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: 58, 34, 53, 55, 17, 53.

n = 6

s =

2.Descriptive statisticsHard

Exam scores are normally distributed with mean 60 and standard deviation 12. Find the z-score z of a score of 48.

μ = 60σ = 12x = 48

z =

3.ProbabilityMedium

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

x: 1, 2, 3, 4, 5p(x): 0.05, 0.35, 0.15, 0.35, 0.1

Var(X) =

4.ProbabilityHard

12 components are tested, and each one fails with probability 0.4, independently. What is the probability P that at most one fails?

n = 12p = 0.4

P =

5.Normal distributionMedium

The bag weights are normally distributed with mean 250 g and standard deviation 5 g. What fraction lies between 245 and 257.5 g? Give P as a decimal.

μ = 250σ = 5x₁ = 245x₂ = 257.5

P =

6.Normal distributionHard

The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. Find the value x below which 10% of them lie (the 10th percentile).

μ = 500σ = 40P = 10%

x = h

7.Confidence intervalsMedium

A 90% confidence interval for a mean, based on 36 observations and a known σ, has a margin of error of 7.7. What is the margin of error E₂ of a 99% interval from the same data?

90%n₁ = 36E₁ = 7.799%

E₂ =

8.Confidence intervalsHard

The population standard deviation is 35. How large a sample n do you need so that a 95% confidence interval for the mean has a margin of error of at most 5?

σ = 35E = 595%

n =

9.Hypothesis tests & confidence intervalsMedium

A sample of 49 students has a mean score of 94.69 and a sample standard deviation of 12. Find the test statistic t for H₀: μ = 100.

μ₀ = 100x̄ = 94.69s = 12n = 49

t =

10.Hypothesis tests & confidence intervalsHard

How large must a sample be so that the 99% confidence interval for a mean has a margin of error of at most 2? The standard deviation is 8. Find the smallest sample size n.

σ = 8E = 299%

n =

11.Linear regression & correlationMedium

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

The table gives 5 paired observations of x = hours studied and y = exam score. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 2.

n = 5x = 2

ŷ =

3Solutions

1.Descriptive statistics

Answer: s = 16.14

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 = 270/6 = 45

Squared deviations

Σ(x − x̄)² = 13² + (−11)² + 8² + 10² + (−28)² + 8² = 1302

Standard deviation

s = √(1302/5) = 16.14

Answer

s = 16.14

2.Descriptive statistics

Answer: z = −1

Standardise

z = (48 − 60)/12 = −1

Answer

z = −1

3.Probability

Answer: Var(X) = 1.29

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 = 1·0.05 + 2·0.35 + 3·0.15 + 4·0.35 + 5·0.1 = 3.1

Mean of the squares

E(X²) = 1·0.05 + 4·0.35 + 9·0.15 + 16·0.35 + 25·0.1 = 10.9

Variance

Var(X) = 10.9 − 3.1² = 1.29

Answer

Var(X) = 1.29

4.Probability

Answer: P = 0.01959

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) = 0.6¹² = 0.002177

P(X = 1) = 12·0.4·0.6¹¹ = 0.01741

Add them

P = 0.002177 + 0.01741 = 0.01959

Answer

P = 0.01959

5.Normal distribution

Answer: P = 0.7745

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 = (245 − 250)/5 = −1

z₂ = (257.5 − 250)/5 = 1.5

Area between

P = Φ(1.5) − Φ(−1) = 0.9332 − 0.1587 = 0.7745

Answer

P = 0.7745

6.Normal distribution

Answer: x = 448.7 h

z for the percentile

Φ(z) = 0.1 → z = −1.282

Back to x

x = 500 + −1.282·40 = 448.74 h

Answer

x = 448.74 h

7.Confidence intervals

Answer: E₂ = 12.06

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.

Ratio of margins

E₂/E₁ = (2.576/1.645)·√(36/36) = 1.566

New margin

E₂ = 7.7·1.566 = 12.06

Answer

E₂ = 12.06

8.Confidence intervals

Answer: n = 189

Solve for n

n = (1.96·35/5)² = 188.2

Round up

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

n = 189

Answer

n = 189

9.Hypothesis tests & confidence intervals

Answer: t = −3.098

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 = 12/√49 = 12/7 = 1.714

Test statistic

t = (94.69 − 100)/1.714 = −3.098

Answer

t = −3.098

t_{0.025; df = 48} = 2.011

|t| > 2.011, so H₀ is rejected at the 5% level (two-sided).

10.Hypothesis tests & confidence intervals

Answer: n = 107

Solve for n

n = (2.576·8/2)² = 106.2

Round up

n = 107

Answer

n = 107

11.Linear regression & correlation

Answer: r = 0.9967

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 = 35/5 = 7

ȳ = 359/5 = 71.8

Sums of squares and products

S_xx = Σ(x − x̄)² = 40

S_xy = Σ(x − x̄)(y − ȳ) = 196

S_yy = Σ(y − ȳ)² = 966.8

Correlation coefficient

r = 196/√(40·966.8) = 0.9967

Answer

r = 0.9967

12.Linear regression & correlation

Answer: ŷ = 39.65

Means

x̄ = 25/5 = 5

ȳ = 238/5 = 47.6

Sums of squares and products

S_xx = Σ(x − x̄)² = 40

S_xy = Σ(x − x̄)(y − ȳ) = 106

Regression line

b = 106/40 = 2.65

a = 47.6 − 2.65·5 = 34.35

ŷ = 34.35 + 2.65·x

Prediction

ŷ = 34.35 + 2.65·2 = 39.65

Answer

ŷ = 39.65

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