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
2530354045505560

Find the sample standard deviation s of the data: 39, 24, 60, 36, 28.

n = 5
s =
2.Descriptive statisticsHard
z −2−1012

A quantity is normally distributed with mean 100 and standard deviation 20. What proportion P of values is below 112?

μ = 100σ = 20x = 112
P =
3.ProbabilityMedium
xp(x)100.2200.25300.35400.2μ = 25.5

For the distribution shown, find the variance Var(X) of X.

x: 10, 20, 30, 40p(x): 0.2, 0.25, 0.35, 0.2
Var(X) =
4.ProbabilityHard
xp(x)00.320.2540.2560.2μ

X has the distribution shown, and Y = −2X + 10. Find the standard deviation σY of Y.

x: 0, 2, 4, 6p(x): 0.3, 0.25, 0.25, 0.2Y = −2X + 10
σY =
5.Normal distributionMedium
−3−2−10123zshaded area between −z and z = 0.8355

Drag z so that the shaded area between −z and z under the standard normal curve is 0.8355. Find z to two decimals.

A = 0.8355
z =
6.Normal distributionHard
μ = 65x = ?95%

The exam scores are normally distributed with mean 65 points and standard deviation 10 points. Find the value x below which 95% of them lie (the 95th percentile).

μ = 65σ = 10P = 95%
x = points
7.Confidence intervalsMedium
n = 12, s = 8.5LU = ?x̄ = 77E95%

A sample of 12 items has mean 77 and standard deviation s = 8.5. The population is roughly normal. Find the upper limit of the 95% confidence interval for the mean.

n = 12x̄ = 77s = 8.595%t(11) = 2.201
upper limit =
8.Confidence intervalsHard
μ₁ − μ₂, t(10) = 1.8120LUx̄₁ − x̄₂E = ?90%

Two fertilisers are tested. Plot group 1: 5 plots, mean yield 38.5 kg, standard deviation 9 kg. Plot group 2: 7 plots, mean 45.5 kg, standard deviation 9 kg. Assume equal variances (pooled t, df = n₁ + n₂ − 2). Find the margin of error E of the 90% confidence interval for μ₁ − μ₂.

n₁ = 5x̄₁ = 38.5s₁ = 9n₂ = 7x̄₂ = 45.5s₂ = 990%t(10) = 1.812
E = kg
9.Hypothesis tests & confidence intervalsMedium
t −2.13102.131

A sample of 16 students has a mean score of 91.75 and a sample standard deviation of 15. Find the test statistic t for H₀: μ = 100.

μ₀ = 100x̄ = 91.75s = 15n = 16
t =
10.Hypothesis tests & confidence intervalsHard
−1.961.96z = ?

In a trial, 75 of 150 patients on treatment A recover, and 46 of 120 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.

x₁ = 75n₁ = 150x₂ = 46n₂ = 120
z =
11.Linear regression & correlationMedium
xy2104176238341042xy51010203040

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

n = 5
r =
12.Linear regression & correlationHard
xy125221319417511xy2410152025ŷ = 20

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. For which value x does the line predict ŷ = 20?

n = 5ŷ = 20
x =

3Solutions

1.Descriptive statistics

Answer: s = 13.99

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 = 187/5 = 37.4

Squared deviations

Σ(x − x̄)² = 1.6² + (−13.4)² + 22.6² + (−1.4)² + (−9.4)² = 783.2

Standard deviation

s = √(783.2/4) = 13.99

Answer

s = 13.99
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: Var(X) = 104.8

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 = 10·0.2 + 20·0.25 + 30·0.35 + 40·0.2 = 25.5

Mean of the squares

E(X²) = 100·0.2 + 400·0.25 + 900·0.35 + 1600·0.2 = 755

Variance

Var(X) = 755 − 25.5² = 104.8

Answer

Var(X) = 104.8
4.Probability

Answer: σY = 4.432

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 = 2.7, Var(X) = 4.91, σX = 2.216

Linear change

σY = |−2|·2.216 = 4.432

Answer

σY = 4.432

The mean does move: E(Y) = −2·2.7 + 10 = 4.6.

5.Normal distribution

Answer: z = 1.39

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 + 0.8355)/2 = 0.9177

Read the z-table backwards

Find 0.9177 inside the table and read off z.

z = 1.39

Answer

z = 1.39
6.Normal distribution

Answer: x = 81.45 points

Normal distribution: convert a value x to a z-score, z = (x − μ)/σ, which counts standard deviations from the mean. A z-table (or Φ(z)) gives the area to the left of z, which is P(X < x). The area to the right is 1 − Φ(z), and the area between two values is the difference of their two areas.

z for the percentile

Φ(z) = 0.95 → z = 1.645

Back to x

x = 65 + 1.645·10 = 81.45 points

Answer

x = 81.45 points
7.Confidence intervals

Answer: upper limit = 82.4

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 = 11, t = 2.201

Margin of error

E = 2.201·8.5/√12 = 5.401

Interval

77 ± 5.401 → [71.6, 82.4]

Answer

upper limit = 82.4
8.Confidence intervals

Answer: E = 9.549 kg

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.

Pooled standard deviation

sp² = (4·9² + 6·9²)/10 = 81
sp = 9

Margin of error

E = 1.812·9·√(1/5 + 1/7) = 9.549

Answer

E = 9.549 kg
9.Hypothesis tests & confidence intervals

Answer: t = −2.2

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/√16 = 15/4 = 3.75

Test statistic

t = (91.75 − 100)/3.75 = −2.2

Answer

t = −2.2
t0.025; df = 15 = 2.131

|t| > 2.131, so H₀ is rejected at the 5% level (two-sided).

10.Hypothesis tests & confidence intervals

Answer: z = 1.915

Hypothesis test: the test statistic = (estimate − value under H₀)/(standard error under H₀). The p-value is the probability, assuming H₀ is true, of a statistic at least as extreme as the one observed: the area in one tail for a one-sided H₁, twice the tail area for a two-sided H₁. Reject H₀ when p ≤ α.

Sample proportions

p̂₁ = 75/150 = 0.5
p̂₂ = 46/120 = 0.3833
p̂ = 121/270 = 0.4481

Standard error under H₀

SE = √(0.4481·0.5519·(1/150 + 1/120)) = 0.06091

Test statistic

z = (0.5 − 0.3833)/0.06091 = 1.915

Answer

z = 1.915

|z| < 1.96, so H₀ is not rejected at the 5% level (two-sided).

11.Linear regression & correlation

Answer: r = 0.9949

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 = 126/5 = 25.2

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 162
Syy = Σ(y − ȳ)² = 662.8

Correlation coefficient

r = 162/√(40·662.8) = 0.9949

Answer

r = 0.9949
12.Linear regression & correlation

Answer: x = 2.563

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 = 93/5 = 18.6

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = −32

Regression line

b = −32/10 = −3.2
a = 18.6 − (−3.2)·3 = 28.2
ŷ = 28.2 − 3.2·x

Solve for x

x = (20 − 28.2)/(−3.2) = 2.563

Answer

x = 2.563

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