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
101520253035404550

Find the sample standard deviation s of the data: 52, 24, 10, 22, 53, 14.

n = 6
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 below 179?

μ = 170σ = 15x = 179
P =
3.ProbabilityMedium
red: 7blue: 5

A bag holds 7 red and 5 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?

7 / 5
P =
4.ProbabilityHard
xp(x)100.35200.1300.35400.2μ

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

x: 10, 20, 30, 40p(x): 0.35, 0.1, 0.35, 0.2Y = 2X + 20
σY =
5.Normal distributionMedium
200 mm200.5 mmX + Y?

X is normal with mean 120 mm and standard deviation 0.3 mm; Y is independent and normal with mean 80 mm and standard deviation 0.4 mm. Find the probability P that X + Y is more than 200.5 mm. Round z to two decimals.

μX = 120, σX = 0.3μY = 80, σY = 0.4X + Y > 200.5 mm
P =
6.Normal distributionHard
235240245250255260265gbottom 40%

The bag weights are normally distributed with mean 250 g and standard deviation 5 g. Drag the cut-off so the shaded bottom 40% is to the left of it. Find the cut-off x.

μ = 250 gσ = 5 g40%
x = g
7.Confidence intervalsMedium
n = 9, sd​ = 2.10LU = ?d̄ = −7.4E90%

The blood pressure of 9 patients is measured before and after a treatment. The differences d = before − after (mmHg) have mean −7.4 and standard deviation 2.1. Find the upper limit of the 90% confidence interval for the mean difference.

n = 9d̄ = −7.4sd = 2.190%t(8) = 1.86
upper limit = mmHg
8.Confidence intervalsHard
σ = 18, n = ?x̄ − Ex̄ + Ex̄E = 290%

The population standard deviation is 18. How large a sample n do you need so that a 90% confidence interval for the mean has a margin of error of at most 2?

σ = 18E = 290%
n =
9.Hypothesis tests & confidence intervalsMedium
t −2.06402.064

A sample of 25 students has a mean score of 70.04 and a sample standard deviation of 8. Find the test statistic t for H₀: μ = 75.

μ₀ = 75x̄ = 70.04s = 8n = 25
t =
10.Hypothesis tests & confidence intervalsHard
H₀H₁μ₀x̄cμ₀ + 4α = 0.05βσ = 30, power 90%, n = ?

A one-sided z test at α = 0.05 must detect a shift of the mean of 4 units with a power of 90%. The standard deviation is 30. Find the minimum sample size n.

α = 0.05power = 90%δ = 4σ = 30zα = 1.645zβ = 1.282
n =
11.Linear regression & correlationMedium
xy248344446549655766xy2465060

The table gives 6 paired observations of x = hours studied and y = exam score. Find the coefficient of determination r² of the least-squares line.

n = 6
r² =
12.Linear regression & correlationHard
xy129226335432535640xy24625303540x = 1.5

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. A new observation (1.5, 24) is made. Find its residual e = y − ŷ.

n = 6(1.5, 24)
e =

3Solutions

1.Descriptive statistics

Answer: s = 18.79

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 = 175/6 = 29.17

Squared deviations

Σ(x − x̄)² = 22.83² + (−5.167)² + (−19.17)² + (−7.167)² + 23.83² + (−15.17)² = 1765

Standard deviation

s = √(1765/5) = 18.79

Answer

s = 18.79
2.Descriptive statistics

Answer: P = 0.7257

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 = 0.6

Standard normal table

Φ(0.6) = 0.7257
P = 0.7257

Answer

P = 0.7257
3.Probability

Answer: P = 0.3182

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 = 7/12 · 6/11 = 42/132 = 0.3182

Answer

P = 0.3182
4.Probability

Answer: σY = 23.15

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 = 24, Var(X) = 134, σX = 11.58

Linear change

σY = |2|·11.58 = 23.15

Answer

σY = 23.15

The mean does move: E(Y) = 2·24 + 20 = 68.

5.Normal distribution

Answer: P = 0.1587

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 sum

μ = 200 mm, σ = √(0.3² + 0.4²) = 0.5 mm

Standardise

z = (200.5 − 200)/0.5 = 1

Upper tail

P = 1 − Φ(1) = 1 − 0.8413 = 0.1587

Answer

P = 0.1587
6.Normal distribution

Answer: x = 248.8 g

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) = 0.4

z from the table

z ≈ −0.25

Back to the original units

x = 250 − 0.25·5 = 248.8 g

Answer

x = 248.8 g
7.Confidence intervals

Answer: upper limit = −6.098 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 = 8, t = 1.86

Margin of error

E = 1.86·2.1/√9 = 1.302

Interval

−7.4 ± 1.302 → [−8.702, −6.098]

Answer

upper limit = −6.098 mmHg
8.Confidence intervals

Answer: n = 220

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 = (1.645·18/2)² = 219.2

Round up

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

n = 220

Answer

n = 220
9.Hypothesis tests & confidence intervals

Answer: t = −3.1

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 = 8/√25 = 8/5 = 1.6

Test statistic

t = (70.04 − 75)/1.6 = −3.1

Answer

t = −3.1
t0.025; df = 24 = 2.064

|t| > 2.064, so H₀ is rejected at the 5% level (two-sided).

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 + 1.282)·30/4)² = 481.9

Round up

n = 482

Answer

n = 482
11.Linear regression & correlation

Answer: r² = 0.6929

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 = 27/6 = 4.5
ȳ = 308/6 = 51.33

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = 63
Syy = Σ(y − ȳ)² = 327.3

Coefficient of determination

r² = 63²/(17.5·327.3) = 0.6929

Answer

r² = 0.6929

About 69.29% of the variation in y is explained by the line.

12.Linear regression & correlation

Answer: e = −4.319

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 = 197/6 = 32.83

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = 39.5

Regression line

b = 39.5/17.5 = 2.257
a = 32.83 − 2.257·3.5 = 24.93
ŷ = 24.93 + 2.257·x

Predicted value

ŷ = 24.93 + 2.257·1.5 = 28.32

Residual

e = 24 − 28.32 = −4.319

Answer

e = −4.319

The point lies below the line.

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