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
10152025303540

Find the sample standard deviation s of the data: 10, 41, 30, 43, 44, 39, 16.

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

Exam scores are normally distributed with mean 100 and standard deviation 20. Find the z-score z of a score of 140.

μ = 100σ = 20x = 140
z =
3.ProbabilityMedium
red: 3blue: 9

A bag holds 3 red and 9 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?

3 / 9
P =
4.ProbabilityHard
n = 12, p = 0.5: P(X = 4) = ?k0123456789101112

A test has 12 multiple-choice questions. A student who guesses gets each one right with probability 0.5, independently. What is the probability P of exactly 4 correct answers?

n = 12p = 0.5k = 4
P =
5.Normal distributionMedium
−3−2−10123zshaded area to the right of z = 0.0475

Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.0475. Find z to two decimals.

A = 0.0475
z =
6.Normal distributionHard
10 g0X − Y?

The mass X of a product from line 1 and Y from line 2 are independent and normal: X has mean 500 g and standard deviation 8 g, Y has mean 490 g and standard deviation 6 g. Find the probability P that X is greater than Y. Round z to two decimals.

μX = 500 g, σX = 8 gμY = 490 g, σY = 6 g
P =
7.Confidence intervalsMedium
n = 15, sd​ = 3.10LU = ?d̄ = 2.1E90%

The blood pressure of 15 patients is measured before and after a treatment. The differences d = before − after (mmHg) have mean 2.1 and standard deviation 3.1. Find the upper limit of the 90% confidence interval for the mean difference.

n = 15d̄ = 2.1sd = 3.190%t(14) = 1.761
upper limit = mmHg
8.Confidence intervalsHard
σ = 34, n = ?x̄ − Ex̄ + Ex̄E = 1.599%

The population standard deviation is 34. 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 1.5?

σ = 34E = 1.599%
n =
9.Hypothesis tests & confidence intervalsMedium
−2.1452.145t = ?df = 14

15 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 11.68 and the standard deviation of the differences is 11.9. Find the paired test statistic t for H₀: μd = 0.

n = 15d̄ = 11.68sd = 11.9
t =
10.Hypothesis tests & confidence intervalsHard
σ = 20, n = ?x̄ − Ex̄ + Ex̄E = 190%

How large must a sample be so that the 90% confidence interval for a mean has a margin of error of at most 1? The standard deviation is 20. Find the smallest sample size n.

σ = 20E = 190%
n =
11.Linear regression & correlationMedium
xy349454565664777xy4650607080

The table gives 5 paired observations of x = advertising spend in k€ and y = sales in k€. Their sums are given. Find Pearson's correlation coefficient r from these sums.

n = 5Σx = 25Σy = 309Σx² = 135Σy² = 19567Σxy = 1611
r =
12.Linear regression & correlationHard
xy156918782188241152712030141xy15202530100ŷ = 73

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. For which value x does the line predict ŷ = 73?

n = 6ŷ = 73
x =

3Solutions

1.Descriptive statistics

Answer: s = 13.78

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 = 223/7 = 31.86

Squared deviations

Σ(x − x̄)² = (−21.86)² + 9.143² + (−1.857)² + 11.14² + 12.14² + 7.143² + (−15.86)² = 1139

Standard deviation

s = √(1139/6) = 13.78

Answer

s = 13.78
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 = (140 − 100)/20 = 2

Answer

z = 2
3.Probability

Answer: P = 0.04545

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 = 3/12 · 2/11 = 6/132 = 0.04545

Answer

P = 0.04545
4.Probability

Answer: P = 0.1208

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 = 4) = 495·0.54·0.58 = 495·0.0625·0.003906 = 0.1208

Answer

P = 0.1208
5.Normal distribution

Answer: z = 1.67

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 − 0.0475 = 0.9525

Read the z-table backwards

Find 0.9525 inside the table and read off z.

z = 1.67

Answer

z = 1.67
6.Normal distribution

Answer: P = 0.8413

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 = 500 − 490 = 10 g, σ = 10 g

Standardise

z = (0 − 10)/10 = −1

Probability

P = 1 − Φ(−1) = 0.8413

Answer

P = 0.8413
7.Confidence intervals

Answer: upper limit = 3.51 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 = 14, t = 1.761

Margin of error

E = 1.761·3.1/√15 = 1.41

Interval

2.1 ± 1.41 → [0.6905, 3.51]

Answer

upper limit = 3.51 mmHg
8.Confidence intervals

Answer: n = 3410

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 = (2.576·34/1.5)² = 3409

Round up

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

n = 3410

Answer

n = 3410
9.Hypothesis tests & confidence intervals

Answer: t = 3.801

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 = 11.9/√15 = 3.073

Test statistic

t = 11.68/3.073 = 3.801

Answer

t = 3.801

With df = 14, the two-sided 5% critical value is 2.145; |t| is larger, so H₀ is rejected.

10.Hypothesis tests & confidence intervals

Answer: n = 1083

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).

Solve for n

n = (1.645·20/1)² = 1082

Round up

n = 1083

Answer

n = 1083
11.Linear regression & correlation

Answer: r = 0.9619

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 = 1611 − 25·309/5 = 66
Sxx = 135 − 25²/5 = 10
Syy = 19567 − 309²/5 = 470.8

Correlation coefficient

r = 66/√(10·470.8) = 0.9619

Answer

r = 0.9619
12.Linear regression & correlation

Answer: x = 16.6

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 = 135/6 = 22.5
ȳ = 611/6 = 101.8

Sums of squares and products

Sxx = Σ(x − x̄)² = 157.5
Sxy = Σ(x − x̄)(y − ȳ) = 769.5

Regression line

b = 769.5/157.5 = 4.886
a = 101.8 − 4.886·22.5 = −8.095
ŷ = −8.095 + 4.886·x

Solve for x

x = (73 − (−8.095))/4.886 = 16.6

Answer

x = 16.6

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