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: 29, 38, 27, 21, 25, 18, 58.
n = 7
s =
2.Descriptive statisticsHard
A quantity is normally distributed with mean 70 and standard deviation 15. What proportion P of values is below 40?
μ = 70σ = 15x = 40
P =
3.ProbabilityMedium
For the distribution shown, find the standard deviation σ of X.
x: 1, 2, 3, 4p(x): 0.15, 0.15, 0.35, 0.35
σ =
4.ProbabilityHard
X has the distribution shown, and Y = −2X − 3. Find the standard deviation σY of Y.
x: 1, 2, 3, 4p(x): 0.2, 0.1, 0.35, 0.35Y = −2X − 3
σY =
5.Normal distributionMedium
The masses of passengers with luggage are independent and normal with mean 85 kg and standard deviation 12 kg. Find the probability P that the total mass of 12 of them is below 978.4 kg. Round z to two decimals.
μ = 85 kgσ = 12 kgn = 12< 978.4 kg
P =
6.Normal distributionHard
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.
μX = 500 g, σX = 8 gμY = 490 g, σY = 6 g
P =
7.Confidence intervalsMedium
A 90% confidence interval for a mean, based on 25 observations and a known σ, has a margin of error of 9.1. What is the margin of error E₂ if the same study uses 225 observations instead?
90%n₁ = 25E₁ = 9.1n₂ = 225
E₂ =
8.Confidence intervalsHard
The population standard deviation is 6. 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 1.5?
σ = 6E = 1.599%
n =
9.Hypothesis tests & confidence intervalsMedium
The mean of a population is claimed to be 120; the population standard deviation is known to be 9. A random sample of 36 observations has a mean of 117.72. Test H₀: μ = 120 against H₁: μ < 120. Find the p-value p.
μ₀ = 120σ = 9n = 36x̄ = 117.72H₁: μ < μ₀
p =
10.Hypothesis tests & confidence intervalsHard
A one-sided z test at α = 0.01 must detect a shift of the mean of 1 units with a power of 90%. The standard deviation is 6. Find the minimum sample size n.
α = 0.01power = 90%δ = 1σ = 6zα = 2.326zβ = 1.282
n =
11.Linear regression & correlationMedium
The table gives 5 paired observations of x = load in kN and y = extension in mm. Find Pearson's correlation coefficient r.
n = 5
r =
12.Linear regression & correlationHard
The table gives 5 paired observations of x = load in kN and y = extension in mm. Use the least-squares line ŷ = a + b·x. A new observation (3, 10) is made. Find its residual e = y − ŷ.
n = 5(3, 10)
e =
3Solutions
1.Descriptive statistics
Answer: s = 13.56
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 = 216/7 = 30.86
Squared deviations
Σ(x − x̄)² = (−1.857)² + 7.143² + (−3.857)² + (−9.857)² + (−5.857)² + (−12.86)² + 27.14² = 1103
Standard deviation
s = √(1103/6) = 13.56
Answer
s = 13.56
2.Descriptive statistics
Answer: P = 0.02275
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 = −2
Standard normal table
Φ(−2) = 0.02275
P = 0.02275
Answer
P = 0.02275
3.Probability
Answer: σ = 1.044
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 = 1·0.15 + 2·0.15 + 3·0.35 + 4·0.35 = 2.9
Mean of the squares
E(X²) = 1·0.15 + 4·0.15 + 9·0.35 + 16·0.35 = 9.5
Variance
Var(X) = 9.5 − 2.9² = 1.09
σ = √1.09 = 1.044
Answer
σ = 1.044
4.Probability
Answer: σY = 2.216
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 = 2.85, Var(X) = 1.228, σX = 1.108
Linear change
σY = |−2|·1.108 = 2.216
Answer
σY = 2.216
The mean does move: E(Y) = −2·2.85 − 3 = −8.7.
5.Normal distribution
Answer: P = 0.1587
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 total
μT = 1020 kg, σT = 12·√12 = 41.57 kg
Standardise
z = (978.4 − 1020)/41.57 = −1
Probability
P = Φ(−1) = 0.1587
Answer
P = 0.1587
6.Normal distribution
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).
Distribution of the difference
μ = 500 − 490 = 10 g, σ = 10 g
Standardise
z = (0 − 10)/10 = −1
Probability
P = 1 − Φ(−1) = 0.8413
Answer
P = 0.8413
7.Confidence intervals
Answer: E₂ = 3.033
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.
Ratio of margins
E₂/E₁ = (1.645/1.645)·√(25/225) = 0.3333
New margin
E₂ = 9.1·0.3333 = 3.033
Answer
E₂ = 3.033
8.Confidence intervals
Answer: n = 107
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 = (2.576·6/1.5)² = 106.2
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 107
Answer
n = 107
9.Hypothesis tests & confidence intervals
Answer: p = 0.06426
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.
Test statistic
SE = σ/√n = 9/√36 = 1.5
z = (117.72 − 120)/1.5 = −1.52
Tail area
Φ(1.52) = 0.9357
1 − Φ(1.52) = 0.0643
p-value
p = 0.06426
Answer
p = 0.06426
p > 0.05, so H₀ is not rejected at the 5% level.
10.Hypothesis tests & confidence intervals
Answer: n = 469
Sample size for a test: to detect a true shift δ with power 1 − β in a one-sided z test at level α, the standard error must satisfy δ = (z_α + z_β)·σ/√n, so n = ((z_α + z_β)·σ/δ)², rounded up.
Solve for n
n = ((2.326 + 1.282)·6/1)² = 468.6
Round up
n = 469
Answer
n = 469
11.Linear regression & correlation
Answer: r = 0.9952
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 = 30/5 = 6
ȳ = 121/5 = 24.2
Sums of squares and products
Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 172
Syy = Σ(y − ȳ)² = 746.8
Correlation coefficient
r = 172/√(40·746.8) = 0.9952
Answer
r = 0.9952
12.Linear regression & correlation
Answer: e = 2.55
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 = 30/5 = 6
ȳ = 59/5 = 11.8
Sums of squares and products
Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 58
Regression line
b = 58/40 = 1.45
a = 11.8 − 1.45·6 = 3.1
ŷ = 3.1 + 1.45·x
Predicted value
ŷ = 3.1 + 1.45·3 = 7.45
Residual
e = 10 − 7.45 = 2.55
Answer
e = 2.55
The point lies above the line.
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