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
253035404550

Find the sample standard deviation s of the data: 48, 51, 48, 23, 50.

n = 5
s =
2.Descriptive statisticsHard
z −2−1012

Exam scores are normally distributed with mean 60 and standard deviation 20. Find the z-score z of a score of 100.

μ = 60σ = 20x = 100
z =
3.ProbabilityMedium
xp(x)00.1520.3540.2560.25μ = 3.2

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

x: 0, 2, 4, 6p(x): 0.15, 0.35, 0.25, 0.25
σ =
4.ProbabilityHard
n = 9, p = 0.25: P(X = 3) = ?k0123456789

A test has 9 multiple-choice questions. A student who guesses gets each one right with probability 0.25, independently. What is the probability P of exactly 3 correct answers?

n = 9p = 0.25k = 3
P =
5.Normal distributionMedium
−3−2−10123zshaded area between −z and z = 0.9729

Drag z so that the shaded area between −z and z under the standard normal curve is 0.9729. Find z to two decimals.

A = 0.9729
z =
6.Normal distributionHard
0.03 mm0D − d?

Hole diameters D are normal with mean 25.03 mm and standard deviation 0.02 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 = 25.03 mm, σ = 0.02 mmshaft: μ = 25 mm, σ = 0.01 mm
P =
7.Confidence intervalsMedium
n = 9, s = 14LUx̄ = 65E = ?99%

A sample of 9 items has mean 65 and standard deviation s = 14. The population is roughly normal. Find the margin of error E of a 99% confidence interval for the mean.

n = 9x̄ = 65s = 1499%t(8) = 3.355
E =
8.Confidence intervalsHard
σ = 24, n = ?x̄ − Ex̄ + Ex̄E = 499%

The population standard deviation is 24. How large a sample n do you need so that a 99% confidence interval for the mean has a margin of error of at most 4?

σ = 24E = 499%
n =
9.Hypothesis tests & confidence intervalsMedium
−2.1452.145t = ?df = 14

Two teaching methods are compared. Group 1: 6 students, mean score 36.5, standard deviation 8. Group 2: 10 students, mean score 30, standard deviation 4.5. Assume equal population variances and use the pooled two-sample t test. Find the test statistic t for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (group 1 minus group 2).

n₁ = 6x̄₁ = 36.5s₁ = 8n₂ = 10x̄₂ = 30s₂ = 4.5
t =
10.Hypothesis tests & confidence intervalsHard
σ = 15, n = ?x̄ − Ex̄ + Ex̄E = 390%

How large must a sample be so that the 90% confidence interval for a mean has a margin of error of at most 3? The standard deviation is 15. Find the smallest sample size n.

σ = 15E = 390%
n =
11.Linear regression & correlationMedium
xy140249353455563xy24405060

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

n = 5Σx = 15Σy = 260Σx² = 55Σy² = 13804Σxy = 832
r =
12.Linear regression & correlationHard
xy16761910022116251392814231162xy15202530100150ŷ = 124

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. For which value x does the line predict ŷ = 124?

n = 6ŷ = 124
x =

3Solutions

1.Descriptive statistics

Answer: s = 11.81

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 = 220/5 = 44

Squared deviations

Σ(x − x̄)² = 4² + 7² + 4² + (−21)² + 6² = 558

Standard deviation

s = √(558/4) = 11.81

Answer

s = 11.81
2.Descriptive statistics

Answer: z = 2

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 = (100 − 60)/20 = 2

Answer

z = 2
3.Probability

Answer: σ = 2.04

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 = 0·0.15 + 2·0.35 + 4·0.25 + 6·0.25 = 3.2

Mean of the squares

E(X²) = 0·0.15 + 4·0.35 + 16·0.25 + 36·0.25 = 14.4

Variance

Var(X) = 14.4 − 3.2² = 4.16
σ = √4.16 = 2.04

Answer

σ = 2.04
4.Probability

Answer: P = 0.2336

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 = 3) = 84·0.253·0.756 = 84·0.01563·0.178 = 0.2336

Answer

P = 0.2336
5.Normal distribution

Answer: z = 2.21

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 + 0.9729)/2 = 0.9864

Read the z-table backwards

Find 0.9864 inside the table and read off z.

z = 2.21

Answer

z = 2.21
6.Normal distribution

Answer: P = 0.9099

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 = 25.03 − 25 = 0.03 mm, σ = 0.02236 mm

Standardise

z = (0 − 0.03)/0.02236 = −1.34

Probability

P = 1 − Φ(−1.34) = 0.9099

Answer

P = 0.9099
7.Confidence intervals

Answer: E = 15.66

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 = 8, t = 3.355

Margin of error

E = 3.355·14/√9 = 15.66

Answer

E = 15.66
8.Confidence intervals

Answer: n = 239

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 = (2.576·24/4)² = 238.9

Round up

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

n = 239

Answer

n = 239
9.Hypothesis tests & confidence intervals

Answer: t = 2.102

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.

Pooled variance

sp² = (5·8² + 9·4.5²)/14 = 35.88
sp = 5.99

Standard error

SE = 5.99·√(1/6 + 1/10) = 3.093

Test statistic

t = (36.5 − 30)/3.093 = 2.102

Answer

t = 2.102

With df = 14, the two-sided 5% critical value is 2.145; |t| is smaller, so H₀ is not rejected.

10.Hypothesis tests & confidence intervals

Answer: n = 68

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 = (1.645·15/3)² = 67.65

Round up

n = 68

Answer

n = 68
11.Linear regression & correlation

Answer: r = 0.9758

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 = 832 − 15·260/5 = 52
Sxx = 55 − 15²/5 = 10
Syy = 13804 − 260²/5 = 284

Correlation coefficient

r = 52/√(10·284) = 0.9758

Answer

r = 0.9758
12.Linear regression & correlation

Answer: x = 23.77

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 = 141/6 = 23.5
ȳ = 735/6 = 122.5

Sums of squares and products

Sxx = Σ(x − x̄)² = 157.5
Sxy = Σ(x − x̄)(y − ȳ) = 868.5

Regression line

b = 868.5/157.5 = 5.514
a = 122.5 − 5.514·23.5 = −7.086
ŷ = −7.086 + 5.514·x

Solve for x

x = (124 − (−7.086))/5.514 = 23.77

Answer

x = 23.77

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