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
202530354045505560

Find the sample standard deviation s of the data: 25, 19, 31, 60, 57, 41, 25.

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

A quantity is normally distributed with mean 500 and standard deviation 40. What proportion P of values is above 560?

μ = 500σ = 40x = 560
P =
3.ProbabilityMedium
xp(x)10.220.3530.140.35μ = 2.6

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

x: 1, 2, 3, 4p(x): 0.2, 0.35, 0.1, 0.35
Var(X) =
4.ProbabilityHard
n = 10, p = 0.6: P(X ≤ 1) = ?k012345678910

10 components are tested, and each one fails with probability 0.6, independently. What is the probability P that at most one fails?

n = 10p = 0.6
P =
5.Normal distributionMedium
240 g245.9 gT = X₁ + … + X₆?

The masses of bolts are independent and normal with mean 40 g and standard deviation 1.2 g. Find the probability P that the total mass of 6 of them exceeds 245.9 g. Round z to two decimals.

μ = 40 gσ = 1.2 gn = 6> 245.9 g
P =
6.Normal distributionHard
0.04 mm0D − d?

Hole diameters D are normal with mean 25.04 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 = 25.04 mm, σ = 0.02 mmshaft: μ = 25 mm, σ = 0.02 mm
P =
7.Confidence intervalsMedium
μ₁ − μ₂0LU = ?x̄₁ − x̄₂E90%

The battery life of two phone models is compared. Model 1: 100 phones, mean 51 h, known standard deviation 17 h. Model 2: 60 phones, mean 39.5 h, known standard deviation 9 h. Find the upper limit of the 90% confidence interval for μ₁ − μ₂.

n₁ = 100x̄₁ = 51 hσ₁ = 17 hn₂ = 60x̄₂ = 39.5 hσ₂ = 9 h90%
upper limit = h
8.Confidence intervalsHard
σ = 7, n = ?x̄ − Ex̄ + Ex̄E = 399%

The population standard deviation is 7. How large a sample n do you need so that a 99% confidence interval for the mean has a margin of error of at most 3?

σ = 7E = 399%
n =
9.Hypothesis tests & confidence intervalsMedium
t −1.99801.998

A sample of 64 students has a mean score of 80.25 and a sample standard deviation of 20. Find the test statistic t for H₀: μ = 75.

μ₀ = 75x̄ = 80.25s = 20n = 64
t =
10.Hypothesis tests & confidence intervalsHard
z −1.64501.645

A sample of 81 measurements has mean 46.7. The population standard deviation is 20. Find the lower limit L of the 90% confidence interval for the mean.

x̄ = 46.7σ = 20n = 8190%
L =
11.Linear regression & correlationMedium
xy126228326422518xy242025

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
xy248356456565664xy2465060ŷ = 56

The table gives 5 paired observations of x = advertising spend in k€ and y = sales in k€. Use the least-squares line ŷ = a + b·x. For which value x does the line predict ŷ = 56?

n = 5ŷ = 56
x =

3Solutions

1.Descriptive statistics

Answer: s = 16.29

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 = 258/7 = 36.86

Squared deviations

Σ(x − x̄)² = (−11.86)² + (−17.86)² + (−5.857)² + 23.14² + 20.14² + 4.143² + (−11.86)² = 1593

Standard deviation

s = √(1593/6) = 16.29

Answer

s = 16.29
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) = 1.34

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 = 1·0.2 + 2·0.35 + 3·0.1 + 4·0.35 = 2.6

Mean of the squares

E(X²) = 1·0.2 + 4·0.35 + 9·0.1 + 16·0.35 = 8.1

Variance

Var(X) = 8.1 − 2.6² = 1.34

Answer

Var(X) = 1.34
4.Probability

Answer: P = 0.001678

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.

Zero and one failure

P(X = 0) = 0.410 = 0.0001049
P(X = 1) = 10·0.6·0.49 = 0.001573

Add them

P = 0.0001049 + 0.001573 = 0.001678

Answer

P = 0.001678
5.Normal distribution

Answer: P = 0.02222

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 total

μT = 240 g, σT = 1.2·√6 = 2.939 g

Standardise

z = (245.9 − 240)/2.939 = 2.01

Probability

P = 1 − Φ(2.01) = 0.02222

Answer

P = 0.02222
6.Normal distribution

Answer: P = 0.9207

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 = 25.04 − 25 = 0.04 mm, σ = 0.02828 mm

Standardise

z = (0 − 0.04)/0.02828 = −1.41

Probability

P = 1 − Φ(−1.41) = 0.9207

Answer

P = 0.9207
7.Confidence intervals

Answer: upper limit = 14.89 h

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.

Point estimate

x̄₁ − x̄₂ = 51 − 39.5 = 11.5

Standard error

SE = √(17²/100 + 9²/60) = 2.059

Margin of error

E = 1.645·2.059 = 3.387

Interval

11.5 ± 3.387 → [8.113, 14.89]

Answer

upper limit = 14.89 h

The interval does not contain 0, so the means differ significantly.

8.Confidence intervals

Answer: n = 37

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.

Solve for n

n = (2.576·7/3)² = 36.13

Round up

A sample size is a whole number, and rounding down would make the margin too large.

n = 37

Answer

n = 37
9.Hypothesis tests & confidence intervals

Answer: t = 2.1

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 = 20/√64 = 20/8 = 2.5

Test statistic

t = (80.25 − 75)/2.5 = 2.1

Answer

t = 2.1
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: L = 43.04

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 = 1.645·20/√81 = 3.656

Interval limit

L = 46.7 − 3.656 = 43.04

Answer

L = 43.04
11.Linear regression & correlation

Answer: r² = 0.7562

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 = 120/5 = 24

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = −22
Syy = Σ(y − ȳ)² = 64

Coefficient of determination

r² = −22²/(10·64) = 0.7562

Answer

r² = 0.7562

About 75.62% of the variation in y is explained by the line.

12.Linear regression & correlation

Answer: x = 3.561

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 = 20/5 = 4
ȳ = 289/5 = 57.8

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = 41

Regression line

b = 41/10 = 4.1
a = 57.8 − 4.1·4 = 41.4
ŷ = 41.4 + 4.1·x

Solve for x

x = (56 − 41.4)/4.1 = 3.561

Answer

x = 3.561

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