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
15202530354045

Find the sample standard deviation s of the data: 49, 34, 35, 30, 14, 27.

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

A quantity is normally distributed with mean 500 and standard deviation 100. What proportion P of values is above 450?

μ = 500σ = 100x = 450
P =
3.ProbabilityMedium
xp(x)00.310.3520.35μ = 1.05

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

x: 0, 1, 2p(x): 0.3, 0.35, 0.35
σ =
4.ProbabilityHard
n = 10, p = 0.4: P(X = 1) = ?k012345678910

A test has 10 multiple-choice questions. A student who guesses gets each one right with probability 0.4, independently. What is the probability P of exactly 1 correct answers?

n = 10p = 0.4k = 1
P =
5.Normal distributionMedium
765 kg729 kgT = X₁ + … + X₉?

The masses of passengers with luggage are independent and normal with mean 85 kg and standard deviation 12 kg. Find the probability P that the total mass of 9 of them exceeds 729 kg. Round z to two decimals.

μ = 85 kgσ = 12 kgn = 9> 729 kg
P =
6.Normal distributionHard
380420460500540580620htop 60%

The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. Drag the cut-off so the shaded top 60% is to the right of it. Find the cut-off x.

μ = 500 hσ = 40 h60%
x = h
7.Confidence intervalsMedium
n = 21, s = 1L = ?Ux̄ = 58E90%

A sample of 21 items has mean 58 and standard deviation s = 1. The population is roughly normal. Find the lower limit of the 90% confidence interval for the mean.

n = 21x̄ = 58s = 190%t(20) = 1.725
lower limit =
8.Confidence intervalsHard
σ = 24, n = ?x̄ − Ex̄ + Ex̄E = 1.599%

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 1.5?

σ = 24E = 1.599%
n =
9.Hypothesis tests & confidence intervalsMedium
−2.2012.201t = ?df = 11

12 employees take a typing test before and after a training course. For each person the difference d = after − before (words per minute) is recorded. The mean difference is 7.37 and the standard deviation of the differences is 11.1. Find the paired test statistic t for H₀: μd = 0.

n = 12d̄ = 7.37sd = 11.1
t =
10.Hypothesis tests & confidence intervalsHard
−1.961.96z = ?

In a trial, 54 of 120 patients on treatment A recover, and 128 of 250 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.

x₁ = 54n₁ = 120x₂ = 128n₂ = 250
z =
11.Linear regression & correlationMedium
xy128226325419518xy242025

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
xy16431846205222502465xy205060ŷ = 58

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. For which value x does the line predict ŷ = 58?

n = 5ŷ = 58
x =

3Solutions

1.Descriptive statistics

Answer: s = 11.43

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 = 189/6 = 31.5

Squared deviations

Σ(x − x̄)² = 17.5² + 2.5² + 3.5² + (−1.5)² + (−17.5)² + (−4.5)² = 653.5

Standard deviation

s = √(653.5/5) = 11.43

Answer

s = 11.43
2.Descriptive statistics

Answer: P = 0.6915

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 = −0.5

Standard normal table

Φ(−0.5) = 0.3085
P = 1 − 0.3085 = 0.6915

Answer

P = 0.6915
3.Probability

Answer: σ = 0.8047

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 + 1·0.35 + 2·0.35 = 1.05

Mean of the squares

E(X²) = 0·0.3 + 1·0.35 + 4·0.35 = 1.75

Variance

Var(X) = 1.75 − 1.05² = 0.6475
σ = √0.6475 = 0.8047

Answer

σ = 0.8047
4.Probability

Answer: P = 0.04031

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 = 1) = 10·0.41·0.69 = 10·0.4·0.01008 = 0.04031

Answer

P = 0.04031
5.Normal distribution

Answer: P = 0.8413

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 total

μT = 765 kg, σT = 12·√9 = 36 kg

Standardise

z = (729 − 765)/36 = −1

Probability

P = 1 − Φ(−1) = 0.8413

Answer

P = 0.8413
6.Normal distribution

Answer: x = 490 h

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) = 0.4

z from the table

z ≈ −0.25

Back to the original units

x = 500 − 0.25·40 = 490 h

Answer

x = 490 h
7.Confidence intervals

Answer: lower limit = 57.62

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 = 20, t = 1.725

Margin of error

E = 1.725·1/√21 = 0.3764

Interval

58 ± 0.3764 → [57.62, 58.38]

Answer

lower limit = 57.62
8.Confidence intervals

Answer: n = 1699

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 = (2.576·24/1.5)² = 1699

Round up

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

n = 1699

Answer

n = 1699
9.Hypothesis tests & confidence intervals

Answer: t = 2.3

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 of the mean difference

SE = 11.1/√12 = 3.204

Test statistic

t = 7.37/3.204 = 2.3

Answer

t = 2.3

With df = 11, the two-sided 5% critical value is 2.201; |t| is larger, so H₀ is rejected.

10.Hypothesis tests & confidence intervals

Answer: z = −1.117

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.

Sample proportions

p̂₁ = 54/120 = 0.45
p̂₂ = 128/250 = 0.512
p̂ = 182/370 = 0.4919

Standard error under H₀

SE = √(0.4919·0.5081·(1/120 + 1/250)) = 0.05552

Test statistic

z = (0.45 − 0.512)/0.05552 = −1.117

Answer

z = −1.117

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

11.Linear regression & correlation

Answer: r² = 0.9251

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 = 116/5 = 23.2

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = −27
Syy = Σ(y − ȳ)² = 78.8

Coefficient of determination

r² = −27²/(10·78.8) = 0.9251

Answer

r² = 0.9251

About 92.51% of the variation in y is explained by the line.

12.Linear regression & correlation

Answer: x = 22.83

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 = 100/5 = 20
ȳ = 256/5 = 51.2

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 96

Regression line

b = 96/40 = 2.4
a = 51.2 − 2.4·20 = 3.2
ŷ = 3.2 + 2.4·x

Solve for x

x = (58 − 3.2)/2.4 = 22.83

Answer

x = 22.83

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