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: 24, 42, 50, 11, 36, 29, 32.

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 460?

μ = 500σ = 40x = 460
P =
3.ProbabilityMedium
xp(x)00.110.320.2530.35μ = 1.85

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

x: 0, 1, 2, 3p(x): 0.1, 0.3, 0.25, 0.35
Var(X) =
4.ProbabilityHard
€p(x)−100.89200.14900.01

A game costs €10 to play. You win €30 with probability 0.1 and €500 with probability 0.01; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.

stake = €10€30: p = 0.1€500: p = 0.01
E(G) = €
5.Normal distributionMedium
32 min35.8 minX + Y?

X is normal with mean 12 min and standard deviation 1.5 min; Y is independent and normal with mean 20 min and standard deviation 2 min. Find the probability P that X + Y is more than 35.8 min. Round z to two decimals.

μX = 12, σX = 1.5μY = 20, σY = 2X + Y > 35.8 min
P =
6.Normal distributionHard
μ = 100x = ?25%

The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. Find the value x below which 25% of them lie (the 25th percentile).

μ = 100σ = 15P = 25%
x = points
7.Confidence intervalsMedium
n = 20, s = 8.5LUx̄ = 24E = ?99%

A sample of 20 items has mean 24 and standard deviation s = 8.5. The population is roughly normal. Find the margin of error E of a 99% confidence interval for the mean.

n = 20x̄ = 24s = 8.599%t(19) = 2.861
E =
8.Confidence intervalsHard
p ≈ 0.2, n = ?p̂ − Ep̂ + Ep̂E = 0.0295%

A pilot study suggests that about 0.2 of products are defective. How many products n must be inspected so that a 95% confidence interval for the proportion has a margin of error of at most 0.02?

p ≈ 0.2E = 0.0295%
n =
9.Hypothesis tests & confidence intervalsMedium
−2.2012.201t = ?df = 11

12 employees take a typing test before and after a training course. For each person the difference d = after − before (words per minute) is recorded. The mean difference is 3.64 and the standard deviation of the differences is 4.2. Find the paired test statistic t for H₀: μd = 0.

n = 12d̄ = 3.64sd = 4.2
t =
10.Hypothesis tests & confidence intervalsHard
μ₀x̄c

A one-sided z test of H₀: μ = 200 against H₁: μ < 200 is done at α = 0.05 with a sample of 100; the standard deviation is known to be 10. In reality the mean is 197.7. Find the power 1 − β of the test in percent.

μ₀ = 200μ = 197.7σ = 10n = 100α = 0.05
1 − β = %
11.Linear regression & correlationMedium
xy2324436508471054xy5104050

The table gives 5 paired observations of x = advertising spend in k€ and y = sales in k€. Find Pearson's correlation coefficient r.

n = 5
r =
12.Linear regression & correlationHard
xy1473495567689761183xy5106080x = 2

The table gives 6 paired observations of x = advertising spend in k€ and y = sales in k€. Use the least-squares line ŷ = a + b·x. A new observation (2, 43) is made. Find its residual e = y − ŷ.

n = 6(2, 43)
e =

3Solutions

1.Descriptive statistics

Answer: s = 12.61

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 = 224/7 = 32

Squared deviations

Σ(x − x̄)² = (−8)² + 10² + 18² + (−21)² + 4² + (−3)² + 0² = 954

Standard deviation

s = √(954/6) = 12.61

Answer

s = 12.61
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) = 0.1587
P = 1 − 0.1587 = 0.8413

Answer

P = 0.8413
3.Probability

Answer: Var(X) = 1.028

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 = 0·0.1 + 1·0.3 + 2·0.25 + 3·0.35 = 1.85

Mean of the squares

E(X²) = 0·0.1 + 1·0.3 + 4·0.25 + 9·0.35 = 4.45

Variance

Var(X) = 4.45 − 1.85² = 1.028

Answer

Var(X) = 1.028
4.Probability

Answer: E(G) = −2 €

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: 20 (p = 0.1), 490 (p = 0.01), −10 (p = 0.89)

Expected value

E(G) = 20·0.1 + 490·0.01 − 10·0.89 = −2 €

Answer

E(G) = −2 €

A negative value means you lose money on average.

5.Normal distribution

Answer: P = 0.06426

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 sum

μ = 32 min, σ = √(1.5² + 2²) = 2.5 min

Standardise

z = (35.8 − 32)/2.5 = 1.52

Upper tail

P = 1 − Φ(1.52) = 1 − 0.9357 = 0.06426

Answer

P = 0.06426
6.Normal distribution

Answer: x = 89.88 points

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) = 0.25 → z = −0.6745

Back to x

x = 100 + −0.6745·15 = 89.88 points

Answer

x = 89.88 points
7.Confidence intervals

Answer: E = 5.438

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 = 19, t = 2.861

Margin of error

E = 2.861·8.5/√20 = 5.438

Answer

E = 5.438
8.Confidence intervals

Answer: n = 1537

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 = 1.96²·0.2·0.8/0.02² = 1537

Round up

n = 1537

Answer

n = 1537
9.Hypothesis tests & confidence intervals

Answer: t = 3.002

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 of the mean difference

SE = 4.2/√12 = 1.212

Test statistic

t = 3.64/1.212 = 3.002

Answer

t = 3.002

With df = 11, the two-sided 5% critical value is 2.201; |t| is larger, so H₀ is rejected.

10.Hypothesis tests & confidence intervals

Answer: 1 − β = 74.38 %

Power of a test: power = 1 − β is the probability of rejecting H₀ when it is false. Find the cutoff x̄c from μ₀ and z_α, then compute the probability of landing beyond x̄c using the true mean: z = (x̄c − μ)/(σ/√n).

Critical sample mean

SE = 10/√100 = 1
x̄c = 200 − 1.645·1 = 198.4

z under the true mean

z = (198.4 − 197.7)/1 = 0.655

Power

power = P(Z < 0.655) = 74.38%
1 − β = 74.38%

Answer

1 − β = 74.38%
11.Linear regression & correlation

Answer: r = 0.9026

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 = 226/5 = 45.2

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 96
Syy = Σ(y − ȳ)² = 282.8

Correlation coefficient

r = 96/√(40·282.8) = 0.9026

Answer

r = 0.9026
12.Linear regression & correlation

Answer: e = −4.567

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 = 379/6 = 63.17

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 273

Regression line

b = 273/70 = 3.9
a = 63.17 − 3.9·6 = 39.77
ŷ = 39.77 + 3.9·x

Predicted value

ŷ = 39.77 + 3.9·2 = 47.57

Residual

e = 43 − 47.57 = −4.567

Answer

e = −4.567

The point lies below the line.

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