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
1520253035404550

Find the sample standard deviation s of the data: 14, 23, 50, 13, 18.

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

A quantity is normally distributed with mean 100 and standard deviation 15. What proportion P of values is above 122.5?

μ = 100σ = 15x = 122.5
P =
3.ProbabilityMedium
xp(x)00.2510.0520.1530.2540.3μ = 2.3

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

x: 0, 1, 2, 3, 4p(x): 0.25, 0.05, 0.15, 0.25, 0.3
Var(X) =
4.ProbabilityHard
n = 11, p = 0.1: P(X = 2) = ?k01234567891011

A test has 11 multiple-choice questions. A student who guesses gets each one right with probability 0.1, independently. What is the probability P of exactly 2 correct answers?

n = 11p = 0.1k = 2
P =
5.Normal distributionMedium
−3−2−10123zshaded area between −z and z = 0.8948

Drag z so that the shaded area between −z and z under the standard normal curve is 0.8948. Find z to two decimals.

A = 0.8948
z =
6.Normal distributionHard
157164171178185192199cmtop 70%

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 70% is to the right of it. Find the cut-off x.

μ = 178 cmσ = 7 cm70%
x = cm
7.Confidence intervalsMedium
n = 20, sd​ = 2.40L = ?Ud̄ = 0.8E90%

The blood pressure of 20 patients is measured before and after a treatment. The differences d = before − after (mmHg) have mean 0.8 and standard deviation 2.4. Find the lower limit of the 90% confidence interval for the mean difference.

n = 20d̄ = 0.8sd = 2.490%t(19) = 1.729
lower limit = mmHg
8.Confidence intervalsHard
p = 0.5, n = ?p̂ − Ep̂ + Ep̂E = 0.0199%

Nothing is known in advance about the proportion of defective products. How many products n must be inspected so that a 99% confidence interval for the proportion has a margin of error of at most 0.01?

p = 0.5E = 0.0199%
n =
9.Hypothesis tests & confidence intervalsMedium
−1.961.96z = ?

Two production lines make the same part. Line 1: a sample of 50 parts has mean length 125.5 mm, and the line's known standard deviation is 8 mm. Line 2: a sample of 80 parts has mean 132.2 mm, known standard deviation 19 mm. Find the test statistic z for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (line 1 minus line 2).

n₁ = 50x̄₁ = 125.5 mmσ₁ = 8 mmn₂ = 80x̄₂ = 132.2 mmσ₂ = 19 mm
z =
10.Hypothesis tests & confidence intervalsHard
z −1.64501.645

A sample of 49 measurements has mean 80.8. The population standard deviation is 15. Find the lower limit L of the 90% confidence interval for the mean.

x̄ = 80.8σ = 15n = 4990%
L =
11.Linear regression & correlationMedium
xy1443465647699821188xy5106080

The table gives 6 paired observations of x = hours studied and y = exam score. Find the coefficient of determination r² of the least-squares line.

n = 6
r² =
12.Linear regression & correlationHard
xy120221314413511xy241520x = 4.5

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. A new observation (4.5, 10) is made. Find its residual e = y − ŷ.

n = 5(4.5, 10)
e =

3Solutions

1.Descriptive statistics

Answer: s = 15.27

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 = 118/5 = 23.6

Squared deviations

Σ(x − x̄)² = (−9.6)² + (−0.6)² + 26.4² + (−10.6)² + (−5.6)² = 933.2

Standard deviation

s = √(933.2/4) = 15.27

Answer

s = 15.27
2.Descriptive statistics

Answer: P = 0.06681

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 = 1.5

Standard normal table

Φ(1.5) = 0.9332
P = 1 − 0.9332 = 0.06681

Answer

P = 0.06681
3.Probability

Answer: Var(X) = 2.41

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 = 0·0.25 + 1·0.05 + 2·0.15 + 3·0.25 + 4·0.3 = 2.3

Mean of the squares

E(X²) = 0·0.25 + 1·0.05 + 4·0.15 + 9·0.25 + 16·0.3 = 7.7

Variance

Var(X) = 7.7 − 2.3² = 2.41

Answer

Var(X) = 2.41
4.Probability

Answer: P = 0.2131

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 = 2) = 55·0.12·0.99 = 55·0.01·0.3874 = 0.2131

Answer

P = 0.2131
5.Normal distribution

Answer: z = 1.62

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 + 0.8948)/2 = 0.9474

Read the z-table backwards

Find 0.9474 inside the table and read off z.

z = 1.62

Answer

z = 1.62
6.Normal distribution

Answer: x = 174.4 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) = 0.3

z from the table

z ≈ −0.52

Back to the original units

x = 178 − 0.52·7 = 174.4 cm

Answer

x = 174.4 cm
7.Confidence intervals

Answer: lower limit = −0.1279 mmHg

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.

Critical value

df = 19, t = 1.729

Margin of error

E = 1.729·2.4/√20 = 0.9279

Interval

0.8 ± 0.9279 → [−0.1279, 1.728]

Answer

lower limit = −0.1279 mmHg
8.Confidence intervals

Answer: n = 16590

Confidence interval for a proportion: p̂ ± z·√(p̂(1 − p̂)/n), where p̂ = x/n. To reach a margin E you need n ≥ z²·p(1 − p)/E², rounded up; without a prior estimate use p = 0.5, which gives the largest n.

Solve for n

n = 2.576²·0.5·0.5/0.01² = 16589

Round up

n = 16590

Answer

n = 16590
9.Hypothesis tests & confidence intervals

Answer: z = −2.784

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.

Standard error of the difference

SE = √(8²/50 + 19²/80) = √5.793 = 2.407

Test statistic

z = (125.5 − 132.2)/2.407 = −2.784

Answer

z = −2.784

|z| > 1.96, so H₀ is rejected at the 5% level (two-sided).

10.Hypothesis tests & confidence intervals

Answer: L = 77.27

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 = 1.645·15/√49 = 3.525

Interval limit

L = 80.8 − 3.525 = 77.27

Answer

L = 77.27
11.Linear regression & correlation

Answer: r² = 0.9686

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 = 36/6 = 6
ȳ = 393/6 = 65.5

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 333
Syy = Σ(y − ȳ)² = 1636

Coefficient of determination

r² = 333²/(70·1636) = 0.9686

Answer

r² = 0.9686

About 96.86% of the variation in y is explained by the line.

12.Linear regression & correlation

Answer: e = −1.9

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 = 79/5 = 15.8

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = −26

Regression line

b = −26/10 = −2.6
a = 15.8 − (−2.6)·3 = 23.6
ŷ = 23.6 − 2.6·x

Predicted value

ŷ = 23.6 + (−2.6)·4.5 = 11.9

Residual

e = 10 − 11.9 = −1.9

Answer

e = −1.9

The point lies below the line.

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