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: 15, 28, 58, 21, 14, 57, 45.
n = 7
s =
2.Descriptive statisticsHard
A quantity is normally distributed with mean 50 and standard deviation 10. What proportion P of values is below 38?
μ = 50σ = 10x = 38
P =
3.ProbabilityMedium
For the distribution shown, find the variance Var(X) of X.
x: 1, 2, 3, 4, 5p(x): 0.15, 0.05, 0.25, 0.35, 0.2
Var(X) =
4.ProbabilityHard
X has the distribution shown, and Y = 0.5X − 3. Find the standard deviation σY of Y.
x: 0, 1, 2, 3p(x): 0.15, 0.3, 0.25, 0.3Y = 0.5X − 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 4 of them exceeds 99.5 kg. Round z to two decimals.
μ = 25 kgσ = 0.5 kgn = 4> 99.5 kg
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 99% of them lie (the 99th percentile).
μ = 65σ = 10P = 99%
x = points
7.Confidence intervalsMedium
A sample of 11 items has mean 87 and standard deviation s = 5. The population is roughly normal. Find the margin of error E of a 90% confidence interval for the mean.
n = 11x̄ = 87s = 590%t(10) = 1.812
E =
8.Confidence intervalsHard
The population standard deviation is 38. 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 2.5?
σ = 38E = 2.590%
n =
9.Hypothesis tests & confidence intervalsMedium
The mean of a population is claimed to be 50; the population standard deviation is known to be 40. A random sample of 100 observations has a mean of 58.8. Test H₀: μ = 50 against H₁: μ > 50. Find the p-value p.
μ₀ = 50σ = 40n = 100x̄ = 58.8H₁: μ > μ₀
p =
10.Hypothesis tests & confidence intervalsHard
In a trial, 75 of 150 patients on treatment A recover, and 62 of 100 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.
x₁ = 75n₁ = 150x₂ = 62n₂ = 100
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 6 paired observations of x = advertising spend in k€ and y = sales in k€. Use the least-squares line ŷ = a + b·x. A new observation (5.5, 60) is made. Find its residual e = y − ŷ.
n = 6(5.5, 60)
e =
3Solutions
1.Descriptive statistics
Answer: s = 19.11
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 = 238/7 = 34
Squared deviations
Σ(x − x̄)² = (−19)² + (−6)² + 24² + (−13)² + (−20)² + 23² + 11² = 2192
Standard deviation
s = √(2192/6) = 19.11
Answer
s = 19.11
2.Descriptive statistics
Answer: P = 0.1151
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 = −1.2
Standard normal table
Φ(−1.2) = 0.1151
P = 0.1151
Answer
P = 0.1151
3.Probability
Answer: Var(X) = 1.64
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 + 2·0.05 + 3·0.25 + 4·0.35 + 5·0.2 = 3.4
Mean of the squares
E(X²) = 1·0.15 + 4·0.05 + 9·0.25 + 16·0.35 + 25·0.2 = 13.2
Variance
Var(X) = 13.2 − 3.4² = 1.64
Answer
Var(X) = 1.64
4.Probability
Answer: σY = 0.5268
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 = 1.7, Var(X) = 1.11, σX = 1.054
Linear change
σY = |0.5|·1.054 = 0.5268
Answer
σY = 0.5268
The mean does move: E(Y) = 0.5·1.7 − 3 = −2.15.
5.Normal distribution
Answer: P = 0.6915
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 = 100 kg, σT = 0.5·√4 = 1 kg
Standardise
z = (99.5 − 100)/1 = −0.5
Probability
P = 1 − Φ(−0.5) = 0.6915
Answer
P = 0.6915
6.Normal distribution
Answer: x = 88.26 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) = 0.99 → z = 2.326
Back to x
x = 65 + 2.326·10 = 88.26 points
Answer
x = 88.26 points
7.Confidence intervals
Answer: E = 2.732
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 = 1.812
Margin of error
E = 1.812·5/√11 = 2.732
Answer
E = 2.732
8.Confidence intervals
Answer: n = 626
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 = (1.645·38/2.5)² = 625.2
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 626
Answer
n = 626
9.Hypothesis tests & confidence intervals
Answer: p = 0.0139
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 = 40/√100 = 4
z = (58.8 − 50)/4 = 2.2
Tail area
Φ(2.2) = 0.9861
1 − Φ(2.2) = 0.0139
p-value
p = 0.0139
Answer
p = 0.0139
p ≤ 0.05, so H₀ is rejected at the 5% level.
10.Hypothesis tests & confidence intervals
Answer: z = −1.868
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 = 75/150 = 0.5
p̂₂ = 62/100 = 0.62
p̂ = 137/250 = 0.548
Standard error under H₀
SE = √(0.548·0.452·(1/150 + 1/100)) = 0.06425
Test statistic
z = (0.5 − 0.62)/0.06425 = −1.868
Answer
z = −1.868
|z| < 1.96, so H₀ is not rejected at the 5% level (two-sided).
11.Linear regression & correlation
Answer: r = −0.9042
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 = 109/5 = 21.8
Sums of squares and products
Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = −11
Syy = Σ(y − ȳ)² = 14.8
Correlation coefficient
r = −11/√(10·14.8) = −0.9042
Answer
r = −0.9042
12.Linear regression & correlation
Answer: e = 4.833
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 = 33/6 = 5.5
ȳ = 331/6 = 55.17
Sums of squares and products
Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = 67.5
Regression line
b = 67.5/17.5 = 3.857
a = 55.17 − 3.857·5.5 = 33.95
ŷ = 33.95 + 3.857·x
Predicted value
ŷ = 33.95 + 3.857·5.5 = 55.17
Residual
e = 60 − 55.17 = 4.833
Answer
e = 4.833
The point lies above the line.
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