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
303540455055

Find the sample standard deviation s of the data: 28, 40, 31, 57, 58, 57, 33.

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

A quantity is normally distributed with mean 50 and standard deviation 12. What proportion P of values is below 68?

μ = 50σ = 12x = 68
P =
3.ProbabilityMedium
xp(x)00.310.1520.230.35μ = 1.6

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

x: 0, 1, 2, 3p(x): 0.3, 0.15, 0.2, 0.35
σ =
4.ProbabilityHard
n = 10, p = 0.1: P(X = 2) = ?k012345678910

A test has 10 multiple-choice questions. A student who guesses gets each one right with probability 0.1, independently. What is the probability P of exactly 2 correct answers?

n = 10p = 0.1k = 2
P =
5.Normal distributionMedium
22.439.5P = ?

The delivery times are normally distributed with mean 32 min and standard deviation 6 min. What fraction lies between 22.4 and 39.5 min? Give P as a decimal.

μ = 32σ = 6x1 = 22.4x2 = 39.5
P =
6.Normal distributionHard
35455565758595pointstop 70%

The exam scores are normally distributed with mean 65 points and standard deviation 10 points. Drag the cut-off so the shaded top 70% is to the right of it. Find the cut-off x.

μ = 65 pointsσ = 10 points70%
x = points
7.Confidence intervalsMedium
n = 25, s = 3.5LUx̄ = 33.5E = ?99%

A sample of 25 items has mean 33.5 and standard deviation s = 3.5. The population is roughly normal. Find the margin of error E of a 99% confidence interval for the mean.

n = 25x̄ = 33.5s = 3.599%t(24) = 2.797
E =
8.Confidence intervalsHard
σ = 33, n = ?x̄ − Ex̄ + Ex̄E = 595%

The population standard deviation is 33. 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 = 33E = 595%
n =
9.Hypothesis tests & confidence intervalsMedium
t −1.9901.99

A sample of 81 students has a mean score of 101.3 and a sample standard deviation of 5. Find the test statistic t for H₀: μ = 100.

μ₀ = 100x̄ = 101.3s = 5n = 81
t =
10.Hypothesis tests & confidence intervalsHard
H₀H₁μ₀x̄cμ₀ + 4α = 0.05βσ = 21, power 90%, n = ?

A one-sided z test at α = 0.05 must detect a shift of the mean of 4 units with a power of 90%. The standard deviation is 21. Find the minimum sample size n.

α = 0.05power = 90%δ = 4σ = 21zα = 1.645zβ = 1.282
n =
11.Linear regression & correlationMedium
xy126221321417518xy242025

The table gives 5 paired observations of x = car age in years and y = price in k€. Their sums are given. Find Pearson's correlation coefficient r from these sums.

n = 5Σx = 15Σy = 103Σx² = 55Σy² = 2171Σxy = 289
r =
12.Linear regression & correlationHard
xy178720106231132614129150xy202530100120140x = 24.5

The table gives 5 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 = 5x = 24.5
ŷ =

3Solutions

1.Descriptive statistics

Answer: s = 13.5

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 = 304/7 = 43.43

Squared deviations

Σ(x − x̄)² = (−15.43)² + (−3.429)² + (−12.43)² + 13.57² + 14.57² + 13.57² + (−10.43)² = 1094

Standard deviation

s = √(1094/6) = 13.5

Answer

s = 13.5
2.Descriptive statistics

Answer: P = 0.9332

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 = 1.5

Standard normal table

Φ(1.5) = 0.9332
P = 0.9332

Answer

P = 0.9332
3.Probability

Answer: σ = 1.241

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 = 0·0.3 + 1·0.15 + 2·0.2 + 3·0.35 = 1.6

Mean of the squares

E(X²) = 0·0.3 + 1·0.15 + 4·0.2 + 9·0.35 = 4.1

Variance

Var(X) = 4.1 − 1.6² = 1.54
σ = √1.54 = 1.241

Answer

σ = 1.241
4.Probability

Answer: P = 0.1937

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 = 2) = 45·0.12·0.98 = 45·0.01·0.4305 = 0.1937

Answer

P = 0.1937
5.Normal distribution

Answer: P = 0.8396

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 = (39.5 − 32)/6 = 1.25

Area between

P = Φ(1.25) − Φ(−1.6) = 0.8944 − 0.0548 = 0.8396

Answer

P = 0.8396
6.Normal distribution

Answer: x = 59.8 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) = 0.3

z from the table

z ≈ −0.52

Back to the original units

x = 65 − 0.52·10 = 59.8 points

Answer

x = 59.8 points
7.Confidence intervals

Answer: E = 1.958

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 = 24, t = 2.797

Margin of error

E = 2.797·3.5/√25 = 1.958

Answer

E = 1.958
8.Confidence intervals

Answer: n = 168

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 = (1.96·33/5)² = 167.3

Round up

A sample size is a whole number, and rounding down would make the margin too large.

n = 168

Answer

n = 168
9.Hypothesis tests & confidence intervals

Answer: t = 2.34

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 = 5/√81 = 5/9 = 0.5556

Test statistic

t = (101.3 − 100)/0.5556 = 2.34

Answer

t = 2.34
t0.025; df = 80 = 1.99

|t| > 1.99, so H₀ is rejected at the 5% level (two-sided).

10.Hypothesis tests & confidence intervals

Answer: n = 237

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 + 1.282)·21/4)² = 236.1

Round up

n = 237

Answer

n = 237
11.Linear regression & correlation

Answer: r = −0.9017

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 = 289 − 15·103/5 = −20
Sxx = 55 − 15²/5 = 10
Syy = 2171 − 103²/5 = 49.2

Correlation coefficient

r = −20/√(10·49.2) = −0.9017

Answer

r = −0.9017
12.Linear regression & correlation

Answer: ŷ = 127.5

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 = 115/5 = 23
ȳ = 597/5 = 119.4

Sums of squares and products

Sxx = Σ(x − x̄)² = 90
Sxy = Σ(x − x̄)(y − ȳ) = 483

Regression line

b = 483/90 = 5.367
a = 119.4 − 5.367·23 = −4.033
ŷ = −4.033 + 5.367·x

Prediction

ŷ = −4.033 + 5.367·24.5 = 127.5

Answer

ŷ = 127.5

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