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: 51, 37, 55, 12, 14.

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

Exam scores are normally distributed with mean 500 and standard deviation 40. Find the z-score z of a score of 468.

μ = 500σ = 40x = 468
z =
3.ProbabilityMedium
xp(x)00.2510.1520.330.3μ = 1.65

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

x: 0, 1, 2, 3p(x): 0.25, 0.15, 0.3, 0.3
Var(X) =
4.ProbabilityHard
n = 11, p = 0.3: P(X ≤ 1) = ?k01234567891011

11 components are tested, and each one fails with probability 0.3, independently. What is the probability P that at most one fails?

n = 11p = 0.3
P =
5.Normal distributionMedium
100 kg102 kgT = X₁ + … + X₄?

The masses of bags of cement are independent and normal with mean 25 kg and standard deviation 0.5 kg. Find the probability P that the total mass of 4 of them exceeds 102 kg. Round z to two decimals.

μ = 25 kgσ = 0.5 kgn = 4> 102 kg
P =
6.Normal distributionHard
557085100115130145pointstop 95%

The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. Drag the cut-off so the shaded top 95% is to the right of it. Find the cut-off x.

μ = 100 pointsσ = 15 points95%
x = points
7.Confidence intervalsMedium
n = 9, s = 2LU = ?x̄ = 91E90%

A sample of 9 items has mean 91 and standard deviation s = 2. The population is roughly normal. Find the upper limit of the 90% confidence interval for the mean.

n = 9x̄ = 91s = 290%t(8) = 1.86
upper limit =
8.Confidence intervalsHard
μ₁ − μ₂, t(10) = 1.8120LUx̄₁ − x̄₂E = ?90%

Two fertilisers are tested. Plot group 1: 6 plots, mean yield 34 kg, standard deviation 8 kg. Plot group 2: 6 plots, mean 30.5 kg, standard deviation 2.5 kg. Assume equal variances (pooled t, df = n₁ + n₂ − 2). Find the margin of error E of the 90% confidence interval for μ₁ − μ₂.

n₁ = 6x̄₁ = 34s₁ = 8n₂ = 6x̄₂ = 30.5s₂ = 2.590%t(10) = 1.812
E = kg
9.Hypothesis tests & confidence intervalsMedium
t −2.06402.064

A sample of 25 students has a mean score of 95.84 and a sample standard deviation of 8. Find the test statistic t for H₀: μ = 100.

μ₀ = 100x̄ = 95.84s = 8n = 25
t =
10.Hypothesis tests & confidence intervalsHard
z −1.64501.645

A sample of 64 measurements has mean 100.8. The population standard deviation is 8. Find the lower limit L of the 90% confidence interval for the mean.

x̄ = 100.8σ = 8n = 6490%
L =
11.Linear regression & correlationMedium
xy15361751195221522359xy1520405060

The table gives 5 paired observations of x = temperature in °C and y = ice creams sold. Their sums are given. Find Pearson's correlation coefficient r from these sums.

n = 5Σx = 95Σy = 250Σx² = 1845Σy² = 12786Σxy = 4844
r =
12.Linear regression & correlationHard
xy125221320416516613xy246152025x = 2.5

The table gives 6 paired observations of x = car age in years and y = price in k€. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 2.5.

n = 6x = 2.5
ŷ =

3Solutions

1.Descriptive statistics

Answer: s = 20.14

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 = 169/5 = 33.8

Squared deviations

Σ(x − x̄)² = 17.2² + 3.2² + 21.2² + (−21.8)² + (−19.8)² = 1623

Standard deviation

s = √(1623/4) = 20.14

Answer

s = 20.14
2.Descriptive statistics

Answer: z = −0.8

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 = (468 − 500)/40 = −0.8

Answer

z = −0.8
3.Probability

Answer: Var(X) = 1.328

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 = 0·0.25 + 1·0.15 + 2·0.3 + 3·0.3 = 1.65

Mean of the squares

E(X²) = 0·0.25 + 1·0.15 + 4·0.3 + 9·0.3 = 4.05

Variance

Var(X) = 4.05 − 1.65² = 1.328

Answer

Var(X) = 1.328
4.Probability

Answer: P = 0.113

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) = 0.711 = 0.01977
P(X = 1) = 11·0.3·0.710 = 0.09322

Add them

P = 0.01977 + 0.09322 = 0.113

Answer

P = 0.113
5.Normal distribution

Answer: P = 0.02275

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 = 100 kg, σT = 0.5·√4 = 1 kg

Standardise

z = (102 − 100)/1 = 2

Probability

P = 1 − Φ(2) = 0.02275

Answer

P = 0.02275
6.Normal distribution

Answer: x = 75.4 points

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) = 0.05

z from the table

z ≈ −1.64

Back to the original units

x = 100 − 1.64·15 = 75.4 points

Answer

x = 75.4 points
7.Confidence intervals

Answer: upper limit = 92.24

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 = 8, t = 1.86

Margin of error

E = 1.86·2/√9 = 1.24

Interval

91 ± 1.24 → [89.76, 92.24]

Answer

upper limit = 92.24
8.Confidence intervals

Answer: E = 6.2 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 = (5·8² + 5·2.5²)/10 = 35.13
sp = 5.927

Margin of error

E = 1.812·5.927·√(1/6 + 1/6) = 6.2

Answer

E = 6.2 kg
9.Hypothesis tests & confidence intervals

Answer: t = −2.6

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 = 8/√25 = 8/5 = 1.6

Test statistic

t = (95.84 − 100)/1.6 = −2.6

Answer

t = −2.6
t0.025; df = 24 = 2.064

|t| > 2.064, so H₀ is rejected at the 5% level (two-sided).

10.Hypothesis tests & confidence intervals

Answer: L = 99.16

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 = 1.645·8/√64 = 1.645

Interval limit

L = 100.8 − 1.645 = 99.16

Answer

L = 99.16
11.Linear regression & correlation

Answer: r = 0.8789

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 = 4844 − 95·250/5 = 94
Sxx = 1845 − 95²/5 = 40
Syy = 12786 − 250²/5 = 286

Correlation coefficient

r = 94/√(40·286) = 0.8789

Answer

r = 0.8789
12.Linear regression & correlation

Answer: ŷ = 20.76

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 = 111/6 = 18.5

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −39.5

Regression line

b = −39.5/17.5 = −2.257
a = 18.5 − (−2.257)·3.5 = 26.4
ŷ = 26.4 − 2.257·x

Prediction

ŷ = 26.4 + (−2.257)·2.5 = 20.76

Answer

ŷ = 20.76

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