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: 59, 48, 20, 36, 41, 30.

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

A quantity is normally distributed with mean 500 and standard deviation 60. What proportion P of values is below 575?

μ = 500σ = 60x = 575
P =
3.ProbabilityMedium
xp(x)00.3510.1520.3530.15μ = 1.3

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

x: 0, 1, 2, 3p(x): 0.35, 0.15, 0.35, 0.15
σ =
4.ProbabilityHard
n = 8, p = 0.5: P(X ≤ 1) = ?k012345678

8 components are tested, and each one fails with probability 0.5, independently. What is the probability P that at most one fails?

n = 8p = 0.5
P =
5.Normal distributionMedium
−3−2−10123zshaded area to the right of z = 0.7881

Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.7881. Find z to two decimals.

A = 0.7881
z =
6.Normal distributionHard
380420460500540580620htop 80%

The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. Drag the cut-off so the shaded top 80% is to the right of it. Find the cut-off x.

μ = 500 hσ = 40 h80%
x = h
7.Confidence intervalsMedium
n = 13, s = 5LUx̄ = 38E = ?99%

A sample of 13 items has mean 38 and standard deviation s = 5. The population is roughly normal. Find the margin of error E of a 99% confidence interval for the mean.

n = 13x̄ = 38s = 599%t(12) = 3.055
E =
8.Confidence intervalsHard
n = 16σ = ?Us = 1.895%

The diameters of 16 machined shafts have a sample standard deviation of 1.8 µm. Assuming the diameters are normally distributed, find the lower limit σ of the 95% confidence interval for the population standard deviation. With df = 15, 2.5% of the χ² distribution lies below 6.262 and 2.5% lies above 27.488.

n = 16s = 1.8 µmχ² = 6.262, 27.488
σ = µm
9.Hypothesis tests & confidence intervalsMedium
t −1.9901.99

A sample of 81 students has a mean score of 43.11 and a sample standard deviation of 20. Find the test statistic t for H₀: μ = 50.

μ₀ = 50x̄ = 43.11s = 20n = 81
t =
10.Hypothesis tests & confidence intervalsHard
H₀H₁μ₀x̄cμ₀ + 3α = 0.01βσ = 12, power 90%, n = ?

A one-sided z test at α = 0.01 must detect a shift of the mean of 3 units with a power of 90%. The standard deviation is 12. Find the minimum sample size n.

α = 0.01power = 90%δ = 3σ = 12zα = 2.326zβ = 1.282
n =
11.Linear regression & correlationMedium
xy326426532636734838xy4683035

The table gives 6 paired observations of x = advertising spend in k€ and y = sales in k€. Find the coefficient of determination r² of the least-squares line.

n = 6
r² =
12.Linear regression & correlationHard
xy122222322419515xy241520x = 2.5

The table gives 5 paired observations of x = car age in years and y = price in k€. Use the least-squares line ŷ = a + b·x. A new observation (2.5, 17) is made. Find its residual e = y − ŷ.

n = 5(2.5, 17)
e =

3Solutions

1.Descriptive statistics

Answer: s = 13.68

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 = 234/6 = 39

Squared deviations

Σ(x − x̄)² = 20² + 9² + (−19)² + (−3)² + 2² + (−9)² = 936

Standard deviation

s = √(936/5) = 13.68

Answer

s = 13.68
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: σ = 1.1

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 = 0·0.35 + 1·0.15 + 2·0.35 + 3·0.15 = 1.3

Mean of the squares

E(X²) = 0·0.35 + 1·0.15 + 4·0.35 + 9·0.15 = 2.9

Variance

Var(X) = 2.9 − 1.3² = 1.21
σ = √1.21 = 1.1

Answer

σ = 1.1
4.Probability

Answer: P = 0.03516

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) = 0.58 = 0.003906
P(X = 1) = 8·0.5·0.57 = 0.03125

Add them

P = 0.003906 + 0.03125 = 0.03516

Answer

P = 0.03516
5.Normal distribution

Answer: z = −0.8

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 − 0.7881 = 0.2119

Read the z-table backwards

Find 0.2119 inside the table and read off z.

z = −0.80

Answer

z = −0.80
6.Normal distribution

Answer: x = 466 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) = 0.2

z from the table

z ≈ −0.84

Back to the original units

x = 500 − 0.84·40 = 466 h

Answer

x = 466 h
7.Confidence intervals

Answer: E = 4.237

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.

Critical value

df = n − 1 = 12, t = 3.055

Margin of error

E = 3.055·5/√13 = 4.237

Answer

E = 4.237
8.Confidence intervals

Answer: σ = 1.33 µm

Interval for a variance: (n − 1)s²/σ² has a χ² distribution with n − 1 degrees of freedom. The 95% interval for σ² runs from (n − 1)s² divided by the upper χ² value (2.5% above it) to (n − 1)s² divided by the lower χ² value (2.5% below it); the interval for σ is the square root of both limits. It is not symmetric around s.

Sum of squares

(n − 1)s² = 15·1.8² = 48.6

Limits for σ²

48.6/27.488 = 1.768
48.6/6.262 = 7.761

Limit for σ

σ = √1.768 = 1.33 µm

Answer

σ = 1.33 µm

95% interval for σ: [1.33, 2.786] µm.

9.Hypothesis tests & confidence intervals

Answer: t = −3.101

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 = 20/√81 = 20/9 = 2.222

Test statistic

t = (43.11 − 50)/2.222 = −3.101

Answer

t = −3.101
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: 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)·12/3)² = 208.3

Round up

n = 209

Answer

n = 209
11.Linear regression & correlation

Answer: r² = 0.8643

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 = 192/6 = 32

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = 44
Syy = Σ(y − ȳ)² = 128

Coefficient of determination

r² = 44²/(17.5·128) = 0.8643

Answer

r² = 0.8643

About 86.43% of the variation in y is explained by the line.

12.Linear regression & correlation

Answer: e = −3.85

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 = 15/5 = 3
ȳ = 100/5 = 20

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = −17

Regression line

b = −17/10 = −1.7
a = 20 − (−1.7)·3 = 25.1
ŷ = 25.1 − 1.7·x

Predicted value

ŷ = 25.1 + (−1.7)·2.5 = 20.85

Residual

e = 17 − 20.85 = −3.85

Answer

e = −3.85

The point lies below the line.

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