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)·p^k(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

Convert the value to a z-score: z = (x − μ)/σ.
Look up Φ(z), the area to the left of z.
For the area to the right use 1 − Φ(z); for an interval subtract the two left areas.
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

Decide between z (population σ known) and t (only the sample s, with n − 1 degrees of freedom).
Find the critical value for the confidence level.
Margin of error: E = critical value × standard deviation/√n.
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

State H₀ and H₁ and decide whether the test is one- or two-sided.
Calculate the test statistic: (x̄ − μ₀)/(σ/√n), or with s and t.
Compare with the critical value, or find the p-value.
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: S_xx = Σ(x − x̄)² = Σx² − (Σx)²/n, S_xy = Σxy − ΣxΣy/n
Slope: b = S_xy/S_xx
Intercept: a = ȳ − b·x̄
Correlation: r = S_xy/√(S_xxS_yy)
Residual: e = y − ŷ

2Practice problems

1.Descriptive statisticsMedium

Find the sample standard deviation s of the data: 38, 11, 41, 46, 43.

n = 5

s =

2.Descriptive statisticsHard

Exam scores are normally distributed with mean 500 and standard deviation 25. Find the z-score z of a score of 531.3.

μ = 500σ = 25x = 531.3

z =

3.ProbabilityMedium

For the distribution shown, find the standard deviation σ of X.

x: 0, 1, 2, 3p(x): 0.15, 0.35, 0.15, 0.35

σ =

4.ProbabilityHard

A game costs €5 to play. You win €15 with probability 0.1 and €250 with probability 0.05; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.

stake = €5€15: p = 0.1€250: p = 0.05

E(G) = €

5.Normal distributionMedium

The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. What fraction lies between 174.5 and 192 cm? Give P as a decimal.

μ = 178σ = 7x₁ = 174.5x₂ = 192

P =

6.Normal distributionHard

The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. Drag the cut-off so the shaded top 20% is to the right of it. Find the cut-off x.

μ = 500 hσ = 40 h20%

x = h

7.Confidence intervalsMedium

A sample of 6 items has mean 66.5 and standard deviation s = 17.5. The population is roughly normal. Find the upper limit of the 99% confidence interval for the mean.

n = 6x̄ = 66.5s = 17.599%t(5) = 4.032

upper limit =

8.Confidence intervalsHard

The population standard deviation is 22. 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 0.5?

σ = 22E = 0.590%

n =

9.Hypothesis tests & confidence intervalsMedium

A sample of 25 students has a mean score of 497.4 and a sample standard deviation of 5. Find the test statistic t for H₀: μ = 500.

μ₀ = 500x̄ = 497.4s = 5n = 25

t =

10.Hypothesis tests & confidence intervalsHard

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 5. Find the smallest sample size n.

σ = 5E = 0.595%

n =

11.Linear regression & correlationMedium

The table gives 6 paired observations of x = car age in years and y = price in k€. Find Pearson's correlation coefficient r.

n = 6

r =

12.Linear regression & correlationHard

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. Predict ŷ for x = 9.

n = 5x = 9

ŷ =

3Solutions

1.Descriptive statistics

Answer: s = 14.17

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 = 179/5 = 35.8

Squared deviations

Σ(x − x̄)² = 2.2² + (−24.8)² + 5.2² + 10.2² + 7.2² = 802.8

Standard deviation

s = √(802.8/4) = 14.17

Answer

s = 14.17

2.Descriptive statistics

Answer: z = 1.25

Standardise

z = (531.3 − 500)/25 = 1.25

Answer

z = 1.25

3.Probability

Answer: σ = 1.1

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 σ_Y = |a|·σ_X; adding a constant does not change the spread.

Mean

μ = 0·0.15 + 1·0.35 + 2·0.15 + 3·0.35 = 1.7

Mean of the squares

E(X²) = 0·0.15 + 1·0.35 + 4·0.15 + 9·0.35 = 4.1

Variance

Var(X) = 4.1 − 1.7² = 1.21

σ = √1.21 = 1.1

Answer

σ = 1.1

4.Probability

Answer: E(G) = 9 €

Net outcomes

G: 10 (p = 0.1), 245 (p = 0.05), −5 (p = 0.85)

Expected value

E(G) = 10·0.1 + 245·0.05 − 5·0.85 = 9 €

Answer

E(G) = 9 €

A positive value means you win money on average.

5.Normal distribution

Answer: P = 0.6687

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-scores

z₁ = (174.5 − 178)/7 = −0.5

z₂ = (192 − 178)/7 = 2

Area between

P = Φ(2) − Φ(−0.5) = 0.9772 − 0.3085 = 0.6687

Answer

P = 0.6687

6.Normal distribution

Answer: x = 534 h

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) = 0.8

z from the table

z ≈ 0.84

Back to the original units

x = 500 + 0.84·40 = 534 h

Answer

x = 534 h

7.Confidence intervals

Answer: upper limit = 95.31

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 = 5, t = 4.032

Margin of error

E = 4.032·17.5/√6 = 28.81

Interval

66.5 ± 28.81 → [37.69, 95.31]

Answer

upper limit = 95.31

8.Confidence intervals

Answer: n = 5239

Solve for n

n = (1.645·22/0.5)² = 5239

Round up

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

n = 5239

Answer

n = 5239

9.Hypothesis tests & confidence intervals

Answer: t = −2.6

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/√25 = 5/5 = 1

Test statistic

t = (497.4 − 500)/1 = −2.6

Answer

t = −2.6

t_{0.025; df = 24} = 2.064

|t| > 2.064, so H₀ is rejected at the 5% level (two-sided).

10.Hypothesis tests & confidence intervals

Answer: n = 385

Solve for n

n = (1.96·5/0.5)² = 384.2

Round up

n = 385

Answer

n = 385

11.Linear regression & correlation

Answer: r = −0.9333

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 = 108/6 = 18

Sums of squares and products

S_xx = Σ(x − x̄)² = 17.5

S_xy = Σ(x − x̄)(y − ȳ) = −50

S_yy = Σ(y − ȳ)² = 164

Correlation coefficient

r = −50/√(17.5·164) = −0.9333

Answer

r = −0.9333

12.Linear regression & correlation

Answer: ŷ = 38.15

Means

x̄ = 30/5 = 6

ȳ = 124/5 = 24.8

Sums of squares and products

S_xx = Σ(x − x̄)² = 40

S_xy = Σ(x − x̄)(y − ȳ) = 178

Regression line

b = 178/40 = 4.45

a = 24.8 − 4.45·6 = −1.9

ŷ = −1.9 + 4.45·x

Prediction

ŷ = −1.9 + 4.45·9 = 38.15

Answer

ŷ = 38.15

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