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
2025303540

Find the sample standard deviation s of the data: 28, 16, 43, 32, 25.

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

A quantity is normally distributed with mean 170 and standard deviation 15. What proportion P of values is above 158?

μ = 170σ = 15x = 158
P =
3.ProbabilityMedium
red: 7blue: 3

A bag holds 7 red and 3 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?

7 / 3
P =
4.ProbabilityHard
€p(x)−10.7310.25490.02

A game costs €1 to play. You win €2 with probability 0.25 and €50 with probability 0.02; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.

stake = €1€2: p = 0.25€50: p = 0.02
E(G) = €
5.Normal distributionMedium
78.25130.75P = ?

The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. What fraction lies between 78.25 and 130.75 points? Give P as a decimal.

μ = 100σ = 15x1 = 78.25x2 = 130.75
P =
6.Normal distributionHard
35455565758595pointsbottom 20%

The exam scores are normally distributed with mean 65 points and standard deviation 10 points. Drag the cut-off so the shaded bottom 20% is to the left of it. Find the cut-off x.

μ = 65 pointsσ = 10 points20%
x = points
7.Confidence intervalsMedium
n = 11, sd​ = 3.40LU = ?d̄ = −8.5E95%

The blood pressure of 11 patients is measured before and after a treatment. The differences d = before − after (mmHg) have mean −8.5 and standard deviation 3.4. Find the upper limit of the 95% confidence interval for the mean difference.

n = 11d̄ = −8.5sd = 3.495%t(10) = 2.228
upper limit = mmHg
8.Confidence intervalsHard
σ = 7, n = ?x̄ − Ex̄ + Ex̄E = 0.595%

The population standard deviation is 7. How large a sample n do you need so that a 95% confidence interval for the mean has a margin of error of at most 0.5?

σ = 7E = 0.595%
n =
9.Hypothesis tests & confidence intervalsMedium
t −1.9901.99

A sample of 81 students has a mean score of 94.83 and a sample standard deviation of 15. Find the test statistic t for H₀: μ = 100.

μ₀ = 100x̄ = 94.83s = 15n = 81
t =
10.Hypothesis tests & confidence intervalsHard
z −1.9601.96

A sample of 16 measurements has mean 97.5. The population standard deviation is 10. Find the upper limit U of the 95% confidence interval for the mean.

x̄ = 97.5σ = 10n = 1695%
U =
11.Linear regression & correlationMedium
xy12221831641351068xy246101520

The table gives 6 paired observations of x = car age in years and y = price in k€. Their sums are given. Find Pearson's correlation coefficient r from these sums.

n = 6Σx = 21Σy = 87Σx² = 91Σy² = 1397Σxy = 256
r =
12.Linear regression & correlationHard
xy1481169018112201112212324135xy15202580100120140x = 15

The table gives 6 paired observations of x = temperature in °C and y = ice creams sold. Use the least-squares line ŷ = a + b·x. A new observation (15, 91) is made. Find its residual e = y − ŷ.

n = 6(15, 91)
e =

3Solutions

1.Descriptive statistics

Answer: s = 9.884

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 = 144/5 = 28.8

Squared deviations

Σ(x − x̄)² = (−0.8)² + (−12.8)² + 14.2² + 3.2² + (−3.8)² = 390.8

Standard deviation

s = √(390.8/4) = 9.884

Answer

s = 9.884
2.Descriptive statistics

Answer: P = 0.7881

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 = −0.8

Standard normal table

Φ(−0.8) = 0.2119
P = 1 − 0.2119 = 0.7881

Answer

P = 0.7881
3.Probability

Answer: P = 0.4667

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 = 7/10 · 6/9 = 42/90 = 0.4667

Answer

P = 0.4667
4.Probability

Answer: E(G) = 0.5 €

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: 1 (p = 0.25), 49 (p = 0.02), −1 (p = 0.73)

Expected value

E(G) = 1·0.25 + 49·0.02 − 1·0.73 = 0.5 €

Answer

E(G) = 0.5 €

A positive value means you win money on average.

5.Normal distribution

Answer: P = 0.9063

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 = (78.25 − 100)/15 = −1.45
z₂ = (130.75 − 100)/15 = 2.05

Area between

P = Φ(2.05) − Φ(−1.45) = 0.9798 − 0.07353 = 0.9063

Answer

P = 0.9063
6.Normal distribution

Answer: x = 56.6 points

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) = 0.2

z from the table

z ≈ −0.84

Back to the original units

x = 65 − 0.84·10 = 56.6 points

Answer

x = 56.6 points
7.Confidence intervals

Answer: upper limit = −6.216 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 = 2.228

Margin of error

E = 2.228·3.4/√11 = 2.284

Interval

−8.5 ± 2.284 → [−10.78, −6.216]

Answer

upper limit = −6.216 mmHg
8.Confidence intervals

Answer: n = 753

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 = (1.96·7/0.5)² = 753

Round up

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

n = 753

Answer

n = 753
9.Hypothesis tests & confidence intervals

Answer: t = −3.102

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

SE = 15/√81 = 15/9 = 1.667

Test statistic

t = (94.83 − 100)/1.667 = −3.102

Answer

t = −3.102
t0.025; df = 80 = 1.99

|t| > 1.99, so H₀ is rejected at the 5% level (two-sided).

10.Hypothesis tests & confidence intervals

Answer: U = 102.4

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 = 1.96·10/√16 = 4.9

Interval limit

U = 97.5 + 4.9 = 102.4

Answer

U = 102.4
11.Linear regression & correlation

Answer: r = −0.996

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 = 256 − 21·87/6 = −48.5
Sxx = 91 − 21²/6 = 17.5
Syy = 1397 − 87²/6 = 135.5

Correlation coefficient

r = −48.5/√(17.5·135.5) = −0.996

Answer

r = −0.996
12.Linear regression & correlation

Answer: e = 3.362

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 = 114/6 = 19
ȳ = 652/6 = 108.7

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 368

Regression line

b = 368/70 = 5.257
a = 108.7 − 5.257·19 = 8.781
ŷ = 8.781 + 5.257·x

Predicted value

ŷ = 8.781 + 5.257·15 = 87.64

Residual

e = 91 − 87.64 = 3.362

Answer

e = 3.362

The point lies above the line.

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