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
15202530354045505560

Find the sample standard deviation s of the data: 17, 14, 24, 60, 18, 17.

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

A quantity is normally distributed with mean 500 and standard deviation 40. What proportion P of values is below 560?

μ = 500σ = 40x = 560
P =
3.ProbabilityMedium
n = 6k = 4N = ?

In how many ways N can a committee of 4 people be chosen from a group of 6? The order does not matter.

n = 6k = 4
N =
4.ProbabilityHard
xp(x)10.0520.130.340.3550.2μ

X has the distribution shown, and Y = 2X + 20. Find the standard deviation σY of Y.

x: 1, 2, 3, 4, 5p(x): 0.05, 0.1, 0.3, 0.35, 0.2Y = 2X + 20
σY =
5.Normal distributionMedium
−3−2−10123zshaded area to the right of z = 0.0526

Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.0526. Find z to two decimals.

A = 0.0526
z =
6.Normal distributionHard
0.05 mm0D − d?

Hole diameters D are normal with mean 25.05 mm and standard deviation 0.01 mm. Shaft diameters d are independent and normal with mean 25 mm and standard deviation 0.02 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 = 0.01 mmshaft: μ = 25 mm, σ = 0.02 mm
P =
7.Confidence intervalsMedium
n = 16, s = 5.5L = ?Ux̄ = 70.5E99%

A sample of 16 items has mean 70.5 and standard deviation s = 5.5. The population is roughly normal. Find the lower limit of the 99% confidence interval for the mean.

n = 16x̄ = 70.5s = 5.599%t(15) = 2.947
lower limit =
8.Confidence intervalsHard
σ = 20, n = ?x̄ − Ex̄ + Ex̄E = 599%

The population standard deviation is 20. 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 5?

σ = 20E = 599%
n =
9.Hypothesis tests & confidence intervalsMedium
−1.961.96z = ?

Two production lines make the same part. Line 1: a sample of 80 parts has mean length 102 mm, and the line's known standard deviation is 14 mm. Line 2: a sample of 50 parts has mean 96.7 mm, known standard deviation 10 mm. Find the test statistic z for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (line 1 minus line 2).

n₁ = 80x̄₁ = 102 mmσ₁ = 14 mmn₂ = 50x̄₂ = 96.7 mmσ₂ = 10 mm
z =
10.Hypothesis tests & confidence intervalsHard
z −1.9601.96

A sample of 100 measurements has mean 47.4. The population standard deviation is 20. Find the upper limit U of the 95% confidence interval for the mean.

x̄ = 47.4σ = 20n = 10095%
U =
11.Linear regression & correlationMedium
xy129227321419521616xy24615202530

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
xy127226331438542648xy246304050ŷ = 44

The table gives 6 paired observations of x = advertising spend in k€ and y = sales in k€. Use the least-squares line ŷ = a + b·x. For which value x does the line predict ŷ = 44?

n = 6ŷ = 44
x =

3Solutions

1.Descriptive statistics

Answer: s = 17.46

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 = 150/6 = 25

Squared deviations

Σ(x − x̄)² = (−8)² + (−11)² + (−1)² + 35² + (−7)² + (−8)² = 1524

Standard deviation

s = √(1524/5) = 17.46

Answer

s = 17.46
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: N = 15

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.

Combinations

C(6, 4) = (6·5·4·3)/(4·3·2·1) = 15

Answer

N = 15
4.Probability

Answer: σY = 2.142

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 = 3.55, Var(X) = 1.148, σX = 1.071

Linear change

σY = |2|·1.071 = 2.142

Answer

σY = 2.142

The mean does move: E(Y) = 2·3.55 + 20 = 27.1.

5.Normal distribution

Answer: z = 1.62

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 − 0.0526 = 0.9474

Read the z-table backwards

Find 0.9474 inside the table and read off z.

z = 1.62

Answer

z = 1.62
6.Normal distribution

Answer: P = 0.9875

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 = 0.02236 mm

Standardise

z = (0 − 0.05)/0.02236 = −2.24

Probability

P = 1 − Φ(−2.24) = 0.9875

Answer

P = 0.9875
7.Confidence intervals

Answer: lower limit = 66.45

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 = 15, t = 2.947

Margin of error

E = 2.947·5.5/√16 = 4.052

Interval

70.5 ± 4.052 → [66.45, 74.55]

Answer

lower limit = 66.45
8.Confidence intervals

Answer: n = 107

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 = (2.576·20/5)² = 106.2

Round up

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

n = 107

Answer

n = 107
9.Hypothesis tests & confidence intervals

Answer: z = 2.512

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.

Standard error of the difference

SE = √(14²/80 + 10²/50) = √4.45 = 2.11

Test statistic

z = (102 − 96.7)/2.11 = 2.512

Answer

z = 2.512

|z| > 1.96, so H₀ is rejected at the 5% level (two-sided).

10.Hypothesis tests & confidence intervals

Answer: U = 51.32

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).

Margin of error

E = 1.96·20/√100 = 3.92

Interval limit

U = 47.4 + 3.92 = 51.32

Answer

U = 51.32
11.Linear regression & correlation

Answer: r = −0.9242

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 = 133/6 = 22.17

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −42.5
Syy = Σ(y − ȳ)² = 120.8

Correlation coefficient

r = −42.5/√(17.5·120.8) = −0.9242

Answer

r = −0.9242
12.Linear regression & correlation

Answer: x = 5.396

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 = 212/6 = 35.33

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = 80

Regression line

b = 80/17.5 = 4.571
a = 35.33 − 4.571·3.5 = 19.33
ŷ = 19.33 + 4.571·x

Solve for x

x = (44 − 19.33)/4.571 = 5.396

Answer

x = 5.396

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