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
152025303540455055

Find the sample standard deviation s of the data: 15, 24, 57, 27, 42, 53, 59.

n = 7
s =
2.Descriptive statisticsHard
z −2−1012

Exam scores are normally distributed with mean 170 and standard deviation 12. Find the z-score z of a score of 188.

μ = 170σ = 12x = 188
z =
3.ProbabilityMedium
xp(x)10.1520.3530.2540.25μ = 2.6

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

x: 1, 2, 3, 4p(x): 0.15, 0.35, 0.25, 0.25
Var(X) =
4.ProbabilityHard
n = 10, p = 0.7: P(X = 2) = ?k012345678910

A test has 10 multiple-choice questions. A student who guesses gets each one right with probability 0.7, independently. What is the probability P of exactly 2 correct answers?

n = 10p = 0.7k = 2
P =
5.Normal distributionMedium
850 kg887.9 kgT = X₁ + … + X₁₀?

The masses of passengers with luggage are independent and normal with mean 85 kg and standard deviation 12 kg. Find the probability P that the total mass of 10 of them exceeds 887.9 kg. Round z to two decimals.

μ = 85 kgσ = 12 kgn = 10> 887.9 kg
P =
6.Normal distributionHard
μ = 32x = ?5%

The delivery times are normally distributed with mean 32 min and standard deviation 6 min. Find the value x below which 5% of them lie (the 5th percentile).

μ = 32σ = 6P = 5%
x = min
7.Confidence intervalsMedium
x̄E₁ = 6.695%, n = 100x̄E₂ = ?90%, n = 25

A 95% confidence interval for a mean, based on 100 observations and a known σ, has a margin of error of 6.6. What is the margin of error E₂ of a 90% interval based on 25 observations?

95%n₁ = 100E₁ = 6.690%n₂ = 25
E₂ =
8.Confidence intervalsHard
p ≈ 0.7, n = ?p̂ − Ep̂ + Ep̂E = 0.0595%

A pilot study suggests that about 0.7 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 0.05?

p ≈ 0.7E = 0.0595%
n =
9.Hypothesis tests & confidence intervalsMedium
t −2.06402.064

A sample of 25 students has a mean score of 197.2 and a sample standard deviation of 10. Find the test statistic t for H₀: μ = 200.

μ₀ = 200x̄ = 197.2s = 10n = 25
t =
10.Hypothesis tests & confidence intervalsHard
μ₀x̄c

A one-sided z test of H₀: μ = 100 against H₁: μ < 100 is done at α = 0.05 with a sample of 100; the standard deviation is known to be 20. In reality the mean is 95.6. Find the probability of a Type II error β in percent.

μ₀ = 100μ = 95.6σ = 20n = 100α = 0.05
β = %
11.Linear regression & correlationMedium
xy121217318414510xy24101520

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

n = 5
r =
12.Linear regression & correlationHard
xy145247351447552656xy246455055x = 5.5

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. Predict ŷ for x = 5.5.

n = 6x = 5.5
ŷ =

3Solutions

1.Descriptive statistics

Answer: s = 17.66

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 = 277/7 = 39.57

Squared deviations

Σ(x − x̄)² = (−24.57)² + (−15.57)² + 17.43² + (−12.57)² + 2.429² + 13.43² + 19.43² = 1872

Standard deviation

s = √(1872/6) = 17.66

Answer

s = 17.66
2.Descriptive statistics

Answer: z = 1.5

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 = (188 − 170)/12 = 1.5

Answer

z = 1.5
3.Probability

Answer: Var(X) = 1.04

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 = 1·0.15 + 2·0.35 + 3·0.25 + 4·0.25 = 2.6

Mean of the squares

E(X²) = 1·0.15 + 4·0.35 + 9·0.25 + 16·0.25 = 7.8

Variance

Var(X) = 7.8 − 2.6² = 1.04

Answer

Var(X) = 1.04
4.Probability

Answer: P = 0.001447

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.

Binomial formula

P(X = 2) = 45·0.72·0.38 = 45·0.49·6.561 × 10⁻⁵ = 0.001447

Answer

P = 0.001447
5.Normal distribution

Answer: P = 0.1587

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 total

μT = 850 kg, σT = 12·√10 = 37.95 kg

Standardise

z = (887.9 − 850)/37.95 = 1

Probability

P = 1 − Φ(1) = 0.1587

Answer

P = 0.1587
6.Normal distribution

Answer: x = 22.13 min

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) = 0.05 → z = −1.645

Back to x

x = 32 + −1.645·6 = 22.13 min

Answer

x = 22.13 min
7.Confidence intervals

Answer: E₂ = 11.08

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 = (1.645/1.96)·√(100/25) = 1.679

New margin

E₂ = 6.6·1.679 = 11.08

Answer

E₂ = 11.08
8.Confidence intervals

Answer: n = 323

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 = 1.96²·0.7·0.3/0.05² = 322.7

Round up

n = 323

Answer

n = 323
9.Hypothesis tests & confidence intervals

Answer: t = −1.4

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 = 10/√25 = 10/5 = 2

Test statistic

t = (197.2 − 200)/2 = −1.4

Answer

t = −1.4
t0.025; df = 24 = 2.064

|t| < 2.064, so H₀ is not rejected at the 5% level (two-sided).

10.Hypothesis tests & confidence intervals

Answer: β = 28.94 %

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 = 20/√100 = 2
x̄c = 100 − 1.645·2 = 96.71

z under the true mean

z = (96.71 − 95.6)/2 = 0.555

Type II error

power = P(Z < 0.555) = 71.06%
β = 100% − 71.06% = 28.94%

Answer

β = 28.94%
11.Linear regression & correlation

Answer: r = −0.9449

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 = 15/5 = 3
ȳ = 80/5 = 16

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = −25
Syy = Σ(y − ȳ)² = 70

Correlation coefficient

r = −25/√(10·70) = −0.9449

Answer

r = −0.9449
12.Linear regression & correlation

Answer: ŷ = 53.44

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 = 298/6 = 49.67

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = 33

Regression line

b = 33/17.5 = 1.886
a = 49.67 − 1.886·3.5 = 43.07
ŷ = 43.07 + 1.886·x

Prediction

ŷ = 43.07 + 1.886·5.5 = 53.44

Answer

ŷ = 53.44

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