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: 58, 58, 53, 17, 18.

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

A quantity is normally distributed with mean 170 and standard deviation 5. What proportion P of values is above 160?

μ = 170σ = 5x = 160
P =
3.ProbabilityMedium
red: 5blue: 4

A bag holds 5 red and 4 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?

5 / 4
P =
4.ProbabilityHard
n = 11, p = 0.3: P(X = 4) = ?k01234567891011

A test has 11 multiple-choice questions. A student who guesses gets each one right with probability 0.3, independently. What is the probability P of exactly 4 correct answers?

n = 11p = 0.3k = 4
P =
5.Normal distributionMedium
−3−2−10123zshaded area to the right of z = 0.6700

Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.6700. Find z to two decimals.

A = 0.6700
z =
6.Normal distributionHard
10 g0X − Y?

The mass X of a product from line 1 and Y from line 2 are independent and normal: X has mean 500 g and standard deviation 8 g, Y has mean 490 g and standard deviation 6 g. Find the probability P that X is greater than Y. Round z to two decimals.

μX = 500 g, σX = 8 gμY = 490 g, σY = 6 g
P =
7.Confidence intervalsMedium
n = 7, s = 11.5L = ?Ux̄ = 62.5E99%

A sample of 7 items has mean 62.5 and standard deviation s = 11.5. The population is roughly normal. Find the lower limit of the 99% confidence interval for the mean.

n = 7x̄ = 62.5s = 11.599%t(6) = 3.707
lower limit =
8.Confidence intervalsHard
μ₁ − μ₂, t(24) = 2.0640LUx̄₁ − x̄₂E = ?95%

Two fertilisers are tested. Plot group 1: 13 plots, mean yield 71.5 kg, standard deviation 8.5 kg. Plot group 2: 13 plots, mean 80.5 kg, standard deviation 7 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 = 71.5s₁ = 8.5n₂ = 13x̄₂ = 80.5s₂ = 795%t(24) = 2.064
E = kg
9.Hypothesis tests & confidence intervalsMedium
t −1.99801.998

A sample of 64 students has a mean score of 501.5 and a sample standard deviation of 5. Find the test statistic t for H₀: μ = 500.

μ₀ = 500x̄ = 501.5s = 5n = 64
t =
10.Hypothesis tests & confidence intervalsHard
σ = 8, n = ?x̄ − Ex̄ + Ex̄E = 0.595%

How large must a sample be so that the 95% confidence interval for a mean has a margin of error of at most 0.5? The standard deviation is 8. Find the smallest sample size n.

σ = 8E = 0.595%
n =
11.Linear regression & correlationMedium
xy2841562182810391244xy5102040

The table gives 6 paired observations of x = load in kN and y = extension in mm. Find the coefficient of determination r² of the least-squares line.

n = 6
r² =
12.Linear regression & correlationHard
xy36057679391011111313132xy510100x = 4

The table gives 6 paired observations of x = hours studied and y = exam score. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 4.

n = 6x = 4
ŷ =

3Solutions

1.Descriptive statistics

Answer: s = 21.37

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 = 204/5 = 40.8

Squared deviations

Σ(x − x̄)² = 17.2² + 17.2² + 12.2² + (−23.8)² + (−22.8)² = 1827

Standard deviation

s = √(1827/4) = 21.37

Answer

s = 21.37
2.Descriptive statistics

Answer: P = 0.9772

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 = 1 − 0.02275 = 0.9772

Answer

P = 0.9772
3.Probability

Answer: P = 0.2778

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.

First and second draw

P = 5/9 · 4/8 = 20/72 = 0.2778

Answer

P = 0.2778
4.Probability

Answer: P = 0.2201

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 = 4) = 330·0.34·0.77 = 330·0.0081·0.08235 = 0.2201

Answer

P = 0.2201
5.Normal distribution

Answer: z = −0.44

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 − 0.6700 = 0.3300

Read the z-table backwards

Find 0.3300 inside the table and read off z.

z = −0.44

Answer

z = −0.44
6.Normal distribution

Answer: P = 0.8413

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 = 500 − 490 = 10 g, σ = 10 g

Standardise

z = (0 − 10)/10 = −1

Probability

P = 1 − Φ(−1) = 0.8413

Answer

P = 0.8413
7.Confidence intervals

Answer: lower limit = 46.39

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 = 3.707

Margin of error

E = 3.707·11.5/√7 = 16.11

Interval

62.5 ± 16.11 → [46.39, 78.61]

Answer

lower limit = 46.39
8.Confidence intervals

Answer: E = 6.303 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 = (12·8.5² + 12·7²)/24 = 60.63
sp = 7.786

Margin of error

E = 2.064·7.786·√(1/13 + 1/13) = 6.303

Answer

E = 6.303 kg
9.Hypothesis tests & confidence intervals

Answer: t = 2.4

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 = 5/√64 = 5/8 = 0.625

Test statistic

t = (501.5 − 500)/0.625 = 2.4

Answer

t = 2.4
t0.025; df = 63 = 1.998

|t| > 1.998, so H₀ is rejected at the 5% level (two-sided).

10.Hypothesis tests & confidence intervals

Answer: n = 984

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).

Solve for n

n = (1.96·8/0.5)² = 983.4

Round up

n = 984

Answer

n = 984
11.Linear regression & correlation

Answer: r² = 0.9912

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 = 155/6 = 25.83

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 259
Syy = Σ(y − ȳ)² = 966.8

Coefficient of determination

r² = 259²/(70·966.8) = 0.9912

Answer

r² = 0.9912

About 99.12% of the variation in y is explained by the line.

12.Linear regression & correlation

Answer: ŷ = 68.46

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 = 48/6 = 8
ȳ = 575/6 = 95.83

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 479

Regression line

b = 479/70 = 6.843
a = 95.83 − 6.843·8 = 41.09
ŷ = 41.09 + 6.843·x

Prediction

ŷ = 41.09 + 6.843·4 = 68.46

Answer

ŷ = 68.46

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