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
24262830323436384042

Find the sample standard deviation s of the data: 24, 36, 27, 24, 31, 42, 24.

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

A quantity is normally distributed with mean 170 and standard deviation 8. What proportion P of values is below 186?

μ = 170σ = 8x = 186
P =
3.ProbabilityMedium
xp(x)10.320.1530.240.35μ = 2.6

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

x: 1, 2, 3, 4p(x): 0.3, 0.15, 0.2, 0.35
σ =
4.ProbabilityHard
n = 6, p = 0.7: P(X = 1) = ?k0123456

A test has 6 multiple-choice questions. A student who guesses gets each one right with probability 0.7, independently. What is the probability P of exactly 1 correct answers?

n = 6p = 0.7k = 1
P =
5.Normal distributionMedium
82103.75P = ?

The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. What fraction lies between 82 and 103.75 points? Give P as a decimal.

μ = 100σ = 15x1 = 82x2 = 103.75
P =
6.Normal distributionHard
0.02 mm0D − d?

Hole diameters D are normal with mean 25.02 mm and standard deviation 0.015 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 = 25.02 mm, σ = 0.015 mmshaft: μ = 25 mm, σ = 0.01 mm
P =
7.Confidence intervalsMedium
μ₁ − μ₂0L = ?Ux̄₁ − x̄₂E90%

The battery life of two phone models is compared. Model 1: 40 phones, mean 61.5 h, known standard deviation 5 h. Model 2: 40 phones, mean 65 h, known standard deviation 13 h. Find the lower limit of the 90% confidence interval for μ₁ − μ₂.

n₁ = 40x̄₁ = 61.5 hσ₁ = 5 hn₂ = 40x̄₂ = 65 hσ₂ = 13 h90%
lower limit = h
8.Confidence intervalsHard
p ≈ 0.15, n = ?p̂ − Ep̂ + Ep̂E = 0.0390%

A pilot study suggests that about 0.15 of products are defective. How many products n must be inspected so that a 90% confidence interval for the proportion has a margin of error of at most 0.03?

p ≈ 0.15E = 0.0390%
n =
9.Hypothesis tests & confidence intervalsMedium
−1.88p = ?

The mean of a population is claimed to be 50; the population standard deviation is known to be 18. A random sample of 144 observations has a mean of 47.18. Test H₀: μ = 50 against H₁: μ < 50. Find the p-value p.

μ₀ = 50σ = 18n = 144x̄ = 47.18H₁: μ < μ₀
p =
10.Hypothesis tests & confidence intervalsHard
H₀H₁μ₀x̄cμ₀ + 1.5α = 0.01βσ = 11, power 80%, n = ?

A one-sided z test at α = 0.01 must detect a shift of the mean of 1.5 units with a power of 80%. The standard deviation is 11. Find the minimum sample size n.

α = 0.01power = 80%δ = 1.5σ = 11zα = 2.326zβ = 0.842
n =
11.Linear regression & correlationMedium
xy264106138191023xy5101020

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
xy122224322416514614xy246152025x = 2.5

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. A new observation (2.5, 25) is made. Find its residual e = y − ŷ.

n = 6(2.5, 25)
e =

3Solutions

1.Descriptive statistics

Answer: s = 7.041

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 = 208/7 = 29.71

Squared deviations

Σ(x − x̄)² = (−5.714)² + 6.286² + (−2.714)² + (−5.714)² + 1.286² + 12.29² + (−5.714)² = 297.4

Standard deviation

s = √(297.4/6) = 7.041

Answer

s = 7.041
2.Descriptive statistics

Answer: P = 0.9772

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 = 2

Standard normal table

Φ(2) = 0.9772
P = 0.9772

Answer

P = 0.9772
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 = 1·0.3 + 2·0.15 + 3·0.2 + 4·0.35 = 2.6

Mean of the squares

E(X²) = 1·0.3 + 4·0.15 + 9·0.2 + 16·0.35 = 8.3

Variance

Var(X) = 8.3 − 2.6² = 1.54
σ = √1.54 = 1.241

Answer

σ = 1.241
4.Probability

Answer: P = 0.01021

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) = 6·0.71·0.35 = 6·0.7·0.00243 = 0.01021

Answer

P = 0.01021
5.Normal distribution

Answer: P = 0.4836

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 = (82 − 100)/15 = −1.2
z₂ = (103.75 − 100)/15 = 0.25

Area between

P = Φ(0.25) − Φ(−1.2) = 0.5987 − 0.1151 = 0.4836

Answer

P = 0.4836
6.Normal distribution

Answer: P = 0.8665

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 = 25.02 − 25 = 0.02 mm, σ = 0.01803 mm

Standardise

z = (0 − 0.02)/0.01803 = −1.11

Probability

P = 1 − Φ(−1.11) = 0.8665

Answer

P = 0.8665
7.Confidence intervals

Answer: lower limit = −7.123 h

Interval for a difference: estimate ± critical value × standard error. For two independent means with known σ: SE = √(σ₁²/n₁ + σ₂²/n₂) and z. With equal unknown variances: pooled s_p and t with n₁ + n₂ − 2 degrees of freedom. For paired data, work with the differences: d̄ ± t·s_d/√n.

Point estimate

x̄₁ − x̄₂ = 61.5 − 65 = −3.5

Standard error

SE = √(5²/40 + 13²/40) = 2.202

Margin of error

E = 1.645·2.202 = 3.623

Interval

−3.5 ± 3.623 → [−7.123, 0.1227]

Answer

lower limit = −7.123 h

The interval contains 0, so the data do not show a significant difference.

8.Confidence intervals

Answer: n = 384

Confidence interval for a proportion: p̂ ± z·√(p̂(1 − p̂)/n), where p̂ = x/n. To reach a margin E you need n ≥ z²·p(1 − p)/E², rounded up; without a prior estimate use p = 0.5, which gives the largest n.

Solve for n

n = 1.645²·0.15·0.85/0.03² = 383.4

Round up

n = 384

Answer

n = 384
9.Hypothesis tests & confidence intervals

Answer: p = 0.03005

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.

Test statistic

SE = σ/√n = 18/√144 = 1.5
z = (47.18 − 50)/1.5 = −1.88

Tail area

Φ(1.88) = 0.9699
1 − Φ(1.88) = 0.0301

p-value

p = 0.03005

Answer

p = 0.03005

p ≤ 0.05, so H₀ is rejected at the 5% level.

10.Hypothesis tests & confidence intervals

Answer: n = 540

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 + 0.842)·11/1.5)² = 539.7

Round up

n = 540

Answer

n = 540
11.Linear regression & correlation

Answer: r² = 0.9898

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 = 71/5 = 14.2

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 86
Syy = Σ(y − ȳ)² = 186.8

Coefficient of determination

r² = 86²/(40·186.8) = 0.9898

Answer

r² = 0.9898

About 98.98% of the variation in y is explained by the line.

12.Linear regression & correlation

Answer: e = 4.162

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 = 112/6 = 18.67

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −38

Regression line

b = −38/17.5 = −2.171
a = 18.67 − (−2.171)·3.5 = 26.27
ŷ = 26.27 − 2.171·x

Predicted value

ŷ = 26.27 + (−2.171)·2.5 = 20.84

Residual

e = 25 − 20.84 = 4.162

Answer

e = 4.162

The point lies above the line.

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