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
2025303540455055

Find the sample standard deviation s of the data: 31, 56, 18, 43, 47, 47.

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

A quantity is normally distributed with mean 170 and standard deviation 5. What proportion P of values is above 175?

μ = 170σ = 5x = 175
P =
3.ProbabilityMedium
xp(x)10.2520.330.1540.150.2μ = 2.7

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

x: 1, 2, 3, 4, 5p(x): 0.25, 0.3, 0.15, 0.1, 0.2
σ =
4.ProbabilityHard
n = 10, p = 0.2: P(X ≤ 1) = ?k012345678910

10 components are tested, and each one fails with probability 0.2, independently. What is the probability P that at most one fails?

n = 10p = 0.2
P =
5.Normal distributionMedium
−3−2−10123zshaded area to the right of z = 0.0351

Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.0351. Find z to two decimals.

A = 0.0351
z =
6.Normal distributionHard
380420460500540580620hbottom 95%

The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. Drag the cut-off so the shaded bottom 95% is to the left of it. Find the cut-off x.

μ = 500 hσ = 40 h95%
x = h
7.Confidence intervalsMedium
n = 16, sd​ = 1.30L = ?Ud̄ = 2.4E99%

The blood pressure of 16 patients is measured before and after a treatment. The differences d = before − after (mmHg) have mean 2.4 and standard deviation 1.3. Find the lower limit of the 99% confidence interval for the mean difference.

n = 16d̄ = 2.4sd = 1.399%t(15) = 2.947
lower limit = mmHg
8.Confidence intervalsHard
p = 0.5, n = ?p̂ − Ep̂ + Ep̂E = 0.0490%

Nothing is known in advance about the proportion of defective products. How many products n must be inspected so that a 90% confidence interval for the proportion has a margin of error of at most 0.04?

p = 0.5E = 0.0490%
n =
9.Hypothesis tests & confidence intervalsMedium
t −1.98401.984

A sample of 100 students has a mean score of 198.6 and a sample standard deviation of 8. Find the test statistic t for H₀: μ = 200.

μ₀ = 200x̄ = 198.6s = 8n = 100
t =
10.Hypothesis tests & confidence intervalsHard
H₀H₁μ₀x̄cμ₀ + 4α = 0.01βσ = 16, power 90%, n = ?

A one-sided z test at α = 0.01 must detect a shift of the mean of 4 units with a power of 90%. The standard deviation is 16. Find the minimum sample size n.

α = 0.01power = 90%δ = 4σ = 16zα = 2.326zβ = 1.282
n =
11.Linear regression & correlationMedium
xy294146228281039xy51010203040

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
xy356467574684788896xy4686080100x = 4.5

The table gives 6 paired observations of x = hours studied and y = exam score. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 4.5.

n = 6x = 4.5
ŷ =

3Solutions

1.Descriptive statistics

Answer: s = 13.62

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 = 242/6 = 40.33

Squared deviations

Σ(x − x̄)² = (−9.333)² + 15.67² + (−22.33)² + 2.667² + 6.667² + 6.667² = 927.3

Standard deviation

s = √(927.3/5) = 13.62

Answer

s = 13.62
2.Descriptive statistics

Answer: P = 0.1587

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

Standard normal table

Φ(1) = 0.8413
P = 1 − 0.8413 = 0.1587

Answer

P = 0.1587
3.Probability

Answer: σ = 1.453

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 = 1·0.25 + 2·0.3 + 3·0.15 + 4·0.1 + 5·0.2 = 2.7

Mean of the squares

E(X²) = 1·0.25 + 4·0.3 + 9·0.15 + 16·0.1 + 25·0.2 = 9.4

Variance

Var(X) = 9.4 − 2.7² = 2.11
σ = √2.11 = 1.453

Answer

σ = 1.453
4.Probability

Answer: P = 0.3758

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.

Zero and one failure

P(X = 0) = 0.810 = 0.1074
P(X = 1) = 10·0.2·0.89 = 0.2684

Add them

P = 0.1074 + 0.2684 = 0.3758

Answer

P = 0.3758
5.Normal distribution

Answer: z = 1.81

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 − 0.0351 = 0.9649

Read the z-table backwards

Find 0.9649 inside the table and read off z.

z = 1.81

Answer

z = 1.81
6.Normal distribution

Answer: x = 566 h

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) = 0.95

z from the table

z ≈ 1.64

Back to the original units

x = 500 + 1.64·40 = 566 h

Answer

x = 566 h
7.Confidence intervals

Answer: lower limit = 1.442 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 = 15, t = 2.947

Margin of error

E = 2.947·1.3/√16 = 0.9578

Interval

2.4 ± 0.9578 → [1.442, 3.358]

Answer

lower limit = 1.442 mmHg
8.Confidence intervals

Answer: n = 423

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 = 1.645²·0.5·0.5/0.04² = 422.8

Round up

n = 423

Answer

n = 423
9.Hypothesis tests & confidence intervals

Answer: t = −1.75

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/√100 = 8/10 = 0.8

Test statistic

t = (198.6 − 200)/0.8 = −1.75

Answer

t = −1.75
t0.025; df = 99 = 1.984

|t| < 1.984, so H₀ is not rejected at the 5% level (two-sided).

10.Hypothesis tests & confidence intervals

Answer: n = 209

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 + 1.282)·16/4)² = 208.3

Round up

n = 209

Answer

n = 209
11.Linear regression & correlation

Answer: r² = 0.9828

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 = 112/5 = 22.4

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 148
Syy = Σ(y − ȳ)² = 557.2

Coefficient of determination

r² = 148²/(40·557.2) = 0.9828

Answer

r² = 0.9828

About 98.28% of the variation in y is explained by the line.

12.Linear regression & correlation

Answer: ŷ = 69.7

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 = 33/6 = 5.5
ȳ = 465/6 = 77.5

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = 136.5

Regression line

b = 136.5/17.5 = 7.8
a = 77.5 − 7.8·5.5 = 34.6
ŷ = 34.6 + 7.8·x

Prediction

ŷ = 34.6 + 7.8·4.5 = 69.7

Answer

ŷ = 69.7

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