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
25303540455055

Find the sample standard deviation s of the data: 45, 21, 52, 56, 51, 44, 54.

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

Exam scores are normally distributed with mean 60 and standard deviation 10. Find the z-score z of a score of 66.

μ = 60σ = 10x = 66
z =
3.ProbabilityMedium
xp(x)10.3520.3530.240.1μ = 2.05

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

x: 1, 2, 3, 4p(x): 0.35, 0.35, 0.2, 0.1
σ =
4.ProbabilityHard
n = 7, p = 0.4: P(X = 1) = ?k01234567

A test has 7 multiple-choice questions. A student who guesses gets each one right with probability 0.4, independently. What is the probability P of exactly 1 correct answers?

n = 7p = 0.4k = 1
P =
5.Normal distributionMedium
−3−2−10123zshaded area between −z and z = 0.9371

Drag z so that the shaded area between −z and z under the standard normal curve is 0.9371. Find z to two decimals.

A = 0.9371
z =
6.Normal distributionHard
0.05 mm0D − d?

Hole diameters D are normal with mean 25.05 mm and standard deviation 0.02 mm. Shaft diameters d are independent and normal with mean 25 mm and standard deviation 0.02 mm. A randomly chosen shaft fits when D − d > 0. Find the probability P that it fits. Round z to two decimals.

hole: μ = 25.05 mm, σ = 0.02 mmshaft: μ = 25 mm, σ = 0.02 mm
P =
7.Confidence intervalsMedium
x̄E₁ = 10.895%, n = 36x̄E₂ = ?90%, n = 36

A 95% confidence interval for a mean, based on 36 observations and a known σ, has a margin of error of 10.8. What is the margin of error E₂ of a 90% interval from the same data?

95%n₁ = 36E₁ = 10.890%
E₂ =
8.Confidence intervalsHard
μ₁ − μ₂, t(29) = 2.0450LU = ?x̄₁ − x̄₂E95%

Two fertilisers are tested. Plot group 1: 9 plots, mean yield 77 kg, standard deviation 1.5 kg. Plot group 2: 22 plots, mean 66 kg, standard deviation 4.5 kg. Assume equal variances (pooled t, df = n₁ + n₂ − 2). Find the upper limit of the 95% confidence interval for μ₁ − μ₂.

n₁ = 9x̄₁ = 77s₁ = 1.5n₂ = 22x̄₂ = 66s₂ = 4.595%t(29) = 2.045
upper limit = kg
9.Hypothesis tests & confidence intervalsMedium
t −2.01102.011

A sample of 49 students has a mean score of 203.4 and a sample standard deviation of 10. Find the test statistic t for H₀: μ = 200.

μ₀ = 200x̄ = 203.4s = 10n = 49
t =
10.Hypothesis tests & confidence intervalsHard
−1.961.96z = ?

In a trial, 55 of 100 patients on treatment A recover, and 118 of 200 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.

x₁ = 55n₁ = 100x₂ = 118n₂ = 200
z =
11.Linear regression & correlationMedium
xy125226321422518xy242025

The table gives 5 paired observations of x = car age in years and y = price in k€. Find the coefficient of determination r² of the least-squares line.

n = 5
r² =
12.Linear regression & correlationHard
xy16781910022112251332814931164xy15202530100150x = 29.5

The table gives 6 paired observations of x = temperature in °C and y = ice creams sold. Use the least-squares line ŷ = a + b·x. A new observation (29.5, 161) is made. Find its residual e = y − ŷ.

n = 6(29.5, 161)
e =

3Solutions

1.Descriptive statistics

Answer: s = 11.94

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 = 323/7 = 46.14

Squared deviations

Σ(x − x̄)² = (−1.143)² + (−25.14)² + 5.857² + 9.857² + 4.857² + (−2.143)² + 7.857² = 854.9

Standard deviation

s = √(854.9/6) = 11.94

Answer

s = 11.94
2.Descriptive statistics

Answer: z = 0.6

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 = (66 − 60)/10 = 0.6

Answer

z = 0.6
3.Probability

Answer: σ = 0.9734

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 = 1·0.35 + 2·0.35 + 3·0.2 + 4·0.1 = 2.05

Mean of the squares

E(X²) = 1·0.35 + 4·0.35 + 9·0.2 + 16·0.1 = 5.15

Variance

Var(X) = 5.15 − 2.05² = 0.9475
σ = √0.9475 = 0.9734

Answer

σ = 0.9734
4.Probability

Answer: P = 0.1306

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.

