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: 17, 53, 38, 47, 40, 14.

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

Exam scores are normally distributed with mean 70 and standard deviation 12. Find the z-score z of a score of 77.2.

μ = 70σ = 12x = 77.2
z =
3.ProbabilityMedium
red: 5blue: 3

A bag holds 5 red and 3 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?

5 / 3
P =
4.ProbabilityHard
n = 12, p = 0.3: P(X = 4) = ?k0123456789101112

A test has 12 multiple-choice questions. A student who guesses gets each one right with probability 0.3, independently. What is the probability P of exactly 4 correct answers?

n = 12p = 0.3k = 4
P =
5.Normal distributionMedium
100115.75P = ?

The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. What fraction lies between 100 and 115.75 points? Give P as a decimal.

μ = 100σ = 15x1 = 100x2 = 115.75
P =
6.Normal distributionHard
380420460500540580620htop 80%

The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. Drag the cut-off so the shaded top 80% is to the right of it. Find the cut-off x.

μ = 500 hσ = 40 h80%
x = h
7.Confidence intervalsMedium
n = 15, s = 9LU = ?x̄ = 108E95%

A sample of 15 items has mean 108 and standard deviation s = 9. The population is roughly normal. Find the upper limit of the 95% confidence interval for the mean.

n = 15x̄ = 108s = 995%t(14) = 2.145
upper limit =
8.Confidence intervalsHard
p ≈ 0.3, n = ?p̂ − Ep̂ + Ep̂E = 0.0390%

A pilot study suggests that about 0.3 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 0.03?

p ≈ 0.3E = 0.0390%
n =
9.Hypothesis tests & confidence intervalsMedium
−2.0862.086t = ?df = 20

Two teaching methods are compared. Group 1: 5 students, mean score 26, standard deviation 3.5. Group 2: 17 students, mean score 34.3, standard deviation 6.5. Assume equal population variances and use the pooled two-sample t test. Find the test statistic t for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (group 1 minus group 2).

n₁ = 5x̄₁ = 26s₁ = 3.5n₂ = 17x̄₂ = 34.3s₂ = 6.5
t =
10.Hypothesis tests & confidence intervalsHard
z −1.9601.96

A sample of 49 measurements has mean 503. The population standard deviation is 8. Find the lower limit L of the 95% confidence interval for the mean.

x̄ = 503σ = 8n = 4995%
L =
11.Linear regression & correlationMedium
xy362463575672784890xy46860708090

The table gives 6 paired observations of x = hours studied and y = exam score. Find Pearson's correlation coefficient r.

n = 6
r =
12.Linear regression & correlationHard
xy149253355462567xy245060x = 4.5

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. A new observation (4.5, 66) is made. Find its residual e = y − ŷ.

n = 5(4.5, 66)
e =

3Solutions

1.Descriptive statistics

Answer: s = 15.92

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 = 209/6 = 34.83

Squared deviations

Σ(x − x̄)² = (−17.83)² + 18.17² + 3.167² + 12.17² + 5.167² + (−20.83)² = 1267

Standard deviation

s = √(1267/5) = 15.92

Answer

s = 15.92
2.Descriptive statistics

Answer: z = 0.6

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 = (77.2 − 70)/12 = 0.6

Answer

z = 0.6
3.Probability

Answer: P = 0.3571

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.

First and second draw

P = 5/8 · 4/7 = 20/56 = 0.3571

Answer

P = 0.3571
4.Probability

Answer: P = 0.2311

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 = 4) = 495·0.34·0.78 = 495·0.0081·0.05765 = 0.2311

Answer

P = 0.2311
5.Normal distribution

Answer: P = 0.3531

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 = (100 − 100)/15 = 0
z₂ = (115.75 − 100)/15 = 1.05

Area between

P = Φ(1.05) − Φ(0) = 0.8531 − 0.5 = 0.3531

Answer

P = 0.3531
6.Normal distribution

Answer: x = 466 h

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) = 0.2

z from the table

z ≈ −0.84

Back to the original units

x = 500 − 0.84·40 = 466 h

Answer

x = 466 h
7.Confidence intervals

Answer: upper limit = 113

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 = 14, t = 2.145

Margin of error

E = 2.145·9/√15 = 4.985

Interval

108 ± 4.985 → [103, 113]

Answer

upper limit = 113
8.Confidence intervals

Answer: n = 632

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 = 1.645²·0.3·0.7/0.03² = 631.4

Round up

n = 632

Answer

n = 632
9.Hypothesis tests & confidence intervals

Answer: t = −2.71

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.

Pooled variance

sp² = (4·3.5² + 16·6.5²)/20 = 36.25
sp = 6.021

Standard error

SE = 6.021·√(1/5 + 1/17) = 3.063

Test statistic

t = (26 − 34.3)/3.063 = −2.71

Answer

t = −2.71

With df = 20, the two-sided 5% critical value is 2.086; |t| is larger, so H₀ is rejected.

10.Hypothesis tests & confidence intervals

Answer: L = 500.8

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/√49 = 2.24

Interval limit

L = 503 − 2.24 = 500.8

Answer

L = 500.8
11.Linear regression & correlation

Answer: r = 0.9559

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 = 33/6 = 5.5
ȳ = 446/6 = 74.33

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = 100
Syy = Σ(y − ȳ)² = 625.3

Correlation coefficient

r = 100/√(17.5·625.3) = 0.9559

Answer

r = 0.9559
12.Linear regression & correlation

Answer: e = 2.05

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 = 286/5 = 57.2

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = 45

Regression line

b = 45/10 = 4.5
a = 57.2 − 4.5·3 = 43.7
ŷ = 43.7 + 4.5·x

Predicted value

ŷ = 43.7 + 4.5·4.5 = 63.95

Residual

e = 66 − 63.95 = 2.05

Answer

e = 2.05

The point lies above the line.

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