 300x₂ = 6n₂ = 100
z =
11.Linear regression & correlationMedium
The table gives 6 paired observations of x = load in kN and y = extension in mm. Their sums are given. Find Pearson's correlation coefficient r from these sums.
n = 6Σx = 42Σy = 157Σx² = 364Σy² = 4963Σxy = 1342
r =
12.Linear regression & correlationHard
The table gives 6 paired observations of x = load in kN and y = extension in mm. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 3.
n = 6x = 3
ŷ =
3Solutions
1.Descriptive statistics
Answer: s = 11.08
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.
Mean
x̄ = 159/6 = 26.5
Squared deviations
Σ(x − x̄)² = (−12.5)² + 15.5² + (−9.5)² + (−6.5)² + 6.5² + 6.5² = 613.5
Standard deviation
s = √(613.5/5) = 11.08
Answer
s = 11.08
2.Descriptive statistics
Answer: z = 0.4
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.
Standardise
z = (72 − 70)/5 = 0.4
Answer
z = 0.4
3.Probability
Answer: Var(X) = 4.96
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.
Mean
μ = 0·0.25 + 2·0.35 + 4·0.15 + 6·0.25 = 2.8
Mean of the squares
E(X²) = 0·0.25 + 4·0.35 + 16·0.15 + 36·0.25 = 12.8
Variance
Var(X) = 12.8 − 2.8² = 4.96
Answer
Var(X) = 4.96
4.Probability
Answer: E(G) = −1.2 €
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.
Net outcomes
G: 3 (p = 0.2), 27 (p = 0.02), −3 (p = 0.78)
Expected value
E(G) = 3·0.2 + 27·0.02 − 3·0.78 = −1.2 €
Answer
E(G) = −1.2 €
A negative value means you lose money on average.
5.Normal distribution
Answer: P = 0.6554
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).
Distribution of the sum
μ = 200 mm, σ = √(0.3² + 0.4²) = 0.5 mm
Standardise
z = (199.8 − 200)/0.5 = −0.4
Upper tail
P = 1 − Φ(−0.4) = 1 − 0.3446 = 0.6554
Answer
P = 0.6554
6.Normal distribution
Answer: x = 81.45 points
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.
z for the percentile
Φ(z) = 0.95 → z = 1.645
Back to x
x = 65 + 1.645·10 = 81.45 points
Answer
x = 81.45 points
7.Confidence intervals
Answer: E = 2.559
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.
Critical value
df = n − 1 = 19, t = 2.861
Margin of error
E = 2.861·4/√20 = 2.559
Answer
E = 2.559
8.Confidence intervals
Answer: n = 68
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.
Solve for n
n = (1.645·25/5)² = 67.65
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 68
Answer
n = 68
9.Hypothesis tests & confidence intervals
Answer: t = −3.09
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.
Pooled variance
sp² = (12·2² + 7·9²)/19 = 32.37
sp = 5.689
Standard error
SE = 5.689·√(1/13 + 1/8) = 2.557
Test statistic
t = (58 − 65.9)/2.557 = −3.09
Answer
t = −3.09
With df = 19, the two-sided 5% critical value is 2.093; |t| is larger, so H₀ is rejected.
10.Hypothesis tests & confidence intervals
Answer: z = 3.266
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.
Sample proportions
p̂₁ = 60/300 = 0.2
p̂₂ = 6/100 = 0.06
p̂ = 66/400 = 0.165
Standard error under H₀
SE = √(0.165·0.835·(1/300 + 1/100)) = 0.04286
Test statistic
z = (0.2 − 0.06)/0.04286 = 3.266
Answer
z = 3.266
|z| > 1.96, so H₀ is rejected at the 5% level (two-sided).
11.Linear regression & correlation
Answer: r = 0.9934
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.
Centred sums
Sxy = 1342 − 42·157/6 = 243
Sxx = 364 − 42²/6 = 70
Syy = 4963 − 157²/6 = 854.8
Correlation coefficient
r = 243/√(70·854.8) = 0.9934
Answer
r = 0.9934
12.Linear regression & correlation
Answer: ŷ = 8.414
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.
Means
x̄ = 42/6 = 7
ȳ = 129/6 = 21.5
Sums of squares and products
Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 229
Regression line
b = 229/70 = 3.271
a = 21.5 − 3.271·7 = −1.4
ŷ = −1.4 + 3.271·x
Prediction
ŷ = −1.4 + 3.271·3 = 8.414
Answer
ŷ = 8.414
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