Introductory Statistics: cram pack

Formula sheets, the key method steps for every topic and 12 practice problems, with full solutions at the back.

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1Formulas and methods

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

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

  1. Convert the value to a z-score: z = (x − μ)/σ.
  2. Look up Φ(z), the area to the left of z.
  3. For the area to the right use 1 − Φ(z); for an interval subtract the two left areas.
  4. 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

  1. Decide between z (population σ known) and t (only the sample s, with n − 1 degrees of freedom).
  2. Find the critical value for the confidence level.
  3. Margin of error: E = critical value × standard deviation/√n.
  4. 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

  1. State H₀ and H₁ and decide whether the test is one- or two-sided.
  2. Calculate the test statistic: (x̄ − μ₀)/(σ/√n), or with s and t.
  3. Compare with the critical value, or find the p-value.
  4. 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
2025303540455055

Find the sample standard deviation s of the data: 16, 37, 19, 59, 32, 25, 25.

n = 7
s =
2.Descriptive statisticsHard
z −2−1012

Exam scores are normally distributed with mean 70 and standard deviation 12. Find the z-score z of a score of 58.

μ = 70σ = 12x = 58
z =
3.ProbabilityMedium
xp(x)10.120.330.2540.35μ = 2.85

For the distribution shown, find the variance Var(X) of X.

x: 1, 2, 3, 4p(x): 0.1, 0.3, 0.25, 0.35
Var(X) =
4.ProbabilityHard
n = 10, p = 0.5: P(X = 5) = ?k012345678910

A test has 10 multiple-choice questions. A student who guesses gets each one right with probability 0.5, independently. What is the probability P of exactly 5 correct answers?

n = 10p = 0.5k = 5
P =
5.Normal distributionMedium
436508P = ?

The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. What fraction lies between 436 and 508 h? Give P as a decimal.

μ = 500σ = 40x1 = 436x2 = 508
P =
6.Normal distributionHard
557085100115130145pointstop 25%

The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. Drag the cut-off so the shaded top 25% is to the right of it. Find the cut-off x.

μ = 100 pointsσ = 15 points25%
x = points
7.Confidence intervalsMedium
n = 16, s = 9L = ?Ux̄ = 16.5E90%

A sample of 16 items has mean 16.5 and standard deviation s = 9. The population is roughly normal. Find the lower limit of the 90% confidence interval for the mean.

n = 16x̄ = 16.5s = 990%t(15) = 1.753
lower limit =
8.Confidence intervalsHard
μ₁ − μ₂, t(12) = 1.7820LUx̄₁ − x̄₂E = ?90%

Two fertilisers are tested. Plot group 1: 7 plots, mean yield 18.5 kg, standard deviation 8.5 kg. Plot group 2: 7 plots, mean 25 kg, standard deviation 6 kg. Assume equal variances (pooled t, df = n₁ + n₂ − 2). Find the margin of error E of the 90% confidence interval for μ₁ − μ₂.

n₁ = 7x̄₁ = 18.5s₁ = 8.5n₂ = 7x̄₂ = 25s₂ = 690%t(12) = 1.782
E = kg
9.Hypothesis tests & confidence intervalsMedium
−1.961.96z = ?

Two production lines make the same part. Line 1: a sample of 100 parts has mean length 55.5 mm, and the line's known standard deviation is 10 mm. Line 2: a sample of 80 parts has mean 56.9 mm, known standard deviation 15 mm. Find the test statistic z for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (line 1 minus line 2).

n₁ = 100x̄₁ = 55.5 mmσ₁ = 10 mmn₂ = 80x̄₂ = 56.9 mmσ₂ = 15 mm
z =
10.Hypothesis tests & confidence intervalsHard
μ₀x̄c

A one-sided z test of H₀: μ = 100 against H₁: μ < 100 is done at α = 0.10 with a sample of 25; the standard deviation is known to be 10. In reality the mean is 98. Find the power 1 − β of the test in percent.