Binomial formula

P(X = 1) = 7·0.41·0.66 = 7·0.4·0.04666 = 0.1306

Answer

P = 0.1306
5.Normal distribution

Answer: z = 1.86

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 + 0.9371)/2 = 0.9686

Read the z-table backwards

Find 0.9686 inside the table and read off z.

z = 1.86

Answer

z = 1.86
6.Normal distribution

Answer: P = 0.9616

Sums and differences of normal variables: if X and Y are independent and normal, X + Y and X − Y are normal too. Means add or subtract: E(X ± Y) = μX ± μY. Variances always add, also for a difference: Var(X ± Y) = σX² + σY², so σ = √(σX² + σY²). Standard deviations never add directly. The total of n independent items with the same distribution has mean n·μ and standard deviation σ·√n. Then standardise as usual: z = (x − mean)/σ and read Φ(z).

Distribution of the difference

μ = 25.05 − 25 = 0.05 mm, σ = 0.02828 mm

Standardise

z = (0 − 0.05)/0.02828 = −1.77

Probability

P = 1 − Φ(−1.77) = 0.9616

Answer

P = 0.9616
7.Confidence intervals

Answer: E₂ = 9.064

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.

Ratio of margins

E₂/E₁ = (1.645/1.96)·√(36/36) = 0.8393

New margin

E₂ = 10.8·0.8393 = 9.064

Answer

E₂ = 9.064
8.Confidence intervals

Answer: upper limit = 14.16 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 = (8·1.5² + 21·4.5²)/29 = 15.28
sp = 3.91

Margin of error

E = 2.045·3.91·√(1/9 + 1/22) = 3.163

Interval

11 ± 3.163 → [7.837, 14.16]

Answer

upper limit = 14.16 kg
9.Hypothesis tests & confidence intervals

Answer: t = 2.38

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 = 10/√49 = 10/7 = 1.429

Test statistic

t = (203.4 − 200)/1.429 = 2.38

Answer

t = 2.38
t0.025; df = 48 = 2.011

|t| > 2.011, so H₀ is rejected at the 5% level (two-sided).

10.Hypothesis tests & confidence intervals

Answer: z = −0.661

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 = 55/100 = 0.55
p̂₂ = 118/200 = 0.59
p̂ = 173/300 = 0.5767

Standard error under H₀

SE = √(0.5767·0.4233·(1/100 + 1/200)) = 0.06051

Test statistic

z = (0.55 − 0.59)/0.06051 = −0.661

Answer

z = −0.661

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

11.Linear regression & correlation

Answer: r² = 0.7864

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 = 112/5 = 22.4

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = −18
Syy = Σ(y − ȳ)² = 41.2

Coefficient of determination

r² = −18²/(10·41.2) = 0.7864

Answer

r² = 0.7864

About 78.64% of the variation in y is explained by the line.

12.Linear regression & correlation

Answer: e = 4.162

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 = 141/6 = 23.5
ȳ = 736/6 = 122.7

Sums of squares and products

Sxx = Σ(x − x̄)² = 157.5
Sxy = Σ(x − x̄)(y − ȳ) = 897

Regression line

b = 897/157.5 = 5.695
a = 122.7 − 5.695·23.5 = −11.17
ŷ = −11.17 + 5.695·x

Predicted value

ŷ = −11.17 + 5.695·29.5 = 156.8

Residual

e = 161 − 156.8 = 4.162

Answer

e = 4.162

The point lies above the line.

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