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: 55, 30, 36, 57, 57.
n = 5
s =
2.Descriptive statisticsHard
Exam scores are normally distributed with mean 60 and standard deviation 15. Find the z-score z of a score of 52.5.
μ = 60σ = 15x = 52.5
z =
3.ProbabilityMedium
For the distribution shown, find the variance Var(X) of X.
x: 0, 1, 2, 3, 4p(x): 0.1, 0.25, 0.15, 0.2, 0.3
Var(X) =
4.ProbabilityHard
X has the distribution shown, and Y = 5X − 3. Find the standard deviation σY of Y.
x: 0, 2, 4, 6p(x): 0.35, 0.25, 0.15, 0.25Y = 5X − 3
σY =
5.Normal distributionMedium
The masses of bags of cement are independent and normal with mean 25 kg and standard deviation 0.5 kg. Find the probability P that the total mass of 10 of them exceeds 251.6 kg. Round z to two decimals.
μ = 25 kgσ = 0.5 kgn = 10> 251.6 kg
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 top 95% is to the right of it. Find the cut-off x.
μ = 500 hσ = 40 h95%
x = h
7.Confidence intervalsMedium
A sample of 15 items has mean 70.5 and standard deviation s = 8.5. The population is roughly normal. Find the upper limit of the 90% confidence interval for the mean.
n = 15x̄ = 70.5s = 8.590%t(14) = 1.761
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 1?
σ = 25E = 199%
n =
9.Hypothesis tests & confidence intervalsMedium
A sample of 81 students has a mean score of 505.3 and a sample standard deviation of 20. Find the test statistic t for H₀: μ = 500.
μ₀ = 500x̄ = 505.3s = 20n = 81
t =
10.Hypothesis tests & confidence intervalsHard
In a trial, 56 of 80 patients on treatment A recover, and 208 of 250 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.
x₁ = 56n₁ = 80x₂ = 208n₂ = 250
z =
11.Linear regression & correlationMedium
The table gives 5 paired observations of x = car age in years and y = price in k€. Find Pearson's correlation coefficient r.
n = 5
r =
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. For which value x does the line predict ŷ = 27?
n = 5ŷ = 27
x =
3Solutions
1.Descriptive statistics
Answer: s = 12.98
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 = 235/5 = 47
Squared deviations
Σ(x − x̄)² = 8² + (−17)² + (−11)² + 10² + 10² = 674
Standard deviation
s = √(674/4) = 12.98
Answer
s = 12.98
2.Descriptive statistics
Answer: z = −0.5
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 = (52.5 − 60)/15 = −0.5
Answer
z = −0.5
3.Probability
Answer: Var(X) = 1.927
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 = 0·0.1 + 1·0.25 + 2·0.15 + 3·0.2 + 4·0.3 = 2.35
Mean of the squares
E(X²) = 0·0.1 + 1·0.25 + 4·0.15 + 9·0.2 + 16·0.3 = 7.45
Variance
Var(X) = 7.45 − 2.35² = 1.927
Answer
Var(X) = 1.927
4.Probability
Answer: σY = 11.87
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 = 2.6, Var(X) = 5.64, σX = 2.375
Linear change
σY = |5|·2.375 = 11.87
Answer
σY = 11.87
The mean does move: E(Y) = 5·2.6 − 3 = 10.
5.Normal distribution
Answer: P = 0.1562
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 = 250 kg, σT = 0.5·√10 = 1.581 kg
Standardise
z = (251.6 − 250)/1.581 = 1.01
Probability
P = 1 − Φ(1.01) = 0.1562
Answer
P = 0.1562
6.Normal distribution
Answer: x = 434 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) = 0.05
z from the table
z ≈ −1.64
Back to the original units
x = 500 − 1.64·40 = 434 h
Answer
x = 434 h
7.Confidence intervals
Answer: upper limit = 74.36
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 = 14, t = 1.761
Margin of error
E = 1.761·8.5/√15 = 3.865
Interval
70.5 ± 3.865 → [66.64, 74.36]
Answer
upper limit = 74.36
8.Confidence intervals
Answer: n = 4148
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 = (2.576·25/1)² = 4147
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 4148
Answer
n = 4148
9.Hypothesis tests & confidence intervals
Answer: t = 2.385
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).
Standard error
SE = 20/√81 = 20/9 = 2.222
Test statistic
t = (505.3 − 500)/2.222 = 2.385
Answer
t = 2.385
t0.025; df = 80 = 1.99
|t| > 1.99, so H₀ is rejected at the 5% level (two-sided).
10.Hypothesis tests & confidence intervals
Answer: z = −2.569
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 = 56/80 = 0.7
p̂₂ = 208/250 = 0.832
p̂ = 264/330 = 0.8
Standard error under H₀
SE = √(0.8·0.2·(1/80 + 1/250)) = 0.05138
Test statistic
z = (0.7 − 0.832)/0.05138 = −2.569
Answer
z = −2.569
|z| > 1.96, so H₀ is rejected at the 5% level (two-sided).
11.Linear regression & correlation
Answer: r = −0.8752
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 = 124/5 = 24.8
Sums of squares and products
Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = −12
Syy = Σ(y − ȳ)² = 18.8
Correlation coefficient
r = −12/√(10·18.8) = −0.8752
Answer
r = −0.8752
12.Linear regression & correlation
Answer: x = 2.417
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 = 128/5 = 25.6
Sums of squares and products
Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = −24
Regression line
b = −24/10 = −2.4
a = 25.6 − (−2.4)·3 = 32.8
ŷ = 32.8 − 2.4·x
Solve for x
x = (27 − 32.8)/(−2.4) = 2.417
Answer
x = 2.417
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