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: 52, 19, 30, 25, 16, 58, 27.

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

A quantity is normally distributed with mean 70 and standard deviation 20. What proportion P of values is below 30?

μ = 70σ = 20x = 30
P =
3.ProbabilityMedium
xp(x)00.310.3520.35μ = 1.05

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

x: 0, 1, 2p(x): 0.3, 0.35, 0.35
σ =
4.ProbabilityHard
xp(x)00.3510.3520.3μ

X has the distribution shown, and Y = −2X + 1. Find the standard deviation σY of Y.

x: 0, 1, 2p(x): 0.35, 0.35, 0.3Y = −2X + 1
σY =
5.Normal distributionMedium
−3−2−10123zshaded area to the right of z = 0.9861

Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.9861. Find z to two decimals.

A = 0.9861
z =
6.Normal distributionHard
157164171178185192199cmtop 25%

The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. Drag the cut-off so the shaded top 25% is to the right of it. Find the cut-off x.

μ = 178 cmσ = 7 cm25%
x = cm
7.Confidence intervalsMedium
n = 7, s = 13LU = ?x̄ = 112E95%

A sample of 7 items has mean 112 and standard deviation s = 13. The population is roughly normal. Find the upper limit of the 95% confidence interval for the mean.

n = 7x̄ = 112s = 1395%t(6) = 2.447
upper limit =
8.Confidence intervalsHard
n = 21σ = ?Us = 10.895%

The diameters of 21 machined shafts have a sample standard deviation of 10.8 µm. Assuming the diameters are normally distributed, find the lower limit σ of the 95% confidence interval for the population standard deviation. With df = 20, 2.5% of the χ² distribution lies below 9.591 and 2.5% lies above 34.17.

n = 21s = 10.8 µmχ² = 9.591, 34.17
σ = µm
9.Hypothesis tests & confidence intervalsMedium
t −2.0302.03

A sample of 36 students has a mean score of 46.5 and a sample standard deviation of 15. Find the test statistic t for H₀: μ = 50.

μ₀ = 50x̄ = 46.5s = 15n = 36
t =
10.Hypothesis tests & confidence intervalsHard
z −1.64501.645

A sample of 81 measurements has mean 200.8. The population standard deviation is 4. Find the upper limit U of the 90% confidence interval for the mean.

x̄ = 200.8σ = 4n = 8190%
U =
11.Linear regression & correlationMedium
xy274146208271032xy510102030

The table gives 5 paired observations of x = load in kN and y = extension in mm. Their sums are given. Find Pearson's correlation coefficient r from these sums.

n = 5Σx = 30Σy = 100Σx² = 220Σy² = 2398Σxy = 726
r =
12.Linear regression & correlationHard
xy2741261681910241224xy5101020x = 5

The table gives 6 paired observations of x = load in kN and y = extension in mm. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 5.

n = 6x = 5
ŷ =

3Solutions

1.Descriptive statistics

Answer: s = 16.22

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 = 227/7 = 32.43

Squared deviations

Σ(x − x̄)² = 19.57² + (−13.43)² + (−2.429)² + (−7.429)² + (−16.43)² + 25.57² + (−5.429)² = 1578

Standard deviation

s = √(1578/6) = 16.22

Answer

s = 16.22
2.Descriptive statistics

Answer: P = 0.02275

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 = −2

Standard normal table

Φ(−2) = 0.02275
P = 0.02275

Answer

P = 0.02275
3.Probability

Answer: σ = 0.8047

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 = 0·0.3 + 1·0.35 + 2·0.35 = 1.05

Mean of the squares

E(X²) = 0·0.3 + 1·0.35 + 4·0.35 = 1.75

Variance

Var(X) = 1.75 − 1.05² = 0.6475
σ = √0.6475 = 0.8047

Answer

σ = 0.8047
4.Probability

Answer: σY = 1.609

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 = 0.95, Var(X) = 0.6475, σX = 0.8047

Linear change

σY = |−2|·0.8047 = 1.609

Answer

σY = 1.609

The mean does move: E(Y) = −2·0.95 + 1 = −0.9.

5.Normal distribution

Answer: z = −2.2

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 − 0.9861 = 0.0139

Read the z-table backwards

Find 0.0139 inside the table and read off z.

z = −2.20

Answer

z = −2.20
6.Normal distribution

Answer: x = 182.7 cm

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) = 0.75

z from the table

z ≈ 0.67

Back to the original units

x = 178 + 0.67·7 = 182.7 cm

Answer

x = 182.7 cm
7.Confidence intervals

Answer: upper limit = 124

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 = 6, t = 2.447

Margin of error

E = 2.447·13/√7 = 12.02

Interval

112 ± 12.02 → [99.98, 124]

Answer

upper limit = 124
8.Confidence intervals

Answer: σ = 8.263 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 = 20·10.8² = 2333

Limits for σ²

2333/34.17 = 68.27
2333/9.591 = 243.2

Limit for σ

σ = √68.27 = 8.263 µm

Answer

σ = 8.263 µm

95% interval for σ: [8.263, 15.6] µm.

9.Hypothesis tests & confidence intervals

Answer: t = −1.4

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).

Standard error

SE = 15/√36 = 15/6 = 2.5

Test statistic

t = (46.5 − 50)/2.5 = −1.4

Answer

t = −1.4
t0.025; df = 35 = 2.03

|t| < 2.03, so H₀ is not rejected at the 5% level (two-sided).

10.Hypothesis tests & confidence intervals

Answer: U = 201.5

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 = 1.645·4/√81 = 0.7311

Interval limit

U = 200.8 + 0.7311 = 201.5

Answer

U = 201.5
11.Linear regression & correlation

Answer: r = 0.9986

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 = 726 − 30·100/5 = 126
Sxx = 220 − 30²/5 = 40
Syy = 2398 − 100²/5 = 398

Correlation coefficient

r = 126/√(40·398) = 0.9986

Answer

r = 0.9986
12.Linear regression & correlation

Answer: ŷ = 13.46

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 = 42/6 = 7
ȳ = 102/6 = 17

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 124

Regression line

b = 124/70 = 1.771
a = 17 − 1.771·7 = 4.6
ŷ = 4.6 + 1.771·x

Prediction

ŷ = 4.6 + 1.771·5 = 13.46

Answer

ŷ = 13.46

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