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: 47, 24, 47, 56, 42, 26, 18.

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

A quantity is normally distributed with mean 500 and standard deviation 25. What proportion P of values is below 510?

μ = 500σ = 25x = 510
P =
3.ProbabilityMedium
xp(x)00.1510.1520.3530.0540.3μ = 2.2

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

x: 0, 1, 2, 3, 4p(x): 0.15, 0.15, 0.35, 0.05, 0.3
σ =
4.ProbabilityHard
xp(x)100.25200.35300.1400.3μ

X has the distribution shown, and Y = 5X + 4. Find the standard deviation σY of Y.

x: 10, 20, 30, 40p(x): 0.25, 0.35, 0.1, 0.3Y = 5X + 4
σY =
5.Normal distributionMedium
−3−2−10123zshaded area to the right of z = 0.9032

Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.9032. Find z to two decimals.

A = 0.9032
z =
6.Normal distributionHard
557085100115130145pointsbottom 20%

The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. Drag the cut-off so the shaded bottom 20% is to the left of it. Find the cut-off x.

μ = 100 pointsσ = 15 points20%
x = points
7.Confidence intervalsMedium
n = 15, s = 1.5LUx̄ = 112E = ?90%

A sample of 15 items has mean 112 and standard deviation s = 1.5. The population is roughly normal. Find the margin of error E of a 90% confidence interval for the mean.

n = 15x̄ = 112s = 1.590%t(14) = 1.761
E =
8.Confidence intervalsHard
μ₁ − μ₂, t(29) = 2.0450LUx̄₁ − x̄₂E = ?95%

Two fertilisers are tested. Plot group 1: 13 plots, mean yield 56.5 kg, standard deviation 6 kg. Plot group 2: 18 plots, mean 47 kg, standard deviation 5 kg. Assume equal variances (pooled t, df = n₁ + n₂ − 2). Find the margin of error E of the 95% confidence interval for μ₁ − μ₂.

n₁ = 13x̄₁ = 56.5s₁ = 6n₂ = 18x̄₂ = 47s₂ = 595%t(29) = 2.045
E = kg
9.Hypothesis tests & confidence intervalsMedium
2.14p = ?

The mean of a population is claimed to be 500; the population standard deviation is known to be 20. A random sample of 25 observations has a mean of 508.56. Test H₀: μ = 500 against H₁: μ > 500. Find the p-value p.

μ₀ = 500σ = 20n = 25x̄ = 508.56H₁: μ > μ₀
p =
10.Hypothesis tests & confidence intervalsHard
H₀H₁μ₀x̄cμ₀ + 5α = 0.05βσ = 25, power 95%, n = ?

A one-sided z test at α = 0.05 must detect a shift of the mean of 5 units with a power of 95%. The standard deviation is 25. Find the minimum sample size n.

α = 0.05power = 95%δ = 5σ = 25zα = 1.645zβ = 1.645
n =
11.Linear regression & correlationMedium
xy149249352451559667xy2465060

The table gives 6 paired observations of x = hours studied and y = exam score. Find Pearson's correlation coefficient r.

n = 6
r =
12.Linear regression & correlationHard
xy123224320417517xy2420ŷ = 23

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 = 23?

n = 5ŷ = 23
x =

3Solutions

1.Descriptive statistics

Answer: s = 14.36

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 = 260/7 = 37.14

Squared deviations

Σ(x − x̄)² = 9.857² + (−13.14)² + 9.857² + 18.86² + 4.857² + (−11.14)² + (−19.14)² = 1237

Standard deviation

s = √(1237/6) = 14.36

Answer

s = 14.36
2.Descriptive statistics

Answer: P = 0.6554

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 = 0.4

Standard normal table

Φ(0.4) = 0.6554
P = 0.6554

Answer

P = 0.6554
3.Probability

Answer: σ = 1.4

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 = 0·0.15 + 1·0.15 + 2·0.35 + 3·0.05 + 4·0.3 = 2.2

Mean of the squares

E(X²) = 0·0.15 + 1·0.15 + 4·0.35 + 9·0.05 + 16·0.3 = 6.8

Variance

Var(X) = 6.8 − 2.2² = 1.96
σ = √1.96 = 1.4

Answer

σ = 1.4
4.Probability

Answer: σY = 58.04

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 = 24.5, Var(X) = 134.8, σX = 11.61

Linear change

σY = |5|·11.61 = 58.04

Answer

σY = 58.04

The mean does move: E(Y) = 5·24.5 + 4 = 126.5.

5.Normal distribution

Answer: z = −1.3

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 − 0.9032 = 0.0968

Read the z-table backwards

Find 0.0968 inside the table and read off z.

z = −1.30

Answer

z = −1.30
6.Normal distribution

Answer: x = 87.4 points

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 = 100 − 0.84·15 = 87.4 points

Answer

x = 87.4 points
7.Confidence intervals

Answer: E = 0.682

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 = 14, t = 1.761

Margin of error

E = 1.761·1.5/√15 = 0.682

Answer

E = 0.682
8.Confidence intervals

Answer: E = 4.046 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 = (12·6² + 17·5²)/29 = 29.55
sp = 5.436

Margin of error

E = 2.045·5.436·√(1/13 + 1/18) = 4.046

Answer

E = 4.046 kg
9.Hypothesis tests & confidence intervals

Answer: p = 0.01618

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.

Test statistic

SE = σ/√n = 20/√25 = 4
z = (508.56 − 500)/4 = 2.14

Tail area

Φ(2.14) = 0.9838
1 − Φ(2.14) = 0.0162

p-value

p = 0.01618

Answer

p = 0.01618

p ≤ 0.05, so H₀ is rejected at the 5% level.

10.Hypothesis tests & confidence intervals

Answer: n = 271

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 + 1.645)·25/5)² = 270.6

Round up

n = 271

Answer

n = 271
11.Linear regression & correlation

Answer: r = 0.8898

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 = 327/6 = 54.5

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = 59.5
Syy = Σ(y − ȳ)² = 255.5

Correlation coefficient

r = 59.5/√(17.5·255.5) = 0.8898

Answer

r = 0.8898
12.Linear regression & correlation

Answer: x = 1.526

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 = 101/5 = 20.2

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = −19

Regression line

b = −19/10 = −1.9
a = 20.2 − (−1.9)·3 = 25.9
ŷ = 25.9 − 1.9·x

Solve for x

x = (23 − 25.9)/(−1.9) = 1.526

Answer

x = 1.526

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