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: 20, 13, 10, 26, 21, 23, 13.
n = 7
s =
2.Descriptive statisticsHard
Exam scores are normally distributed with mean 170 and standard deviation 12. Find the z-score z of a score of 185.
μ = 170σ = 12x = 185
z =
3.ProbabilityMedium
For the distribution shown, find the standard deviation σ of X.
x: 10, 20, 30, 40p(x): 0.35, 0.05, 0.25, 0.35
σ =
4.ProbabilityHard
X has the distribution shown, and Y = 3X + 20. Find the standard deviation σY of Y.
x: 1, 2, 3, 4, 5p(x): 0.05, 0.25, 0.15, 0.2, 0.35Y = 3X + 20
σY =
5.Normal distributionMedium
Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.0764. Find z to two decimals.
A = 0.0764
z =
6.Normal distributionHard
Hole diameters D are normal with mean 25.04 mm and standard deviation 0.015 mm. Shaft diameters d are independent and normal with mean 25 mm and standard deviation 0.01 mm. A randomly chosen shaft fits when D − d > 0. Find the probability P that it fits. Round z to two decimals.
hole: μ = 25.04 mm, σ = 0.015 mmshaft: μ = 25 mm, σ = 0.01 mm
P =
7.Confidence intervalsMedium
A sample of 10 items has mean 68.5 and standard deviation s = 20. The population is roughly normal. Find the lower limit of the 95% confidence interval for the mean.
n = 10x̄ = 68.5s = 2095%t(9) = 2.262
lower limit =
8.Confidence intervalsHard
The population standard deviation is 39. How large a sample n do you need so that a 95% confidence interval for the mean has a margin of error of at most 5?
σ = 39E = 595%
n =
9.Hypothesis tests & confidence intervalsMedium
A sample of 25 students has a mean score of 41.2 and a sample standard deviation of 20. Find the test statistic t for H₀: μ = 50.
μ₀ = 50x̄ = 41.2s = 20n = 25
t =
10.Hypothesis tests & confidence intervalsHard
A one-sided z test of H₀: μ = 50 against H₁: μ < 50 is done at α = 0.05 with a sample of 36; the standard deviation is known to be 15. In reality the mean is 47.25. Find the power 1 − β of the test in percent.
μ₀ = 50μ = 47.25σ = 15n = 36α = 0.05
1 − β = %
11.Linear regression & correlationMedium
The table gives 6 paired observations of x = temperature in °C and y = ice creams sold. Their sums are given. Find Pearson's correlation coefficient r from these sums.
n = 6Σx = 153Σy = 696Σx² = 4059Σy² = 84712Σxy = 18531
r =
12.Linear regression & correlationHard
The table gives 6 paired observations of x = car age in years and y = price in k€. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 5.5.
n = 6x = 5.5
ŷ =
3Solutions
1.Descriptive statistics
Answer: s = 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.
Mean
x̄ = 126/7 = 18
Squared deviations
Σ(x − x̄)² = 2² + (−5)² + (−8)² + 8² + 3² + 5² + (−5)² = 216
Standard deviation
s = √(216/6) = 6
Answer
s = 6
2.Descriptive statistics
Answer: z = 1.25
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 = (185 − 170)/12 = 1.25
Answer
z = 1.25
3.Probability
Answer: σ = 12.81
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 = 10·0.35 + 20·0.05 + 30·0.25 + 40·0.35 = 26
Mean of the squares
E(X²) = 100·0.35 + 400·0.05 + 900·0.25 + 1600·0.35 = 840
Variance
Var(X) = 840 − 26² = 164
σ = √164 = 12.81
Answer
σ = 12.81
4.Probability
Answer: σY = 3.966
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 = 3.55, Var(X) = 1.748, σX = 1.322
Linear change
σY = |3|·1.322 = 3.966
Answer
σY = 3.966
The mean does move: E(Y) = 3·3.55 + 20 = 30.65.
5.Normal distribution
Answer: z = 1.43
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 − 0.0764 = 0.9236
Read the z-table backwards
Find 0.9236 inside the table and read off z.
z = 1.43
Answer
z = 1.43
6.Normal distribution
Answer: P = 0.9868
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 difference
μ = 25.04 − 25 = 0.04 mm, σ = 0.01803 mm
Standardise
z = (0 − 0.04)/0.01803 = −2.22
Probability
P = 1 − Φ(−2.22) = 0.9868
Answer
P = 0.9868
7.Confidence intervals
Answer: lower limit = 54.19
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 = 9, t = 2.262
Margin of error
E = 2.262·20/√10 = 14.31
Interval
68.5 ± 14.31 → [54.19, 82.81]
Answer
lower limit = 54.19
8.Confidence intervals
Answer: n = 234
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 = (1.96·39/5)² = 233.7
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 234
Answer
n = 234
9.Hypothesis tests & confidence intervals
Answer: t = −2.2
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 = (41.2 − 50)/4 = −2.2
Answer
t = −2.2
t0.025; df = 24 = 2.064
|t| > 2.064, so H₀ is rejected at the 5% level (two-sided).
10.Hypothesis tests & confidence intervals
Answer: 1 − β = 29.29 %
Power of a test: power = 1 − β is the probability of rejecting H₀ when it is false. Find the cutoff x̄c from μ₀ and z_α, then compute the probability of landing beyond x̄c using the true mean: z = (x̄c − μ)/(σ/√n).
Critical sample mean
SE = 15/√36 = 2.5
x̄c = 50 − 1.645·2.5 = 45.89
z under the true mean
z = (45.89 − 47.25)/2.5 = −0.545
Power
power = P(Z < −0.545) = 29.29%
1 − β = 29.29%
Answer
1 − β = 29.29%
11.Linear regression & correlation
Answer: r = 0.9895
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.
Centred sums
Sxy = 18531 − 153·696/6 = 783
Sxx = 4059 − 153²/6 = 157.5
Syy = 84712 − 696²/6 = 3976
Correlation coefficient
r = 783/√(157.5·3976) = 0.9895
Answer
r = 0.9895
12.Linear regression & correlation
Answer: ŷ = 11.83
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 = 107/6 = 17.83
Sums of squares and products
Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −52.5
Regression line
b = −52.5/17.5 = −3
a = 17.83 − (−3)·3.5 = 28.33
ŷ = 28.33 − 3·x
Prediction
ŷ = 28.33 + (−3)·5.5 = 11.83
Answer
ŷ = 11.83
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