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
152025303540455055

Find the sample standard deviation s of the data: 46, 48, 56, 35, 11, 14.

n = 6
s =
2.Descriptive statisticsHard
z −2−1012

A quantity is normally distributed with mean 60 and standard deviation 8. What proportion P of values is below 68?

μ = 60σ = 8x = 68
P =
3.ProbabilityMedium
xp(x)10.3520.230.0540.3550.05μ = 2.55

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

x: 1, 2, 3, 4, 5p(x): 0.35, 0.2, 0.05, 0.35, 0.05
Var(X) =
4.ProbabilityHard
€p(x)−100.88400.14900.02

A game costs €10 to play. You win €50 with probability 0.1 and €500 with probability 0.02; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.

stake = €10€50: p = 0.1€500: p = 0.02
E(G) = €
5.Normal distributionMedium
100 kg99.5 kgT = X₁ + … + X₄?

The masses of bags of cement are independent and normal with mean 25 kg and standard deviation 0.5 kg. Find the probability P that the total mass of 4 of them exceeds 99.5 kg. Round z to two decimals.

μ = 25 kgσ = 0.5 kgn = 4> 99.5 kg
P =
6.Normal distributionHard
μ = 250x = ?95%

The bag weights are normally distributed with mean 250 g and standard deviation 5 g. Find the value x below which 95% of them lie (the 95th percentile).

μ = 250σ = 5P = 95%
x = g
7.Confidence intervalsMedium
n = 11, s = 13LUx̄ = 92E = ?90%

A sample of 11 items has mean 92 and standard deviation s = 13. The population is roughly normal. Find the margin of error E of a 90% confidence interval for the mean.

n = 11x̄ = 92s = 1390%t(10) = 1.812
E =
8.Confidence intervalsHard
σ = 23, n = ?x̄ − Ex̄ + Ex̄E = 590%

The population standard deviation is 23. How large a sample n do you need so that a 90% confidence interval for the mean has a margin of error of at most 5?

σ = 23E = 590%
n =
9.Hypothesis tests & confidence intervalsMedium
1.18p = ?

The mean of a population is claimed to be 80; the population standard deviation is known to be 10. A random sample of 25 observations has a mean of 82.36. Test H₀: μ = 80 against H₁: μ > 80. Find the p-value p.

μ₀ = 80σ = 10n = 25x̄ = 82.36H₁: μ > μ₀
p =
10.Hypothesis tests & confidence intervalsHard
z −2.57602.576

A sample of 100 measurements has mean 500.6. The population standard deviation is 10. Find the lower limit L of the 99% confidence interval for the mean.

x̄ = 500.6σ = 10n = 10099%
L =
11.Linear regression & correlationMedium
xy146250364471576686xy2466080

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
xy284146198261035xy510102030ŷ = 30

The table gives 5 paired observations of x = load in kN and y = extension in mm. Use the least-squares line ŷ = a + b·x. For which value x does the line predict ŷ = 30?

n = 5ŷ = 30
x =

3Solutions

1.Descriptive statistics

Answer: s = 18.7

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 = 210/6 = 35

Squared deviations

Σ(x − x̄)² = 11² + 13² + 21² + 0² + (−24)² + (−21)² = 1748

Standard deviation

s = √(1748/5) = 18.7

Answer

s = 18.7
2.Descriptive statistics

Answer: P = 0.8413

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

Standard normal table

Φ(1) = 0.8413
P = 0.8413

Answer

P = 0.8413
3.Probability

Answer: Var(X) = 1.947

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 + 2·0.2 + 3·0.05 + 4·0.35 + 5·0.05 = 2.55

Mean of the squares

E(X²) = 1·0.35 + 4·0.2 + 9·0.05 + 16·0.35 + 25·0.05 = 8.45

Variance

Var(X) = 8.45 − 2.55² = 1.947

Answer

Var(X) = 1.947
4.Probability

Answer: E(G) = 5 €

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.

Net outcomes

G: 40 (p = 0.1), 490 (p = 0.02), −10 (p = 0.88)

Expected value

E(G) = 40·0.1 + 490·0.02 − 10·0.88 = 5 €

Answer

E(G) = 5 €

A positive value means you win money on average.

5.Normal distribution

Answer: P = 0.6915

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 = 100 kg, σT = 0.5·√4 = 1 kg

Standardise

z = (99.5 − 100)/1 = −0.5

Probability

P = 1 − Φ(−0.5) = 0.6915

Answer

P = 0.6915
6.Normal distribution

Answer: x = 258.2 g

Normal distribution: convert a value x to a z-score, z = (x − μ)/σ, which counts standard deviations from the mean. A z-table (or Φ(z)) gives the area to the left of z, which is P(X < x). The area to the right is 1 − Φ(z), and the area between two values is the difference of their two areas.

z for the percentile

Φ(z) = 0.95 → z = 1.645

Back to x

x = 250 + 1.645·5 = 258.22 g

Answer

x = 258.22 g
7.Confidence intervals

Answer: E = 7.102

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 = 10, t = 1.812

Margin of error

E = 1.812·13/√11 = 7.102

Answer

E = 7.102
8.Confidence intervals

Answer: n = 58

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 = (1.645·23/5)² = 57.26

Round up

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

n = 58

Answer

n = 58
9.Hypothesis tests & confidence intervals

Answer: p = 0.119

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 = 10/√25 = 2
z = (82.36 − 80)/2 = 1.18

Tail area

Φ(1.18) = 0.8810
1 − Φ(1.18) = 0.1190

p-value

p = 0.119

Answer

p = 0.119

p > 0.05, so H₀ is not rejected at the 5% level.

10.Hypothesis tests & confidence intervals

Answer: L = 498

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 = 2.576·10/√100 = 2.576

Interval limit

L = 500.6 − 2.576 = 498

Answer

L = 498
11.Linear regression & correlation

Answer: r² = 0.9804

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 = 393/6 = 65.5

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = 142.5
Syy = Σ(y − ȳ)² = 1184

Coefficient of determination

r² = 142.5²/(17.5·1184) = 0.9804

Answer

r² = 0.9804

About 98.04% of the variation in y is explained by the line.

12.Linear regression & correlation

Answer: x = 8.909

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 = 30/5 = 6
ȳ = 102/5 = 20.4

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 132

Regression line

b = 132/40 = 3.3
a = 20.4 − 3.3·6 = 0.6
ŷ = 0.6 + 3.3·x

Solve for x

x = (30 − 0.6)/3.3 = 8.909

Answer

x = 8.909

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