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
25303540455055

Find the sample standard deviation s of the data: 44, 50, 24, 55, 50, 26.

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

Exam scores are normally distributed with mean 100 and standard deviation 15. Find the z-score z of a score of 106.

μ = 100σ = 15x = 106
z =
3.ProbabilityMedium
P(at least one six) = ??1?2?3?4?5P(six) = 1/6 on each roll

A fair die is rolled 5 times. What is the probability P of getting at least one six?

n = 5
P =
4.ProbabilityHard
n = 7, p = 0.5: P(X ≤ 1) = ?k01234567

7 components are tested, and each one fails with probability 0.5, independently. What is the probability P that at most one fails?

n = 7p = 0.5
P =
5.Normal distributionMedium
400 kg398 kgT = X₁ + … + X₁₆?

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 16 of them exceeds 398 kg. Round z to two decimals.

μ = 25 kgσ = 0.5 kgn = 16> 398 kg
P =
6.Normal distributionHard
35455565758595pointstop 90%

The exam scores are normally distributed with mean 65 points and standard deviation 10 points. Drag the cut-off so the shaded top 90% is to the right of it. Find the cut-off x.

μ = 65 pointsσ = 10 points90%
x = points
7.Confidence intervalsMedium
μ₁ − μ₂0LUx̄₁ − x̄₂E = ?99%

The battery life of two phone models is compared. Model 1: 30 phones, mean 52.5 h, known standard deviation 3 h. Model 2: 100 phones, mean 51.5 h, known standard deviation 10 h. Find the margin of error E of the 99% confidence interval for μ₁ − μ₂.

n₁ = 30x̄₁ = 52.5 hσ₁ = 3 hn₂ = 100x̄₂ = 51.5 hσ₂ = 10 h99%
E = h
8.Confidence intervalsHard
p ≈ 0.1, n = ?p̂ − Ep̂ + Ep̂E = 0.0590%

A pilot study suggests that about 0.1 of products are defective. How many products n must be inspected so that a 90% confidence interval for the proportion has a margin of error of at most 0.05?

p ≈ 0.1E = 0.0590%
n =
9.Hypothesis tests & confidence intervalsMedium
−1.961.96z = ?

Two production lines make the same part. Line 1: a sample of 40 parts has mean length 63 mm, and the line's known standard deviation is 11 mm. Line 2: a sample of 30 parts has mean 70.7 mm, known standard deviation 13 mm. Find the test statistic z for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (line 1 minus line 2).

n₁ = 40x̄₁ = 63 mmσ₁ = 11 mmn₂ = 30x̄₂ = 70.7 mmσ₂ = 13 mm
z =
10.Hypothesis tests & confidence intervalsHard
z −1.9601.96

A sample of 36 measurements has mean 46.4. The population standard deviation is 8. Find the upper limit U of the 95% confidence interval for the mean.

x̄ = 46.4σ = 8n = 3695%
U =
11.Linear regression & correlationMedium
xy24486118131018xy51051015

The table gives 5 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 = 5Σx = 30Σy = 54Σx² = 220Σy² = 694Σxy = 390
r =
12.Linear regression & correlationHard
xy124219321420515xy24152025x = 3.5

The table gives 5 paired observations of x = car age in years and y = price in k€. Use the least-squares line ŷ = a + b·x. A new observation (3.5, 23) is made. Find its residual e = y − ŷ.

n = 5(3.5, 23)
e =

3Solutions

1.Descriptive statistics

Answer: s = 13.26

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 = 249/6 = 41.5

Squared deviations

Σ(x − x̄)² = 2.5² + 8.5² + (−17.5)² + 13.5² + 8.5² + (−15.5)² = 879.5

Standard deviation

s = √(879.5/5) = 13.26

Answer

s = 13.26
2.Descriptive statistics

Answer: z = 0.4

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 = (106 − 100)/15 = 0.4

Answer

z = 0.4
3.Probability

Answer: P = 0.5981

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.

Complement

P(no six) = (5/6)5 = 0.4019
P = 1 − 0.4019 = 0.5981

Answer

P = 0.5981
4.Probability

Answer: P = 0.0625

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) = 0.57 = 0.007813
P(X = 1) = 7·0.5·0.56 = 0.05469

Add them

P = 0.007813 + 0.05469 = 0.0625

Answer

P = 0.0625
5.Normal distribution

Answer: P = 0.8413

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 = 400 kg, σT = 0.5·√16 = 2 kg

Standardise

z = (398 − 400)/2 = −1

Probability

P = 1 − Φ(−1) = 0.8413

Answer

P = 0.8413
6.Normal distribution

Answer: x = 52.2 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 = 65 − 1.28·10 = 52.2 points

Answer

x = 52.2 points
7.Confidence intervals

Answer: E = 2.937 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 = 52.5 − 51.5 = 1

Standard error

SE = √(3²/30 + 10²/100) = 1.14

Margin of error

E = 2.576·1.14 = 2.937

Answer

E = 2.937 h

The interval contains 0, so the data do not show a significant difference.

8.Confidence intervals

Answer: n = 98

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 = 1.645²·0.1·0.9/0.05² = 97.42

Round up

n = 98

Answer

n = 98
9.Hypothesis tests & confidence intervals

Answer: z = −2.617

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 = √(11²/40 + 13²/30) = √8.658 = 2.943

Test statistic

z = (63 − 70.7)/2.943 = −2.617

Answer

z = −2.617

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

10.Hypothesis tests & confidence intervals

Answer: U = 49.01

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 = 1.96·8/√36 = 2.613

Interval limit

U = 46.4 + 2.613 = 49.01

Answer

U = 49.01
11.Linear regression & correlation

Answer: r = 0.9914

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 = 390 − 30·54/5 = 66
Sxx = 220 − 30²/5 = 40
Syy = 694 − 54²/5 = 110.8

Correlation coefficient

r = 66/√(40·110.8) = 0.9914

Answer

r = 0.9914
12.Linear regression & correlation

Answer: e = 4.05

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 = 99/5 = 19.8

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = −17

Regression line

b = −17/10 = −1.7
a = 19.8 − (−1.7)·3 = 24.9
ŷ = 24.9 − 1.7·x

Predicted value

ŷ = 24.9 + (−1.7)·3.5 = 18.95

Residual

e = 23 − 18.95 = 4.05

Answer

e = 4.05

The point lies above the line.

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