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
1520253035404550

Find the sample standard deviation s of the data: 50, 51, 24, 15, 28, 52.

n = 6
s =
2.Descriptive statisticsHard
z −2−1012

A quantity is normally distributed with mean 170 and standard deviation 10. What proportion P of values is below 160?

μ = 170σ = 10x = 160
P =
3.ProbabilityMedium
xp(x)100.1200.25300.3400.35μ = 29

For the distribution shown, find the standard deviation σ of X.

x: 10, 20, 30, 40p(x): 0.1, 0.25, 0.3, 0.35
σ =
4.ProbabilityHard
n = 8, p = 0.25: P(X = 5) = ?k012345678

A test has 8 multiple-choice questions. A student who guesses gets each one right with probability 0.25, independently. What is the probability P of exactly 5 correct answers?

n = 8p = 0.25k = 5
P =
5.Normal distributionMedium
248.5254P = ?

The bag weights are normally distributed with mean 250 g and standard deviation 5 g. What fraction lies between 248.5 and 254 g? Give P as a decimal.

μ = 250σ = 5x1 = 248.5x2 = 254
P =
6.Normal distributionHard
0.02 mm0D − d?

Hole diameters D are normal with mean 25.02 mm and standard deviation 0.02 mm. Shaft diameters d are independent and normal with mean 25 mm and standard deviation 0.015 mm. A randomly chosen shaft fits when D − d > 0. Find the probability P that it fits. Round z to two decimals.

hole: μ = 25.02 mm, σ = 0.02 mmshaft: μ = 25 mm, σ = 0.015 mm
P =
7.Confidence intervalsMedium
μ₁ − μ₂0LUx̄₁ − x̄₂E = ?90%

The battery life of two phone models is compared. Model 1: 30 phones, mean 85 h, known standard deviation 4 h. Model 2: 50 phones, mean 100 h, known standard deviation 11 h. Find the margin of error E of the 90% confidence interval for μ₁ − μ₂.

n₁ = 30x̄₁ = 85 hσ₁ = 4 hn₂ = 50x̄₂ = 100 hσ₂ = 11 h90%
E = h
8.Confidence intervalsHard
p = 0.5, n = ?p̂ − Ep̂ + Ep̂E = 0.02595%

Nothing is known in advance about the proportion of defective products. How many products n must be inspected so that a 95% confidence interval for the proportion has a margin of error of at most 0.025?

p = 0.5E = 0.02595%
n =
9.Hypothesis tests & confidence intervalsMedium
t −1.99801.998

A sample of 64 students has a mean score of 73.88 and a sample standard deviation of 10. Find the test statistic t for H₀: μ = 75.

μ₀ = 75x̄ = 73.88s = 10n = 64
t =
10.Hypothesis tests & confidence intervalsHard
H₀H₁μ₀x̄cμ₀ + 1α = 0.05βσ = 8, power 80%, n = ?

A one-sided z test at α = 0.05 must detect a shift of the mean of 1 units with a power of 80%. The standard deviation is 8. Find the minimum sample size n.

α = 0.05power = 80%δ = 1σ = 8zα = 1.645zβ = 0.842
n =
11.Linear regression & correlationMedium
xy125221322416512613xy246152025

The table gives 6 paired observations of x = car age in years and y = price in k€. Find Pearson's correlation coefficient r.

n = 6
r =
12.Linear regression & correlationHard
xy124217319416514610xy24610152025x = 3.5

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. A new observation (3.5, 12) is made. Find its residual e = y − ŷ.

n = 6(3.5, 12)
e =

3Solutions

1.Descriptive statistics

Answer: s = 16.27

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 = 220/6 = 36.67

Squared deviations

Σ(x − x̄)² = 13.33² + 14.33² + (−12.67)² + (−21.67)² + (−8.667)² + 15.33² = 1323

Standard deviation

s = √(1323/5) = 16.27

Answer

s = 16.27
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: σ = 9.95

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 = 10·0.1 + 20·0.25 + 30·0.3 + 40·0.35 = 29

Mean of the squares

E(X²) = 100·0.1 + 400·0.25 + 900·0.3 + 1600·0.35 = 940

Variance

Var(X) = 940 − 29² = 99
σ = √99 = 9.95

Answer

σ = 9.95
4.Probability

Answer: P = 0.02307

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) = 56·0.255·0.753 = 56·0.0009766·0.4219 = 0.02307

Answer

P = 0.02307
5.Normal distribution

Answer: P = 0.4061

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 = (254 − 250)/5 = 0.8

Area between

P = Φ(0.8) − Φ(−0.3) = 0.7881 − 0.3821 = 0.4061

Answer

P = 0.4061
6.Normal distribution

Answer: P = 0.7881

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 = 25.02 − 25 = 0.02 mm, σ = 0.025 mm

Standardise

z = (0 − 0.02)/0.025 = −0.8

Probability

P = 1 − Φ(−0.8) = 0.7881

Answer

P = 0.7881
7.Confidence intervals

Answer: E = 2.827 h

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.

Point estimate

x̄₁ − x̄₂ = 85 − 100 = −15

Standard error

SE = √(4²/30 + 11²/50) = 1.719

Margin of error

E = 1.645·1.719 = 2.827

Answer

E = 2.827 h

The interval does not contain 0, so the means differ significantly.

8.Confidence intervals

Answer: n = 1537

Confidence interval for a proportion: p̂ ± z·√(p̂(1 − p̂)/n), where p̂ = x/n. To reach a margin E you need n ≥ z²·p(1 − p)/E², rounded up; without a prior estimate use p = 0.5, which gives the largest n.

Solve for n

n = 1.96²·0.5·0.5/0.025² = 1537

Round up

n = 1537

Answer

n = 1537
9.Hypothesis tests & confidence intervals

Answer: t = −0.896

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 = 10/√64 = 10/8 = 1.25

Test statistic

t = (73.88 − 75)/1.25 = −0.896

Answer

t = −0.896
t0.025; df = 63 = 1.998

|t| < 1.998, so H₀ is not rejected at the 5% level (two-sided).

10.Hypothesis tests & confidence intervals

Answer: n = 396

Sample size for a test: to detect a true shift δ with power 1 − β in a one-sided z test at level α, the standard error must satisfy δ = (z_α + z_β)·σ/√n, so n = ((z_α + z_β)·σ/δ)², rounded up.

Solve for n

n = ((1.645 + 0.842)·8/1)² = 395.9

Round up

n = 396

Answer

n = 396
11.Linear regression & correlation

Answer: r = −0.9434

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 = 109/6 = 18.17

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −46.5
Syy = Σ(y − ȳ)² = 138.8

Correlation coefficient

r = −46.5/√(17.5·138.8) = −0.9434

Answer

r = −0.9434
12.Linear regression & correlation

Answer: e = −4.667

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 = 100/6 = 16.67

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −41

Regression line

b = −41/17.5 = −2.343
a = 16.67 − (−2.343)·3.5 = 24.87
ŷ = 24.87 − 2.343·x

Predicted value

ŷ = 24.87 + (−2.343)·3.5 = 16.67

Residual

e = 12 − 16.67 = −4.667

Answer

e = −4.667

The point lies below the line.

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