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: 41, 56, 56, 55, 14, 33, 23.
n = 7
s =
2.Descriptive statisticsHard
A quantity is normally distributed with mean 500 and standard deviation 100. What proportion P of values is above 625?
μ = 500σ = 100x = 625
P =
3.ProbabilityMedium
For the distribution shown, find the variance Var(X) of X.
x: 0, 1, 2, 3p(x): 0.2, 0.3, 0.35, 0.15
Var(X) =
4.ProbabilityHard
X has the distribution shown, and Y = 3X + 20. Find the standard deviation σY of Y.
x: 0, 2, 4, 6p(x): 0.25, 0.25, 0.2, 0.3Y = 3X + 20
σY =
5.Normal distributionMedium
Drag z so that the shaded area between −z and z under the standard normal curve is 0.3255. Find z to two decimals.
A = 0.3255
z =
6.Normal distributionHard
The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. Find the value x below which 20% of them lie (the 20th percentile).
μ = 178σ = 7P = 20%
x = cm
7.Confidence intervalsMedium
A sample of 11 items has mean 56 and standard deviation s = 8.5. The population is roughly normal. Find the upper limit of the 95% confidence interval for the mean.
n = 11x̄ = 56s = 8.595%t(10) = 2.228
upper limit =
8.Confidence intervalsHard
The population standard deviation is 40. How large a sample n do you need so that a 95% confidence interval for the mean has a margin of error of at most 2?
σ = 40E = 295%
n =
9.Hypothesis tests & confidence intervalsMedium
Two production lines make the same part. Line 1: a sample of 60 parts has mean length 55.5 mm, and the line's known standard deviation is 15 mm. Line 2: a sample of 30 parts has mean 61.6 mm, known standard deviation 14 mm. Find the test statistic z for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (line 1 minus line 2).
n₁ = 60x̄₁ = 55.5 mmσ₁ = 15 mmn₂ = 30x̄₂ = 61.6 mmσ₂ = 14 mm
z =
10.Hypothesis tests & confidence intervalsHard
How large must a sample be so that the 99% confidence interval for a mean has a margin of error of at most 3? The standard deviation is 5. Find the smallest sample size n.
σ = 5E = 399%
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 6 paired observations of x = car age in years and y = price in k€. Use the least-squares line ŷ = a + b·x. For which value x does the line predict ŷ = 27?
n = 6ŷ = 27
x =
3Solutions
1.Descriptive statistics
Answer: s = 17.09
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 = 278/7 = 39.71
Squared deviations
Σ(x − x̄)² = 1.286² + 16.29² + 16.29² + 15.29² + (−25.71)² + (−6.714)² + (−16.71)² = 1751
Standard deviation
s = √(1751/6) = 17.09
Answer
s = 17.09
2.Descriptive statistics
Answer: P = 0.1056
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 = 1.25
Standard normal table
Φ(1.25) = 0.8944
P = 1 − 0.8944 = 0.1056
Answer
P = 0.1056
3.Probability
Answer: Var(X) = 0.9475
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 = 0·0.2 + 1·0.3 + 2·0.35 + 3·0.15 = 1.45
Mean of the squares
E(X²) = 0·0.2 + 1·0.3 + 4·0.35 + 9·0.15 = 3.05
Variance
Var(X) = 3.05 − 1.45² = 0.9475
Answer
Var(X) = 0.9475
4.Probability
Answer: σY = 6.965
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 = 3.1, Var(X) = 5.39, σX = 2.322
Linear change
σY = |3|·2.322 = 6.965
Answer
σY = 6.965
The mean does move: E(Y) = 3·3.1 + 20 = 29.3.
5.Normal distribution
Answer: z = 0.42
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 of z
Φ(z) = (1 + 0.3255)/2 = 0.6628
Read the z-table backwards
Find 0.6628 inside the table and read off z.
z = 0.42
Answer
z = 0.42
6.Normal distribution
Answer: x = 172.1 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.
z for the percentile
Φ(z) = 0.2 → z = −0.8416
Back to x
x = 178 + −0.8416·7 = 172.11 cm
Answer
x = 172.11 cm
7.Confidence intervals
Answer: upper limit = 61.71
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 = 10, t = 2.228
Margin of error
E = 2.228·8.5/√11 = 5.71
Interval
56 ± 5.71 → [50.29, 61.71]
Answer
upper limit = 61.71
8.Confidence intervals
Answer: n = 1537
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 = (1.96·40/2)² = 1537
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 1537
Answer
n = 1537
9.Hypothesis tests & confidence intervals
Answer: z = −1.902
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.
Standard error of the difference
SE = √(15²/60 + 14²/30) = √10.28 = 3.207
Test statistic
z = (55.5 − 61.6)/3.207 = −1.902
Answer
z = −1.902
|z| < 1.96, so H₀ is not rejected at the 5% level (two-sided).
10.Hypothesis tests & confidence intervals
Answer: n = 19
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 = (2.576·5/3)² = 18.43
Round up
n = 19
Answer
n = 19
11.Linear regression & correlation
Answer: r² = 0.6207
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 = 98/5 = 19.6
Sums of squares and products
Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = −12
Syy = Σ(y − ȳ)² = 23.2
Coefficient of determination
r² = −12²/(10·23.2) = 0.6207
Answer
r² = 0.6207
About 62.07% of the variation in y is explained by the line.
12.Linear regression & correlation
Answer: x = 1.4
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 = 21/6 = 3.5
ȳ = 144/6 = 24
Sums of squares and products
Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −25
Regression line
b = −25/17.5 = −1.429
a = 24 − (−1.429)·3.5 = 29
ŷ = 29 − 1.429·x
Solve for x
x = (27 − 29)/(−1.429) = 1.4
Answer
x = 1.4
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