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
40424446485052545658

Find the sample standard deviation s of the data: 39, 43, 43, 58, 41, 41.

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

A quantity is normally distributed with mean 70 and standard deviation 20. What proportion P of values is below 95?

μ = 70σ = 20x = 95
P =
3.ProbabilityMedium
n = 13k = 3N = ?

In how many ways N can a committee of 3 people be chosen from a group of 13? The order does not matter.

n = 13k = 3
N =
4.ProbabilityHard
€p(x)−20.780.25180.05

A game costs €2 to play. You win €10 with probability 0.25 and €20 with probability 0.05; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.

stake = €2€10: p = 0.25€20: p = 0.05
E(G) = €
5.Normal distributionMedium
1020 kg957.6 kgT = X₁ + … + X₁₂?

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 12 of them exceeds 957.6 kg. Round z to two decimals.

μ = 85 kgσ = 12 kgn = 12> 957.6 kg
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 0.02 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 = 0.02 mmshaft: μ = 25 mm, σ = 0.02 mm
P =
7.Confidence intervalsMedium
n = 11, sd​ = 3.90L = ?Ud̄ = 9.5E99%

The blood pressure of 11 patients is measured before and after a treatment. The differences d = before − after (mmHg) have mean 9.5 and standard deviation 3.9. Find the lower limit of the 99% confidence interval for the mean difference.

n = 11d̄ = 9.5sd = 3.999%t(10) = 3.169
lower limit = mmHg
8.Confidence intervalsHard
μ₁ − μ₂, t(12) = 1.7820L = ?Ux̄₁ − x̄₂E90%

Two fertilisers are tested. Plot group 1: 7 plots, mean yield 38 kg, standard deviation 6.5 kg. Plot group 2: 7 plots, mean 31.5 kg, standard deviation 7 kg. Assume equal variances (pooled t, df = n₁ + n₂ − 2). Find the lower limit of the 90% confidence interval for μ₁ − μ₂.

n₁ = 7x̄₁ = 38s₁ = 6.5n₂ = 7x̄₂ = 31.5s₂ = 790%t(12) = 1.782
lower limit = kg
9.Hypothesis tests & confidence intervalsMedium
0.94p = ?

The mean of a population is claimed to be 500; the population standard deviation is known to be 24. A random sample of 144 observations has a mean of 501.88. Test H₀: μ = 500 against H₁: μ > 500. Find the p-value p.

μ₀ = 500σ = 24n = 144x̄ = 501.88H₁: μ > μ₀
p =
10.Hypothesis tests & confidence intervalsHard
H₀H₁μ₀x̄cμ₀ + 3α = 0.05βσ = 20, power 95%, n = ?

A one-sided z test at α = 0.05 must detect a shift of the mean of 3 units with a power of 95%. The standard deviation is 20. Find the minimum sample size n.

α = 0.05power = 95%δ = 3σ = 20zα = 1.645zβ = 1.645
n =
11.Linear regression & correlationMedium
xy152255366472578673xy24650607080

The table gives 6 paired observations of x = hours studied and y = exam score. Their sums are given. Find Pearson's correlation coefficient r from these sums.

n = 6Σx = 21Σy = 396Σx² = 91Σy² = 26682Σxy = 1476
r =
12.Linear regression & correlationHard
xy264126168211029xy510102030x = 9

The table gives 5 paired observations of x = load in kN and y = extension in mm. Use the least-squares line ŷ = a + b·x. A new observation (9, 23) is made. Find its residual e = y − ŷ.

n = 5(9, 23)
e =

3Solutions

1.Descriptive statistics

Answer: s = 6.94

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 = 265/6 = 44.17

Squared deviations

Σ(x − x̄)² = (−5.167)² + (−1.167)² + (−1.167)² + 13.83² + (−3.167)² + (−3.167)² = 240.8

Standard deviation

s = √(240.8/5) = 6.94

Answer

s = 6.94
2.Descriptive statistics

Answer: P = 0.8944

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 = 1.25

Standard normal table

Φ(1.25) = 0.8944
P = 0.8944

Answer

P = 0.8944
3.Probability

Answer: N = 286

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(13, 3) = (13·12·11)/(3·2·1) = 286

Answer

N = 286
4.Probability

Answer: E(G) = 1.5 €

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.

Net outcomes

G: 8 (p = 0.25), 18 (p = 0.05), −2 (p = 0.7)

Expected value

E(G) = 8·0.25 + 18·0.05 − 2·0.7 = 1.5 €

Answer

E(G) = 1.5 €

A positive value means you win money on average.

5.Normal distribution

Answer: P = 0.9332

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 = 1020 kg, σT = 12·√12 = 41.57 kg

Standardise

z = (957.6 − 1020)/41.57 = −1.5

Probability

P = 1 − Φ(−1.5) = 0.9332

Answer

P = 0.9332
6.Normal distribution

Answer: P = 0.9616

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 = 0.02828 mm

Standardise

z = (0 − 0.05)/0.02828 = −1.77

Probability

P = 1 − Φ(−1.77) = 0.9616

Answer

P = 0.9616
7.Confidence intervals

Answer: lower limit = 5.774 mmHg

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.

Critical value

df = 10, t = 3.169

Margin of error

E = 3.169·3.9/√11 = 3.726

Interval

9.5 ± 3.726 → [5.774, 13.23]

Answer

lower limit = 5.774 mmHg
8.Confidence intervals

Answer: lower limit = 0.06609 kg

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.

Pooled standard deviation

sp² = (6·6.5² + 6·7²)/12 = 45.63
sp = 6.755

Margin of error

E = 1.782·6.755·√(1/7 + 1/7) = 6.434

Interval

6.5 ± 6.434 → [0.06609, 12.93]

Answer

lower limit = 0.06609 kg
9.Hypothesis tests & confidence intervals

Answer: p = 0.1736

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 = 24/√144 = 2
z = (501.88 − 500)/2 = 0.94

Tail area

Φ(0.94) = 0.8264
1 − Φ(0.94) = 0.1736

p-value

p = 0.1736

Answer

p = 0.1736

p > 0.05, so H₀ is not rejected at the 5% level.

10.Hypothesis tests & confidence intervals

Answer: n = 482

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 = ((1.645 + 1.645)·20/3)² = 481.1

Round up

n = 482

Answer

n = 482
11.Linear regression & correlation

Answer: r = 0.9207

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 = 1476 − 21·396/6 = 90
Sxx = 91 − 21²/6 = 17.5
Syy = 26682 − 396²/6 = 546

Correlation coefficient

r = 90/√(17.5·546) = 0.9207

Answer

r = 0.9207
12.Linear regression & correlation

Answer: e = −2.05

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 = 84/5 = 16.8

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 110

Regression line

b = 110/40 = 2.75
a = 16.8 − 2.75·6 = 0.3
ŷ = 0.3 + 2.75·x

Predicted value

ŷ = 0.3 + 2.75·9 = 25.05

Residual

e = 23 − 25.05 = −2.05

Answer

e = −2.05

The point lies below the line.

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