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
1520253035404550

Find the sample standard deviation s of the data: 49, 13, 28, 12, 40, 50.

n = 6
s =
2.Descriptive statisticsHard
z −2−1012

A quantity is normally distributed with mean 70 and standard deviation 20. What proportion P of values is above 95?

μ = 70σ = 20x = 95
P =
3.ProbabilityMedium
xp(x)00.310.1520.230.35μ = 1.6

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

x: 0, 1, 2, 3p(x): 0.3, 0.15, 0.2, 0.35
σ =
4.ProbabilityHard
n = 9, p = 0.6: P(X ≤ 1) = ?k0123456789

9 components are tested, and each one fails with probability 0.6, independently. What is the probability P that at most one fails?

n = 9p = 0.6
P =
5.Normal distributionMedium
2333.2P = ?

The delivery times are normally distributed with mean 32 min and standard deviation 6 min. What fraction lies between 23 and 33.2 min? Give P as a decimal.

μ = 32σ = 6x1 = 23x2 = 33.2
P =
6.Normal distributionHard
μ = 100x = ?99%

The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. Find the value x below which 99% of them lie (the 99th percentile).

μ = 100σ = 15P = 99%
x = points
7.Confidence intervalsMedium
x̄E₁ = 1.799%, n = 100x̄E₂ = ?99%, n = 900

A 99% confidence interval for a mean, based on 100 observations and a known σ, has a margin of error of 1.7. What is the margin of error E₂ if the same study uses 900 observations instead?

99%n₁ = 100E₁ = 1.7n₂ = 900
E₂ =
8.Confidence intervalsHard
μ₁ − μ₂, t(29) = 2.0450L = ?Ux̄₁ − x̄₂E95%

Two fertilisers are tested. Plot group 1: 12 plots, mean yield 94.5 kg, standard deviation 2 kg. Plot group 2: 19 plots, mean 106.5 kg, standard deviation 8.5 kg. Assume equal variances (pooled t, df = n₁ + n₂ − 2). Find the lower limit of the 95% confidence interval for μ₁ − μ₂.

n₁ = 12x̄₁ = 94.5s₁ = 2n₂ = 19x̄₂ = 106.5s₂ = 8.595%t(29) = 2.045
lower limit = kg
9.Hypothesis tests & confidence intervalsMedium
t −2.01102.011

A sample of 49 students has a mean score of 101.7 and a sample standard deviation of 5. Find the test statistic t for H₀: μ = 100.

μ₀ = 100x̄ = 101.7s = 5n = 49
t =
10.Hypothesis tests & confidence intervalsHard
μ₀x̄c

A one-sided z test of H₀: μ = 100 against H₁: μ > 100 is done at α = 0.01 with a sample of 36; the standard deviation is known to be 18. In reality the mean is 108.1. Find the power 1 − β of the test in percent.

μ₀ = 100μ = 108.1σ = 18n = 36α = 0.01
1 − β = %
11.Linear regression & correlationMedium
xy244961381610181221xy5101020

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

n = 6
r =
12.Linear regression & correlationHard
xy3465587639741184xy5106080x = 8

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. Predict ŷ for x = 8.

n = 5x = 8
ŷ =

3Solutions

1.Descriptive statistics

Answer: s = 17.05

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 = 192/6 = 32

Squared deviations

Σ(x − x̄)² = 17² + (−19)² + (−4)² + (−20)² + 8² + 18² = 1454

Standard deviation

s = √(1454/5) = 17.05

Answer

s = 17.05
2.Descriptive statistics

Answer: P = 0.1056

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 = 1.25

Standard normal table

Φ(1.25) = 0.8944
P = 1 − 0.8944 = 0.1056

Answer

P = 0.1056
3.Probability

Answer: σ = 1.241

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 + 1·0.15 + 2·0.2 + 3·0.35 = 1.6

Mean of the squares

E(X²) = 0·0.3 + 1·0.15 + 4·0.2 + 9·0.35 = 4.1

Variance

Var(X) = 4.1 − 1.6² = 1.54
σ = √1.54 = 1.241

Answer

σ = 1.241
4.Probability

Answer: P = 0.003801

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) = 0.49 = 0.0002621
P(X = 1) = 9·0.6·0.48 = 0.003539

Add them

P = 0.0002621 + 0.003539 = 0.003801

Answer

P = 0.003801
5.Normal distribution

Answer: P = 0.5125

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 = (23 − 32)/6 = −1.5
z₂ = (33.2 − 32)/6 = 0.2

Area between

P = Φ(0.2) − Φ(−1.5) = 0.5793 − 0.06681 = 0.5125

Answer

P = 0.5125
6.Normal distribution

Answer: x = 134.9 points

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) = 0.99 → z = 2.326

Back to x

x = 100 + 2.326·15 = 134.89 points

Answer

x = 134.89 points
7.Confidence intervals

Answer: E₂ = 0.5667

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 = (2.576/2.576)·√(100/900) = 0.3333

New margin

E₂ = 1.7·0.3333 = 0.5667

Answer

E₂ = 0.5667
8.Confidence intervals

Answer: lower limit = −17.13 kg

Interval for a difference: estimate ± critical value × standard error. For two independent means with known σ: SE = √(σ₁²/n₁ + σ₂²/n₂) and z. With equal unknown variances: pooled s_p and t with n₁ + n₂ − 2 degrees of freedom. For paired data, work with the differences: d̄ ± t·s_d/√n.

Pooled standard deviation

sp² = (11·2² + 18·8.5²)/29 = 46.36
sp = 6.809

Margin of error

E = 2.045·6.809·√(1/12 + 1/19) = 5.134

Interval

−12 ± 5.134 → [−17.13, −6.866]

Answer

lower limit = −17.13 kg
9.Hypothesis tests & confidence intervals

Answer: t = 2.38

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/√49 = 5/7 = 0.7143

Test statistic

t = (101.7 − 100)/0.7143 = 2.38

Answer

t = 2.38
t0.025; df = 48 = 2.011

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

10.Hypothesis tests & confidence intervals

Answer: 1 − β = 64.58 %

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 = 18/√36 = 3
x̄c = 100 + 2.326·3 = 107

z under the true mean

z = (107 − 108.1)/3 = −0.374

Power

power = P(Z > −0.374) = 64.58%
1 − β = 64.58%

Answer

1 − β = 64.58%
11.Linear regression & correlation

Answer: r = 0.9881

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 = 42/6 = 7
ȳ = 81/6 = 13.5

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 115
Syy = Σ(y − ȳ)² = 193.5

Correlation coefficient

r = 115/√(70·193.5) = 0.9881

Answer

r = 0.9881
12.Linear regression & correlation

Answer: ŷ = 69.6

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 = 35/5 = 7
ȳ = 325/5 = 65

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 184

Regression line

b = 184/40 = 4.6
a = 65 − 4.6·7 = 32.8
ŷ = 32.8 + 4.6·x

Prediction

ŷ = 32.8 + 4.6·8 = 69.6

Answer

ŷ = 69.6

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