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
253035404550

Find the sample standard deviation s of the data: 37, 21, 32, 52, 38.

n = 5
s =
2.Descriptive statisticsHard
z −2−1012

Exam scores are normally distributed with mean 170 and standard deviation 20. Find the z-score z of a score of 195.

μ = 170σ = 20x = 195
z =
3.ProbabilityMedium
xp(x)00.310.2520.0530.240.2μ = 1.75

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

x: 0, 1, 2, 3, 4p(x): 0.3, 0.25, 0.05, 0.2, 0.2
Var(X) =
4.ProbabilityHard
n = 5, p = 0.4: P(X ≤ 1) = ?k012345

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

n = 5p = 0.4
P =
5.Normal distributionMedium
454574P = ?

The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. What fraction lies between 454 and 574 h? Give P as a decimal.

μ = 500σ = 40x1 = 454x2 = 574
P =
6.Normal distributionHard
μ = 500x = ?90%

The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. Find the value x below which 90% of them lie (the 90th percentile).

μ = 500σ = 40P = 90%
x = h
7.Confidence intervalsMedium
n = 7, s = 18LU = ?x̄ = 21.5E95%

A sample of 7 items has mean 21.5 and standard deviation s = 18. The population is roughly normal. Find the upper limit of the 95% confidence interval for the mean.

n = 7x̄ = 21.5s = 1895%t(6) = 2.447
upper limit =
8.Confidence intervalsHard
σ = 11, n = ?x̄ − Ex̄ + Ex̄E = 199%

The population standard deviation is 11. 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 = 11E = 199%
n =
9.Hypothesis tests & confidence intervalsMedium
−2.1452.145t = ?df = 14

Two teaching methods are compared. Group 1: 6 students, mean score 82, standard deviation 10. Group 2: 10 students, mean score 91.2, standard deviation 7.5. 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 = 82s₁ = 10n₂ = 10x̄₂ = 91.2s₂ = 7.5
t =
10.Hypothesis tests & confidence intervalsHard
z −1.9601.96

A sample of 36 measurements has mean 84.3. The population standard deviation is 20. Find the lower limit L of the 95% confidence interval for the mean.

x̄ = 84.3σ = 20n = 3695%
L =
11.Linear regression & correlationMedium
xy2441061081410211223xy5101020

The table gives 6 paired observations of x = load in kN and y = extension in mm. Their sums are given. Find Pearson's correlation coefficient r from these sums.

n = 6Σx = 42Σy = 82Σx² = 364Σy² = 1382Σxy = 706
r =
12.Linear regression & correlationHard
xy1468178020922310226129xy15202580100120x = 15.5

The table gives 5 paired observations of x = temperature in °C and y = ice creams sold. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 15.5.

n = 5x = 15.5
ŷ =

3Solutions

1.Descriptive statistics

Answer: s = 11.2

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 = 180/5 = 36

Squared deviations

Σ(x − x̄)² = 1² + (−15)² + (−4)² + 16² + 2² = 502

Standard deviation

s = √(502/4) = 11.2

Answer

s = 11.2
2.Descriptive statistics

Answer: z = 1.25

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 = (195 − 170)/20 = 1.25

Answer

z = 1.25
3.Probability

Answer: Var(X) = 2.388

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 = 0·0.3 + 1·0.25 + 2·0.05 + 3·0.2 + 4·0.2 = 1.75

Mean of the squares

E(X²) = 0·0.3 + 1·0.25 + 4·0.05 + 9·0.2 + 16·0.2 = 5.45

Variance

Var(X) = 5.45 − 1.75² = 2.388

Answer

Var(X) = 2.388
4.Probability

Answer: P = 0.337

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) = 0.65 = 0.07776
P(X = 1) = 5·0.4·0.64 = 0.2592

Add them

P = 0.07776 + 0.2592 = 0.337

Answer

P = 0.337
5.Normal distribution

Answer: P = 0.8428

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 = (454 − 500)/40 = −1.15
z₂ = (574 − 500)/40 = 1.85

Area between

P = Φ(1.85) − Φ(−1.15) = 0.9678 − 0.1251 = 0.8428

Answer

P = 0.8428
6.Normal distribution

Answer: x = 551.3 h

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) = 0.9 → z = 1.282

Back to x

x = 500 + 1.282·40 = 551.26 h

Answer

x = 551.26 h
7.Confidence intervals

Answer: upper limit = 38.15

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 = 6, t = 2.447

Margin of error

E = 2.447·18/√7 = 16.65

Interval

21.5 ± 16.65 → [4.852, 38.15]

Answer

upper limit = 38.15
8.Confidence intervals

Answer: n = 803

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 = (2.576·11/1)² = 802.9

Round up

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

n = 803

Answer

n = 803
9.Hypothesis tests & confidence intervals

Answer: t = −2.101

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

sp² = (5·10² + 9·7.5²)/14 = 71.88
sp = 8.478

Standard error

SE = 8.478·√(1/6 + 1/10) = 4.378

Test statistic

t = (82 − 91.2)/4.378 = −2.101

Answer

t = −2.101

With df = 14, the two-sided 5% critical value is 2.145; |t| is smaller, so H₀ is not rejected.

10.Hypothesis tests & confidence intervals

Answer: L = 77.77

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 = 1.96·20/√36 = 6.533

Interval limit

L = 84.3 − 6.533 = 77.77

Answer

L = 77.77
11.Linear regression & correlation

Answer: r = 0.9759

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

Sxy = 706 − 42·82/6 = 132
Sxx = 364 − 42²/6 = 70
Syy = 1382 − 82²/6 = 261.3

Correlation coefficient

r = 132/√(70·261.3) = 0.9759

Answer

r = 0.9759
12.Linear regression & correlation

Answer: ŷ = 72.6

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 = 100/5 = 20
ȳ = 471/5 = 94.2

Sums of squares and products

Sxx = Σ(x − x̄)² = 90
Sxy = Σ(x − x̄)(y − ȳ) = 432

Regression line

b = 432/90 = 4.8
a = 94.2 − 4.8·20 = −1.8
ŷ = −1.8 + 4.8·x

Prediction

ŷ = −1.8 + 4.8·15.5 = 72.6

Answer

ŷ = 72.6

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