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: 53, 27, 49, 13, 36, 59.
n = 6
s =
2.Descriptive statisticsHard
A quantity is normally distributed with mean 70 and standard deviation 8. What proportion P of values is below 60.4?
μ = 70σ = 8x = 60.4
P =
3.ProbabilityMedium
For the distribution shown, find the variance Var(X) of X.
x: 0, 1, 2, 3, 4p(x): 0.15, 0.25, 0.2, 0.05, 0.35
Var(X) =
4.ProbabilityHard
A game costs €2 to play. You win €6 with probability 0.1 and €40 with probability 0.01; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.
stake = €2€6: p = 0.1€40: p = 0.01
E(G) = €
5.Normal distributionMedium
The bag weights are normally distributed with mean 250 g and standard deviation 5 g. What fraction lies between 248.5 and 252.75 g? Give P as a decimal.
μ = 250σ = 5x1 = 248.5x2 = 252.75
P =
6.Normal distributionHard
The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. Find the value x below which 97.5% of them lie (the 97.5th percentile).
μ = 500σ = 40P = 97.5%
x = h
7.Confidence intervalsMedium
A sample of 20 items has mean 64.5 and standard deviation s = 20. The population is roughly normal. Find the lower limit of the 90% confidence interval for the mean.
n = 20x̄ = 64.5s = 2090%t(19) = 1.729
lower limit =
8.Confidence intervalsHard
The diameters of 30 machined shafts have a sample standard deviation of 3.9 µm. Assuming the diameters are normally distributed, find the lower limit σ of the 95% confidence interval for the population standard deviation. With df = 29, 2.5% of the χ² distribution lies below 16.047 and 2.5% lies above 45.722.
n = 30s = 3.9 µmχ² = 16.047, 45.722
σ = µm
9.Hypothesis tests & confidence intervalsMedium
20 employees take a typing test before and after a training course. For each person the difference d = after − before (words per minute) is recorded. The mean difference is 0.29 and the standard deviation of the differences is 1. Find the paired test statistic t for H₀: μd = 0.
n = 20d̄ = 0.29sd = 1
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 15. Find the smallest sample size n.
σ = 15E = 190%
n =
11.Linear regression & correlationMedium
The table gives 6 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 = 6
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. For which value x does the line predict ŷ = 43?
n = 6ŷ = 43
x =
3Solutions
1.Descriptive statistics
Answer: s = 17.46
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 = 237/6 = 39.5
Squared deviations
Σ(x − x̄)² = 13.5² + (−12.5)² + 9.5² + (−26.5)² + (−3.5)² + 19.5² = 1524
Standard deviation
s = √(1524/5) = 17.46
Answer
s = 17.46
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) = 2.26
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 = 0·0.15 + 1·0.25 + 2·0.2 + 3·0.05 + 4·0.35 = 2.2
Mean of the squares
E(X²) = 0·0.15 + 1·0.25 + 4·0.2 + 9·0.05 + 16·0.35 = 7.1
Variance
Var(X) = 7.1 − 2.2² = 2.26
Answer
Var(X) = 2.26
4.Probability
Answer: E(G) = −1 €
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.
Net outcomes
G: 4 (p = 0.1), 38 (p = 0.01), −2 (p = 0.89)
Expected value
E(G) = 4·0.1 + 38·0.01 − 2·0.89 = −1 €
Answer
E(G) = −1 €
A negative value means you lose money on average.
5.Normal distribution
Answer: P = 0.3268
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 = (248.5 − 250)/5 = −0.3
z₂ = (252.75 − 250)/5 = 0.55
Area between
P = Φ(0.55) − Φ(−0.3) = 0.7088 − 0.3821 = 0.3268
Answer
P = 0.3268
6.Normal distribution
Answer: x = 578.4 h
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) = 0.975 → z = 1.96
Back to x
x = 500 + 1.96·40 = 578.4 h
Answer
x = 578.4 h
7.Confidence intervals
Answer: lower limit = 56.77
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 = 19, t = 1.729
Margin of error
E = 1.729·20/√20 = 7.732
Interval
64.5 ± 7.732 → [56.77, 72.23]
Answer
lower limit = 56.77
8.Confidence intervals
Answer: σ = 3.106 µm
Interval for a variance: (n − 1)s²/σ² has a χ² distribution with n − 1 degrees of freedom. The 95% interval for σ² runs from (n − 1)s² divided by the upper χ² value (2.5% above it) to (n − 1)s² divided by the lower χ² value (2.5% below it); the interval for σ is the square root of both limits. It is not symmetric around s.
Sum of squares
(n − 1)s² = 29·3.9² = 441.1
Limits for σ²
441.1/45.722 = 9.647
441.1/16.047 = 27.49
Limit for σ
σ = √9.647 = 3.106 µm
Answer
σ = 3.106 µm
95% interval for σ: [3.106, 5.243] µm.
9.Hypothesis tests & confidence intervals
Answer: t = 1.297
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 of the mean difference
SE = 1/√20 = 0.2236
Test statistic
t = 0.29/0.2236 = 1.297
Answer
t = 1.297
With df = 19, the two-sided 5% critical value is 2.093; |t| is smaller, so H₀ is not rejected.
10.Hypothesis tests & confidence intervals
Answer: n = 609
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 = (1.645·15/1)² = 608.9
Round up
n = 609
Answer
n = 609
11.Linear regression & correlation
Answer: r² = 0.7313
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 = 130/6 = 21.67
Sums of squares and products
Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −23
Syy = Σ(y − ȳ)² = 41.33
Coefficient of determination
r² = −23²/(17.5·41.33) = 0.7313
Answer
r² = 0.7313
About 73.13% of the variation in y is explained by the line.
12.Linear regression & correlation
Answer: x = 10.92
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 = 173/6 = 28.83
Sums of squares and products
Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 253
Regression line
b = 253/70 = 3.614
a = 28.83 − 3.614·7 = 3.533
ŷ = 3.533 + 3.614·x
Solve for x
x = (43 − 3.533)/3.614 = 10.92
Answer
x = 10.92
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