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
1015202530354045

Find the sample standard deviation s of the data: 41, 47, 34, 27, 10.

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

Exam scores are normally distributed with mean 70 and standard deviation 8. Find the z-score z of a score of 86.

μ = 70σ = 8x = 86
z =
3.ProbabilityMedium
red: 6blue: 6

A bag holds 6 red and 6 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?

6 / 6
P =
4.ProbabilityHard
xp(x)10.1520.2530.3540.250.05μ

X has the distribution shown, and Y = 10X + 1. Find the standard deviation σY of Y.

x: 1, 2, 3, 4, 5p(x): 0.15, 0.25, 0.35, 0.2, 0.05Y = 10X + 1
σY =
5.Normal distributionMedium
3242.8P = ?

The delivery times are normally distributed with mean 32 min and standard deviation 6 min. What fraction lies between 32 and 42.8 min? Give P as a decimal.

μ = 32σ = 6x1 = 32x2 = 42.8
P =
6.Normal distributionHard
0.05 mm0D − d?

Hole diameters D are normal with mean 25.05 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 = 25.05 mm, σ = 0.02 mmshaft: μ = 25 mm, σ = 0.015 mm
P =
7.Confidence intervalsMedium
n = 12, s = 4LUx̄ = 94E = ?95%

A sample of 12 items has mean 94 and standard deviation s = 4. The population is roughly normal. Find the margin of error E of a 95% confidence interval for the mean.

n = 12x̄ = 94s = 495%t(11) = 2.201
E =
8.Confidence intervalsHard
n = 21σ = ?Us = 2.595%

The diameters of 21 machined shafts have a sample standard deviation of 2.5 µm. Assuming the diameters are normally distributed, find the lower limit σ of the 95% confidence interval for the population standard deviation. With df = 20, 2.5% of the χ² distribution lies below 9.591 and 2.5% lies above 34.17.

n = 21s = 2.5 µmχ² = 9.591, 34.17
σ = µm
9.Hypothesis tests & confidence intervalsMedium
−2.1312.131t = ?df = 15

16 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 −1.49 and the standard deviation of the differences is 4.6. Find the paired test statistic t for H₀: μd = 0.

n = 16d̄ = −1.49sd = 4.6
t =
10.Hypothesis tests & confidence intervalsHard
H₀H₁μ₀x̄cμ₀ + 2α = 0.01βσ = 5, power 90%, n = ?

A one-sided z test at α = 0.01 must detect a shift of the mean of 2 units with a power of 90%. The standard deviation is 5. Find the minimum sample size n.

α = 0.01power = 90%δ = 2σ = 5zα = 2.326zβ = 1.282
n =
11.Linear regression & correlationMedium
xy2104166248331043xy5102040

The table gives 5 paired observations of x = load in kN and y = extension in mm. Find the coefficient of determination r² of the least-squares line.

n = 5
r² =
12.Linear regression & correlationHard
xy129229328425523621xy24625ŷ = 28

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. For which value x does the line predict ŷ = 28?

n = 6ŷ = 28
x =

3Solutions

1.Descriptive statistics

Answer: s = 14.31

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 = 159/5 = 31.8

Squared deviations

Σ(x − x̄)² = 9.2² + 15.2² + 2.2² + (−4.8)² + (−21.8)² = 818.8

Standard deviation

s = √(818.8/4) = 14.31

Answer

s = 14.31
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 = (86 − 70)/8 = 2

Answer

z = 2
3.Probability

Answer: P = 0.2273

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.

First and second draw

P = 6/12 · 5/11 = 30/132 = 0.2273

Answer

P = 0.2273
4.Probability

Answer: σY = 10.9

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.

Spread of X

μ = 2.75, Var(X) = 1.188, σX = 1.09

Linear change

σY = |10|·1.09 = 10.9

Answer

σY = 10.9

The mean does move: E(Y) = 10·2.75 + 1 = 28.5.

5.Normal distribution

Answer: P = 0.4641

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 = (32 − 32)/6 = 0
z₂ = (42.8 − 32)/6 = 1.8

Area between

P = Φ(1.8) − Φ(0) = 0.9641 − 0.5 = 0.4641

Answer

P = 0.4641
6.Normal distribution

Answer: P = 0.9772

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 = 25.05 − 25 = 0.05 mm, σ = 0.025 mm

Standardise

z = (0 − 0.05)/0.025 = −2

Probability

P = 1 − Φ(−2) = 0.9772

Answer

P = 0.9772
7.Confidence intervals

Answer: E = 2.541

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 = 11, t = 2.201

Margin of error

E = 2.201·4/√12 = 2.541

Answer

E = 2.541
8.Confidence intervals

Answer: σ = 1.913 µm

Interval for a variance: (n − 1)s²/σ² has a χ² distribution with n − 1 degrees of freedom. The 95% interval for σ² runs from (n − 1)s² divided by the upper χ² value (2.5% above it) to (n − 1)s² divided by the lower χ² value (2.5% below it); the interval for σ is the square root of both limits. It is not symmetric around s.

Sum of squares

(n − 1)s² = 20·2.5² = 125

Limits for σ²

125/34.17 = 3.658
125/9.591 = 13.03

Limit for σ

σ = √3.658 = 1.913 µm

Answer

σ = 1.913 µm

95% interval for σ: [1.913, 3.61] µm.

9.Hypothesis tests & confidence intervals

Answer: t = −1.296

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 = 4.6/√16 = 1.15

Test statistic

t = −1.49/1.15 = −1.296

Answer

t = −1.296

With df = 15, the two-sided 5% critical value is 2.131; |t| is smaller, so H₀ is not rejected.

10.Hypothesis tests & confidence intervals

Answer: n = 82

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 = ((2.326 + 1.282)·5/2)² = 81.36

Round up

n = 82

Answer

n = 82
11.Linear regression & correlation

Answer: r² = 0.9915

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 = 30/5 = 6
ȳ = 126/5 = 25.2

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 166
Syy = Σ(y − ȳ)² = 694.8

Coefficient of determination

r² = 166²/(40·694.8) = 0.9915

Answer

r² = 0.9915

About 99.15% of the variation in y is explained by the line.

12.Linear regression & correlation

Answer: x = 2.257

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 = 155/6 = 25.83

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −30.5

Regression line

b = −30.5/17.5 = −1.743
a = 25.83 − (−1.743)·3.5 = 31.93
ŷ = 31.93 − 1.743·x

Solve for x

x = (28 − 31.93)/(−1.743) = 2.257

Answer

x = 2.257

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