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: 59, 49, 42, 57, 41, 16.
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 62?
μ = 70σ = 8x = 62
P =
3.ProbabilityMedium
For the distribution shown, find the variance Var(X) of X.
x: 1, 2, 3, 4p(x): 0.2, 0.25, 0.35, 0.2
Var(X) =
4.ProbabilityHard
X has the distribution shown, and Y = 5X + 20. Find the standard deviation σY of Y.
x: 0, 1, 2, 3p(x): 0.35, 0.35, 0.25, 0.05Y = 5X + 20
σY =
5.Normal distributionMedium
The delivery times are normally distributed with mean 32 min and standard deviation 6 min. What fraction lies between 22.4 and 42.8 min? Give P as a decimal.
μ = 32σ = 6x1 = 22.4x2 = 42.8
P =
6.Normal distributionHard
The delivery times are normally distributed with mean 32 min and standard deviation 6 min. Find the value x below which 20% of them lie (the 20th percentile).
μ = 32σ = 6P = 20%
x = min
7.Confidence intervalsMedium
A sample of 11 items has mean 82 and standard deviation s = 12.5. The population is roughly normal. Find the upper limit of the 99% confidence interval for the mean.
n = 11x̄ = 82s = 12.599%t(10) = 3.169
upper limit =
8.Confidence intervalsHard
Two fertilisers are tested. Plot group 1: 9 plots, mean yield 32 kg, standard deviation 2.5 kg. Plot group 2: 7 plots, mean 33.5 kg, standard deviation 7 kg. Assume equal variances (pooled t, df = n₁ + n₂ − 2). Find the lower limit of the 90% confidence interval for μ₁ − μ₂.
n₁ = 9x̄₁ = 32s₁ = 2.5n₂ = 7x̄₂ = 33.5s₂ = 790%t(14) = 1.761
lower limit = kg
9.Hypothesis tests & confidence intervalsMedium
A sample of 64 students has a mean score of 52.9 and a sample standard deviation of 8. Find the test statistic t for H₀: μ = 50.
μ₀ = 50x̄ = 52.9s = 8n = 64
t =
10.Hypothesis tests & confidence intervalsHard
In a trial, 68 of 150 patients on treatment A recover, and 110 of 200 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.
x₁ = 68n₁ = 150x₂ = 110n₂ = 200
z =
11.Linear regression & correlationMedium
The table gives 6 paired observations of x = hours studied and y = exam score. 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. A new observation (11, 28) is made. Find its residual e = y − ŷ.
n = 6(11, 28)
e =
3Solutions
1.Descriptive statistics
Answer: s = 15.59
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 = 264/6 = 44
Squared deviations
Σ(x − x̄)² = 15² + 5² + (−2)² + 13² + (−3)² + (−28)² = 1216
Standard deviation
s = √(1216/5) = 15.59
Answer
s = 15.59
2.Descriptive statistics
Answer: P = 0.1587
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
Standard normal table
Φ(−1) = 0.1587
P = 0.1587
Answer
P = 0.1587
3.Probability
Answer: Var(X) = 1.048
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 = 1·0.2 + 2·0.25 + 3·0.35 + 4·0.2 = 2.55
Mean of the squares
E(X²) = 1·0.2 + 4·0.25 + 9·0.35 + 16·0.2 = 7.55
Variance
Var(X) = 7.55 − 2.55² = 1.048
Answer
Var(X) = 1.048
4.Probability
Answer: σY = 4.472
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, Var(X) = 0.8, σX = 0.8944
Linear change
σY = |5|·0.8944 = 4.472
Answer
σY = 4.472
The mean does move: E(Y) = 5·1 + 20 = 25.
5.Normal distribution
Answer: P = 0.9093
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 = (22.4 − 32)/6 = −1.6
z₂ = (42.8 − 32)/6 = 1.8
Area between
P = Φ(1.8) − Φ(−1.6) = 0.9641 − 0.0548 = 0.9093
Answer
P = 0.9093
6.Normal distribution
Answer: x = 26.95 min
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) = 0.2 → z = −0.8416
Back to x
x = 32 + −0.8416·6 = 26.95 min
Answer
x = 26.95 min
7.Confidence intervals
Answer: upper limit = 93.94
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 = 3.169
Margin of error
E = 3.169·12.5/√11 = 11.94
Interval
82 ± 11.94 → [70.06, 93.94]
Answer
upper limit = 93.94
8.Confidence intervals
Answer: lower limit = −5.899 kg
Interval for a difference: estimate ± critical value × standard error. For two independent means with known σ: SE = √(σ₁²/n₁ + σ₂²/n₂) and z. With equal unknown variances: pooled s_p and t with n₁ + n₂ − 2 degrees of freedom. For paired data, work with the differences: d̄ ± t·s_d/√n.
Pooled standard deviation
sp² = (8·2.5² + 6·7²)/14 = 24.57
sp = 4.957
Margin of error
E = 1.761·4.957·√(1/9 + 1/7) = 4.399
Interval
−1.5 ± 4.399 → [−5.899, 2.899]
Answer
lower limit = −5.899 kg
9.Hypothesis tests & confidence intervals
Answer: t = 2.9
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 = 8/√64 = 8/8 = 1
Test statistic
t = (52.9 − 50)/1 = 2.9
Answer
t = 2.9
t0.025; df = 63 = 1.998
|t| > 1.998, so H₀ is rejected at the 5% level (two-sided).
10.Hypothesis tests & confidence intervals
Answer: z = −1.79
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 = 68/150 = 0.4533
p̂₂ = 110/200 = 0.55
p̂ = 178/350 = 0.5086
Standard error under H₀
SE = √(0.5086·0.4914·(1/150 + 1/200)) = 0.054
Test statistic
z = (0.4533 − 0.55)/0.054 = −1.79
Answer
z = −1.79
|z| < 1.96, so H₀ is not rejected at the 5% level (two-sided).
11.Linear regression & correlation
Answer: r² = 0.9463
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 = 358/6 = 59.67
Sums of squares and products
Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = 107
Syy = Σ(y − ȳ)² = 691.3
Coefficient of determination
r² = 107²/(17.5·691.3) = 0.9463
Answer
r² = 0.9463
About 94.63% of the variation in y is explained by the line.
12.Linear regression & correlation
Answer: e = −4.771
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 = 126/6 = 21
Sums of squares and products
Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 206
Regression line
b = 206/70 = 2.943
a = 21 − 2.943·7 = 0.4
ŷ = 0.4 + 2.943·x
Predicted value
ŷ = 0.4 + 2.943·11 = 32.77
Residual
e = 28 − 32.77 = −4.771
Answer
e = −4.771
The point lies below the line.
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