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: 30, 14, 48, 54, 34, 35, 60.
n = 7
s =
2.Descriptive statisticsHard
Exam scores are normally distributed with mean 70 and standard deviation 15. Find the z-score z of a score of 79.
μ = 70σ = 15x = 79
z =
3.ProbabilityMedium
For the distribution shown, find the variance Var(X) of X.
x: 0, 1, 2p(x): 0.35, 0.3, 0.35
Var(X) =
4.ProbabilityHard
X has the distribution shown, and Y = 0.5X + 1. Find the standard deviation σY of Y.
x: 0, 1, 2p(x): 0.3, 0.35, 0.35Y = 0.5X + 1
σY =
5.Normal distributionMedium
Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.9332. Find z to two decimals.
A = 0.9332
z =
6.Normal distributionHard
The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. Drag the cut-off so the shaded bottom 70% is to the left of it. Find the cut-off x.
μ = 100 pointsσ = 15 points70%
x = points
7.Confidence intervalsMedium
A 90% confidence interval for a mean, based on 100 observations and a known σ, has a margin of error of 2.8. What is the margin of error E₂ of a 95% interval from the same data?
90%n₁ = 100E₁ = 2.895%
E₂ =
8.Confidence intervalsHard
The population standard deviation is 35. 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 3?
σ = 35E = 390%
n =
9.Hypothesis tests & confidence intervalsMedium
Two production lines make the same part. Line 1: a sample of 30 parts has mean length 69 mm, and the line's known standard deviation is 13 mm. Line 2: a sample of 60 parts has mean 72 mm, known standard deviation 10 mm. Find the test statistic z for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (line 1 minus line 2).
n₁ = 30x̄₁ = 69 mmσ₁ = 13 mmn₂ = 60x̄₂ = 72 mmσ₂ = 10 mm
z =
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 2.5? The standard deviation is 5. Find the smallest sample size n.
σ = 5E = 2.590%
n =
11.Linear regression & correlationMedium
The table gives 6 paired observations of x = load in kN and y = extension in mm. 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 = temperature in °C and y = ice creams sold. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 24.5.
n = 6x = 24.5
ŷ =
3Solutions
1.Descriptive statistics
Answer: s = 15.78
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 = 275/7 = 39.29
Squared deviations
Σ(x − x̄)² = (−9.286)² + (−25.29)² + 8.714² + 14.71² + (−5.286)² + (−4.286)² + 20.71² = 1493
Standard deviation
s = √(1493/6) = 15.78
Answer
s = 15.78
2.Descriptive statistics
Answer: z = 0.6
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 = (79 − 70)/15 = 0.6
Answer
z = 0.6
3.Probability
Answer: Var(X) = 0.7
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 = 0·0.35 + 1·0.3 + 2·0.35 = 1
Mean of the squares
E(X²) = 0·0.35 + 1·0.3 + 4·0.35 = 1.7
Variance
Var(X) = 1.7 − 1² = 0.7
Answer
Var(X) = 0.7
4.Probability
Answer: σY = 0.4023
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 = 1.05, Var(X) = 0.6475, σX = 0.8047
Linear change
σY = |0.5|·0.8047 = 0.4023
Answer
σY = 0.4023
The mean does move: E(Y) = 0.5·1.05 + 1 = 1.525.
5.Normal distribution
Answer: z = −1.5
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 of z
Φ(z) = 1 − 0.9332 = 0.0668
Read the z-table backwards
Find 0.0668 inside the table and read off z.
z = −1.50
Answer
z = −1.50
6.Normal distribution
Answer: x = 107.8 points
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) = 0.7
z from the table
z ≈ 0.52
Back to the original units
x = 100 + 0.52·15 = 107.8 points
Answer
x = 107.8 points
7.Confidence intervals
Answer: E₂ = 3.336
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 = (1.96/1.645)·√(100/100) = 1.191
New margin
E₂ = 2.8·1.191 = 3.336
Answer
E₂ = 3.336
8.Confidence intervals
Answer: n = 369
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 = (1.645·35/3)² = 368.3
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 369
Answer
n = 369
9.Hypothesis tests & confidence intervals
Answer: z = −1.11
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.
Standard error of the difference
SE = √(13²/30 + 10²/60) = √7.3 = 2.702
Test statistic
z = (69 − 72)/2.702 = −1.11
Answer
z = −1.11
|z| < 1.96, so H₀ is not rejected at the 5% level (two-sided).
10.Hypothesis tests & confidence intervals
Answer: n = 11
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 = (1.645·5/2.5)² = 10.82
Round up
n = 11
Answer
n = 11
11.Linear regression & correlation
Answer: r² = 0.9741
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 = 104/6 = 17.33
Sums of squares and products
Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 134
Syy = Σ(y − ȳ)² = 263.3
Coefficient of determination
r² = 134²/(70·263.3) = 0.9741
Answer
r² = 0.9741
About 97.41% of the variation in y is explained by the line.
12.Linear regression & correlation
Answer: ŷ = 103.2
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 = 147/6 = 24.5
ȳ = 619/6 = 103.2
Sums of squares and products
Sxx = Σ(x − x̄)² = 157.5
Sxy = Σ(x − x̄)(y − ȳ) = 856.5
Regression line
b = 856.5/157.5 = 5.438
a = 103.2 − 5.438·24.5 = −30.07
ŷ = −30.07 + 5.438·x
Prediction
ŷ = −30.07 + 5.438·24.5 = 103.2
Answer
ŷ = 103.2
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