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: 23, 11, 11, 21, 20, 32.
n = 6
s =
2.Descriptive statisticsHard
A quantity is normally distributed with mean 50 and standard deviation 12. What proportion P of values is below 38?
μ = 50σ = 12x = 38
P =
3.ProbabilityMedium
For the distribution shown, find the variance Var(X) of X.
x: 0, 1, 2, 3p(x): 0.25, 0.35, 0.25, 0.15
Var(X) =
4.ProbabilityHard
X has the distribution shown, and Y = 2X + 20. Find the standard deviation σY of Y.
x: 0, 1, 2, 3, 4p(x): 0.15, 0.1, 0.25, 0.2, 0.3Y = 2X + 20
σY =
5.Normal distributionMedium
The exam scores are normally distributed with mean 65 points and standard deviation 10 points. What fraction lies between 57.5 and 67.5 points? Give P as a decimal.
μ = 65σ = 10x1 = 57.5x2 = 67.5
P =
6.Normal distributionHard
The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. Drag the cut-off so the shaded bottom 40% is to the left of it. Find the cut-off x.
μ = 500 hσ = 40 h40%
x = h
7.Confidence intervalsMedium
A sample of 7 items has mean 95.5 and standard deviation s = 16. The population is roughly normal. Find the upper limit of the 90% confidence interval for the mean.
n = 7x̄ = 95.5s = 1690%t(6) = 1.943
upper limit =
8.Confidence intervalsHard
The population standard deviation is 25. How large a sample n do you need so that a 99% confidence interval for the mean has a margin of error of at most 0.5?
σ = 25E = 0.599%
n =
9.Hypothesis tests & confidence intervalsMedium
Two teaching methods are compared. Group 1: 14 students, mean score 101, standard deviation 12. Group 2: 7 students, mean score 104.3, standard deviation 12. 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 = 14x̄₁ = 101s₁ = 12n₂ = 7x̄₂ = 104.3s₂ = 12
t =
10.Hypothesis tests & confidence intervalsHard
How large must a sample be so that the 90% confidence interval for a mean has a margin of error of at most 0.5? The standard deviation is 15. Find the smallest sample size n.
σ = 15E = 0.590%
n =
11.Linear regression & correlationMedium
The table gives 5 paired observations of x = car age in years and y = price in k€. Find the coefficient of determination r² of the least-squares line.
n = 5
r² =
12.Linear regression & correlationHard
The table gives 5 paired observations of x = car age in years and y = price in k€. Use the least-squares line ŷ = a + b·x. A new observation (1.5, 24) is made. Find its residual e = y − ŷ.
n = 5(1.5, 24)
e =
3Solutions
1.Descriptive statistics
Answer: s = 7.941
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 = 118/6 = 19.67
Squared deviations
Σ(x − x̄)² = 3.333² + (−8.667)² + (−8.667)² + 1.333² + 0.3333² + 12.33² = 315.3
Standard deviation
s = √(315.3/5) = 7.941
Answer
s = 7.941
2.Descriptive statistics
Answer: P = 0.1587
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 = −1
Standard normal table
Φ(−1) = 0.1587
P = 0.1587
Answer
P = 0.1587
3.Probability
Answer: Var(X) = 1.01
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 + 1·0.35 + 2·0.25 + 3·0.15 = 1.3
Mean of the squares
E(X²) = 0·0.25 + 1·0.35 + 4·0.25 + 9·0.15 = 2.7
Variance
Var(X) = 2.7 − 1.3² = 1.01
Answer
Var(X) = 1.01
4.Probability
Answer: σY = 2.786
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.
Spread of X
μ = 2.4, Var(X) = 1.94, σX = 1.393
Linear change
σY = |2|·1.393 = 2.786
Answer
σY = 2.786
The mean does move: E(Y) = 2·2.4 + 20 = 24.8.
5.Normal distribution
Answer: P = 0.3721
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-scores
z₁ = (57.5 − 65)/10 = −0.75
z₂ = (67.5 − 65)/10 = 0.25
Area between
P = Φ(0.25) − Φ(−0.75) = 0.5987 − 0.2266 = 0.3721
Answer
P = 0.3721
6.Normal distribution
Answer: x = 490 h
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.
Area to the left
Φ(z) = 0.4
z from the table
z ≈ −0.25
Back to the original units
x = 500 − 0.25·40 = 490 h
Answer
x = 490 h
7.Confidence intervals
Answer: upper limit = 107.3
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 = 6, t = 1.943
Margin of error
E = 1.943·16/√7 = 11.75
Interval
95.5 ± 11.75 → [83.75, 107.3]
Answer
upper limit = 107.3
8.Confidence intervals
Answer: n = 16590
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 = (2.576·25/0.5)² = 16589
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 16590
Answer
n = 16590
9.Hypothesis tests & confidence intervals
Answer: t = −0.5941
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 = (13·12² + 6·12²)/19 = 144
sp = 12
Standard error
SE = 12·√(1/14 + 1/7) = 5.555
Test statistic
t = (101 − 104.3)/5.555 = −0.5941
Answer
t = −0.5941
With df = 19, the two-sided 5% critical value is 2.093; |t| is smaller, so H₀ is not rejected.
10.Hypothesis tests & confidence intervals
Answer: n = 2436
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).
Solve for n
n = (1.645·15/0.5)² = 2435
Round up
n = 2436
Answer
n = 2436
11.Linear regression & correlation
Answer: r² = 0.9356
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 = 15/5 = 3
ȳ = 81/5 = 16.2
Sums of squares and products
Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = −25
Syy = Σ(y − ȳ)² = 66.8
Coefficient of determination
r² = −25²/(10·66.8) = 0.9356
Answer
r² = 0.9356
About 93.56% of the variation in y is explained by the line.
12.Linear regression & correlation
Answer: e = −3.35
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 = 15/5 = 3
ȳ = 121/5 = 24.2
Sums of squares and products
Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = −21
Regression line
b = −21/10 = −2.1
a = 24.2 − (−2.1)·3 = 30.5
ŷ = 30.5 − 2.1·x
Predicted value
ŷ = 30.5 + (−2.1)·1.5 = 27.35
Residual
e = 24 − 27.35 = −3.35
Answer
e = −3.35
The point lies below the line.
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