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: 32, 33, 60, 31, 47, 58.
n = 6
s =
2.Descriptive statisticsHard
Exam scores are normally distributed with mean 70 and standard deviation 10. Find the z-score z of a score of 65.
μ = 70σ = 10x = 65
z =
3.ProbabilityMedium
For the distribution shown, find the standard deviation σ of X.
x: 1, 2, 3, 4p(x): 0.3, 0.25, 0.1, 0.35
σ =
4.ProbabilityHard
A test has 6 multiple-choice questions. A student who guesses gets each one right with probability 0.3, independently. What is the probability P of exactly 2 correct answers?
n = 6p = 0.3k = 2
P =
5.Normal distributionMedium
The delivery times are normally distributed with mean 32 min and standard deviation 6 min. What fraction lies between 25.1 and 38 min? Give P as a decimal.
μ = 32σ = 6x1 = 25.1x2 = 38
P =
6.Normal distributionHard
The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. Drag the cut-off so the shaded bottom 25% is to the left of it. Find the cut-off x.
μ = 178 cmσ = 7 cm25%
x = cm
7.Confidence intervalsMedium
A 95% confidence interval for a mean, based on 50 observations and a known σ, has a margin of error of 8.7. What is the margin of error E₂ of a 99% interval based on 450 observations?
95%n₁ = 50E₁ = 8.799%n₂ = 450
E₂ =
8.Confidence intervalsHard
The population standard deviation is 12. 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 1.5?
σ = 12E = 1.590%
n =
9.Hypothesis tests & confidence intervalsMedium
A sample of 25 students has a mean score of 94.4 and a sample standard deviation of 20. Find the test statistic t for H₀: μ = 100.
μ₀ = 100x̄ = 94.4s = 20n = 25
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 1? The standard deviation is 4. Find the smallest sample size n.
σ = 4E = 190%
n =
11.Linear regression & correlationMedium
The table gives 5 paired observations of x = advertising spend in k€ and y = sales 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 = load in kN and y = extension in mm. Use the least-squares line ŷ = a + b·x. A new observation (3, 18) is made. Find its residual e = y − ŷ.
n = 6(3, 18)
e =
3Solutions
1.Descriptive statistics
Answer: s = 13.37
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 = 261/6 = 43.5
Squared deviations
Σ(x − x̄)² = (−11.5)² + (−10.5)² + 16.5² + (−12.5)² + 3.5² + 14.5² = 893.5
Standard deviation
s = √(893.5/5) = 13.37
Answer
s = 13.37
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 = (65 − 70)/10 = −0.5
Answer
z = −0.5
3.Probability
Answer: σ = 1.245
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 = 1·0.3 + 2·0.25 + 3·0.1 + 4·0.35 = 2.5
Mean of the squares
E(X²) = 1·0.3 + 4·0.25 + 9·0.1 + 16·0.35 = 7.8
Variance
Var(X) = 7.8 − 2.5² = 1.55
σ = √1.55 = 1.245
Answer
σ = 1.245
4.Probability
Answer: P = 0.3241
Probability: with equally likely outcomes, P(A) = favourable/total. For independent events multiply: P(A and B) = P(A)·P(B). The complement often helps: P(at least one) = 1 − P(none). Binomial: in n independent trials with success probability p, P(X = k) = C(n, k)·pᵏ·(1 − p)ⁿ⁻ᵏ.
Binomial formula
P(X = 2) = 15·0.32·0.74 = 15·0.09·0.2401 = 0.3241
Answer
P = 0.3241
5.Normal distribution
Answer: P = 0.7163
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 = (25.1 − 32)/6 = −1.15
z₂ = (38 − 32)/6 = 1
Area between
P = Φ(1) − Φ(−1.15) = 0.8413 − 0.1251 = 0.7163
Answer
P = 0.7163
6.Normal distribution
Answer: x = 173.3 cm
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) = 0.25
z from the table
z ≈ −0.67
Back to the original units
x = 178 − 0.67·7 = 173.3 cm
Answer
x = 173.3 cm
7.Confidence intervals
Answer: E₂ = 3.811
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 = (2.576/1.96)·√(50/450) = 0.4381
New margin
E₂ = 8.7·0.4381 = 3.811
Answer
E₂ = 3.811
8.Confidence intervals
Answer: n = 174
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 = (1.645·12/1.5)² = 173.2
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 174
Answer
n = 174
9.Hypothesis tests & confidence intervals
Answer: t = −1.4
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/√25 = 20/5 = 4
Test statistic
t = (94.4 − 100)/4 = −1.4
Answer
t = −1.4
t0.025; df = 24 = 2.064
|t| < 2.064, so H₀ is not rejected at the 5% level (two-sided).
10.Hypothesis tests & confidence intervals
Answer: n = 44
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 = (1.645·4/1)² = 43.3
Round up
n = 44
Answer
n = 44
11.Linear regression & correlation
Answer: r² = 0.9322
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 = 191/5 = 38.2
Sums of squares and products
Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = 54
Syy = Σ(y − ȳ)² = 312.8
Coefficient of determination
r² = 54²/(10·312.8) = 0.9322
Answer
r² = 0.9322
About 93.22% of the variation in y is explained by the line.
12.Linear regression & correlation
Answer: e = 4.176
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 = 42/6 = 7
ȳ = 169/6 = 28.17
Sums of squares and products
Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 251
Regression line
b = 251/70 = 3.586
a = 28.17 − 3.586·7 = 3.067
ŷ = 3.067 + 3.586·x
Predicted value
ŷ = 3.067 + 3.586·3 = 13.82
Residual
e = 18 − 13.82 = 4.176
Answer
e = 4.176
The point lies above the line.
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