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: 15, 47, 28, 41, 28, 10, 24.

n = 7

s =

2.Descriptive statisticsHard

Exam scores are normally distributed with mean 50 and standard deviation 15. Find the z-score z of a score of 27.5.

μ = 50σ = 15x = 27.5

z =

3.ProbabilityMedium

In how many ways N can a committee of 4 people be chosen from a group of 12? The order does not matter.

n = 12k = 4

N =

4.ProbabilityHard

A test has 5 multiple-choice questions. A student who guesses gets each one right with probability 0.4, independently. What is the probability P of exactly 1 correct answers?

n = 5p = 0.4k = 1

P =

5.Normal distributionMedium

X is normal with mean 120 mm and standard deviation 0.3 mm; Y is independent and normal with mean 80 mm and standard deviation 0.4 mm. Find the probability P that X + Y is more than 199.8 mm. Round z to two decimals.

μ_X = 120, σ_X = 0.3μ_Y = 80, σ_Y = 0.4X + Y > 199.8 mm

P =

6.Normal distributionHard

The bag weights are normally distributed with mean 250 g and standard deviation 5 g. Drag the cut-off so the shaded top 40% is to the right of it. Find the cut-off x.

μ = 250 gσ = 5 g40%

x = g

7.Confidence intervalsMedium

A sample of 30 items has mean 74 and standard deviation s = 11.5. The population is roughly normal. Find the lower limit of the 99% confidence interval for the mean.

n = 30x̄ = 74s = 11.599%t(29) = 2.756

lower limit =

8.Confidence intervalsHard

The population standard deviation is 20. 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 1?

σ = 20E = 199%

n =

9.Hypothesis tests & confidence intervalsMedium

Two teaching methods are compared. Group 1: 6 students, mean score 59, standard deviation 4. Group 2: 6 students, mean score 55.2, standard deviation 6. Assume equal population variances and use the pooled two-sample t test. Find the test statistic t for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (group 1 minus group 2).

n₁ = 6x̄₁ = 59s₁ = 4n₂ = 6x̄₂ = 55.2s₂ = 6

t =

10.Hypothesis tests & confidence intervalsHard

A sample of 16 measurements has mean 201.4. The population standard deviation is 5. Find the lower limit L of the 95% confidence interval for the mean.

x̄ = 201.4σ = 5n = 1695%

L =

11.Linear regression & correlationMedium

The table gives 5 paired observations of x = car age in years and y = price in k€. Their sums are given. Find Pearson's correlation coefficient r from these sums.

n = 5Σx = 15Σy = 84Σx² = 55Σy² = 1486Σxy = 226

r =

12.Linear regression & correlationHard

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. A new observation (10, 72) is made. Find its residual e = y − ŷ.

n = 5(10, 72)

e =

3Solutions

1.Descriptive statistics

Answer: s = 13.15

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 = 193/7 = 27.57

Squared deviations

Σ(x − x̄)² = (−12.57)² + 19.43² + 0.4286² + 13.43² + 0.4286² + (−17.57)² + (−3.571)² = 1038

Standard deviation

s = √(1038/6) = 13.15

Answer

s = 13.15

2.Descriptive statistics

Answer: z = −1.5

Standardise

z = (27.5 − 50)/15 = −1.5

Answer

z = −1.5

3.Probability

Answer: N = 495

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.

Combinations

C(12, 4) = (12·11·10·9)/(4·3·2·1) = 495

Answer

N = 495

4.Probability

Answer: P = 0.2592

Binomial formula

P(X = 1) = 5·0.4¹·0.6⁴ = 5·0.4·0.1296 = 0.2592

Answer

P = 0.2592

5.Normal distribution

Answer: P = 0.6554

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) = μ_X ± μ_Y. Variances always add, also for a difference: Var(X ± Y) = σ_X² + σ_Y², so σ = √(σ_X² + σ_Y²). 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 = 200 mm, σ = √(0.3² + 0.4²) = 0.5 mm

Standardise

z = (199.8 − 200)/0.5 = −0.4

Upper tail

P = 1 − Φ(−0.4) = 1 − 0.3446 = 0.6554

Answer

P = 0.6554

6.Normal distribution

Answer: x = 251.3 g

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) = 0.6

z from the table

z ≈ 0.25

Back to the original units

x = 250 + 0.25·5 = 251.3 g

Answer

x = 251.3 g

7.Confidence intervals

Answer: lower limit = 68.21

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 = 29, t = 2.756

Margin of error

E = 2.756·11.5/√30 = 5.787

Interval

74 ± 5.787 → [68.21, 79.79]

Answer

lower limit = 68.21

8.Confidence intervals

Answer: n = 2655

Solve for n

n = (2.576·20/1)² = 2654

Round up

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

n = 2655

Answer

n = 2655

9.Hypothesis tests & confidence intervals

Answer: t = 1.291

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.

Pooled variance

s_p² = (5·4² + 5·6²)/10 = 26

s_p = 5.099

Standard error

SE = 5.099·√(1/6 + 1/6) = 2.944

Test statistic

t = (59 − 55.2)/2.944 = 1.291

Answer

t = 1.291

With df = 10, the two-sided 5% critical value is 2.228; |t| is smaller, so H₀ is not rejected.

10.Hypothesis tests & confidence intervals

Answer: L = 199

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 = 1.96·5/√16 = 2.45

Interval limit

L = 201.4 − 2.45 = 199

Answer

L = 199

11.Linear regression & correlation

Answer: r = −0.9507

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.

Centred sums

S_xy = 226 − 15·84/5 = −26

S_xx = 55 − 15²/5 = 10

S_yy = 1486 − 84²/5 = 74.8

Correlation coefficient

r = −26/√(10·74.8) = −0.9507

Answer

r = −0.9507

12.Linear regression & correlation

Answer: e = 5.45

Means

x̄ = 35/5 = 7

ȳ = 272/5 = 54.4

Sums of squares and products

S_xx = Σ(x − x̄)² = 40

S_xy = Σ(x − x̄)(y − ȳ) = 162

Regression line

b = 162/40 = 4.05

a = 54.4 − 4.05·7 = 26.05

ŷ = 26.05 + 4.05·x

Predicted value

ŷ = 26.05 + 4.05·10 = 66.55

Residual

e = 72 − 66.55 = 5.45

Answer

e = 5.45

The point lies above the line.

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