μ₀ = 100μ = 98σ = 10n = 25α = 0.10
1 − β = %
11.Linear regression & correlationMedium
xy123221315412511xy241520

The table gives 5 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 = 5
r² =
12.Linear regression & correlationHard
xy357461570669780xy46607080ŷ = 59

The table gives 5 paired observations of x = hours studied and y = exam score. Use the least-squares line ŷ = a + b·x. For which value x does the line predict ŷ = 59?

n = 5ŷ = 59
x =

3Solutions

1.Descriptive statistics

Answer: s = 14.49

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 = 213/7 = 30.43

Squared deviations

Σ(x − x̄)² = (−14.43)² + 6.571² + (−11.43)² + 28.57² + 1.571² + (−5.429)² + (−5.429)² = 1260

Standard deviation

s = √(1260/6) = 14.49

Answer

s = 14.49
2.Descriptive statistics

Answer: z = −1

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 = (58 − 70)/12 = −1

Answer

z = −1
3.Probability

Answer: Var(X) = 1.027

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 = 1·0.1 + 2·0.3 + 3·0.25 + 4·0.35 = 2.85

Mean of the squares

E(X²) = 1·0.1 + 4·0.3 + 9·0.25 + 16·0.35 = 9.15

Variance

Var(X) = 9.15 − 2.85² = 1.027

Answer

Var(X) = 1.027
4.Probability

Answer: P = 0.2461

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 = 5) = 252·0.55·0.55 = 252·0.03125·0.03125 = 0.2461

Answer

P = 0.2461
5.Normal distribution

Answer: P = 0.5245

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 = (436 − 500)/40 = −1.6
z₂ = (508 − 500)/40 = 0.2

Area between

P = Φ(0.2) − Φ(−1.6) = 0.5793 − 0.0548 = 0.5245

Answer

P = 0.5245
6.Normal distribution

Answer: x = 110.1 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) = 0.75

z from the table

z ≈ 0.67

Back to the original units

x = 100 + 0.67·15 = 110.1 points

Answer

x = 110.1 points
7.Confidence intervals

Answer: lower limit = 12.56

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 = 15, t = 1.753

Margin of error

E = 1.753·9/√16 = 3.944

Interval

16.5 ± 3.944 → [12.56, 20.44]

Answer

lower limit = 12.56
8.Confidence intervals

Answer: E = 7.008 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 = (6·8.5² + 6·6²)/12 = 54.13
sp = 7.357

Margin of error

E = 1.782·7.357·√(1/7 + 1/7) = 7.008

Answer

E = 7.008 kg
9.Hypothesis tests & confidence intervals

Answer: z = −0.717

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 = √(10²/100 + 15²/80) = √3.812 = 1.953

Test statistic

z = (55.5 − 56.9)/1.953 = −0.717

Answer

z = −0.717

|z| < 1.96, so H₀ is not rejected at the 5% level (two-sided).

10.Hypothesis tests & confidence intervals

Answer: 1 − β = 38.9 %

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 = 10/√25 = 2
x̄c = 100 − 1.282·2 = 97.44

z under the true mean

z = (97.44 − 98)/2 = −0.282

Power

power = P(Z < −0.282) = 38.9%
1 − β = 38.9%

Answer

1 − β = 38.9%
11.Linear regression & correlation

Answer: r² = 0.9453

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 = 82/5 = 16.4

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = −33
Syy = Σ(y − ȳ)² = 115.2

Coefficient of determination

r² = −33²/(10·115.2) = 0.9453

Answer

r² = 0.9453

About 94.53% of the variation in y is explained by the line.

12.Linear regression & correlation

Answer: x = 3.444

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 = 25/5 = 5
ȳ = 337/5 = 67.4

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = 54

Regression line

b = 54/10 = 5.4
a = 67.4 − 5.4·5 = 40.4
ŷ = 40.4 + 5.4·x

Solve for x

x = (59 − 40.4)/5.4 = 3.444

Answer

x = 3.444

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