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: 26, 13, 33, 24, 25, 57.

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

A quantity is normally distributed with mean 500 and standard deviation 75. What proportion P of values is below 350?

μ = 500σ = 75x = 350
P =
3.ProbabilityMedium
n = 8k = 5N = ?

In how many ways N can a committee of 5 people be chosen from a group of 8? The order does not matter.

n = 8k = 5
N =
4.ProbabilityHard
n = 9, p = 0.4: P(X ≤ 1) = ?k0123456789

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

n = 9p = 0.4
P =
5.Normal distributionMedium
77.5123.25P = ?

The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. What fraction lies between 77.5 and 123.25 points? Give P as a decimal.

μ = 100σ = 15x1 = 77.5x2 = 123.25
P =
6.Normal distributionHard
557085100115130145pointsbottom 10%

The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. Drag the cut-off so the shaded bottom 10% is to the left of it. Find the cut-off x.

μ = 100 pointsσ = 15 points10%
x = points
7.Confidence intervalsMedium
μ₁ − μ₂0L = ?Ux̄₁ − x̄₂E95%

The battery life of two phone models is compared. Model 1: 80 phones, mean 129 h, known standard deviation 5 h. Model 2: 40 phones, mean 135.5 h, known standard deviation 12 h. Find the lower limit of the 95% confidence interval for μ₁ − μ₂.

n₁ = 80x̄₁ = 129 hσ₁ = 5 hn₂ = 40x̄₂ = 135.5 hσ₂ = 12 h95%
lower limit = h
8.Confidence intervalsHard
σ = 11, n = ?x̄ − Ex̄ + Ex̄E = 2.595%

The population standard deviation is 11. How large a sample n do you need so that a 95% confidence interval for the mean has a margin of error of at most 2.5?

σ = 11E = 2.595%
n =
9.Hypothesis tests & confidence intervalsMedium
−1.961.96z = ?

Two production lines make the same part. Line 1: a sample of 80 parts has mean length 142.5 mm, and the line's known standard deviation is 7 mm. Line 2: a sample of 60 parts has mean 145.9 mm, known standard deviation 9 mm. Find the test statistic z for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (line 1 minus line 2).

n₁ = 80x̄₁ = 142.5 mmσ₁ = 7 mmn₂ = 60x̄₂ = 145.9 mmσ₂ = 9 mm
z =
10.Hypothesis tests & confidence intervalsHard
H₀H₁μ₀x̄cμ₀ + 1α = 0.05βσ = 10, power 80%, n = ?

A one-sided z test at α = 0.05 must detect a shift of the mean of 1 units with a power of 80%. The standard deviation is 10. Find the minimum sample size n.

α = 0.05power = 80%δ = 1σ = 10zα = 1.645zβ = 0.842
n =
11.Linear regression & correlationMedium
xy3525627649751191xy5106080

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

n = 5Σx = 35Σy = 344Σx² = 285Σy² = 24550Σxy = 2590
r =
12.Linear regression & correlationHard
xy24496148131016xy51051015x = 7

The table gives 5 paired observations of x = load in kN and y = extension in mm. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 7.

n = 5x = 7
ŷ =

3Solutions

1.Descriptive statistics

Answer: s = 14.85

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 = 178/6 = 29.67

Squared deviations

Σ(x − x̄)² = (−3.667)² + (−16.67)² + 3.333² + (−5.667)² + (−4.667)² + 27.33² = 1103

Standard deviation

s = √(1103/5) = 14.85

Answer

s = 14.85
2.Descriptive statistics

Answer: P = 0.02275

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 = −2

Standard normal table

Φ(−2) = 0.02275
P = 0.02275

Answer

P = 0.02275
3.Probability

Answer: N = 56

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(8, 5) = (8·7·6·5·4)/(5·4·3·2·1) = 56

Answer

N = 56
4.Probability

Answer: P = 0.07054

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) = 0.69 = 0.01008
P(X = 1) = 9·0.4·0.68 = 0.06047

Add them

P = 0.01008 + 0.06047 = 0.07054

Answer

P = 0.07054
5.Normal distribution

Answer: P = 0.8726

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 = (77.5 − 100)/15 = −1.5
z₂ = (123.25 − 100)/15 = 1.55

Area between

P = Φ(1.55) − Φ(−1.5) = 0.9394 − 0.06681 = 0.8726

Answer

P = 0.8726
6.Normal distribution

Answer: x = 80.8 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) = 0.1

z from the table

z ≈ −1.28

Back to the original units

x = 100 − 1.28·15 = 80.8 points

Answer

x = 80.8 points
7.Confidence intervals

Answer: lower limit = −10.38 h

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.

Point estimate

x̄₁ − x̄₂ = 129 − 135.5 = −6.5

Standard error

SE = √(5²/80 + 12²/40) = 1.978

Margin of error

E = 1.96·1.978 = 3.877

Interval

−6.5 ± 3.877 → [−10.38, −2.623]

Answer

lower limit = −10.38 h

The interval does not contain 0, so the means differ significantly.

8.Confidence intervals

Answer: n = 75

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 = (1.96·11/2.5)² = 74.37

Round up

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

n = 75

Answer

n = 75
9.Hypothesis tests & confidence intervals

Answer: z = −2.427

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.

Standard error of the difference

SE = √(7²/80 + 9²/60) = √1.963 = 1.401

Test statistic

z = (142.5 − 145.9)/1.401 = −2.427

Answer

z = −2.427

|z| > 1.96, so H₀ is rejected at the 5% level (two-sided).

10.Hypothesis tests & confidence intervals

Answer: n = 619

Sample size for a test: to detect a true shift δ with power 1 − β in a one-sided z test at level α, the standard error must satisfy δ = (z_α + z_β)·σ/√n, so n = ((z_α + z_β)·σ/δ)², rounded up.

Solve for n

n = ((1.645 + 0.842)·10/1)² = 618.5

Round up

n = 619

Answer

n = 619
11.Linear regression & correlation

Answer: r = 0.9685

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 = 2590 − 35·344/5 = 182
Sxx = 285 − 35²/5 = 40
Syy = 24550 − 344²/5 = 882.8

Correlation coefficient

r = 182/√(40·882.8) = 0.9685

Answer

r = 0.9685
12.Linear regression & correlation

Answer: ŷ = 12.6

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 = 30/5 = 6
ȳ = 56/5 = 11.2

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 56

Regression line

b = 56/40 = 1.4
a = 11.2 − 1.4·6 = 2.8
ŷ = 2.8 + 1.4·x

Prediction

ŷ = 2.8 + 1.4·7 = 12.6

Answer

ŷ = 12.6

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