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
2025303540455055

Find the sample standard deviation s of the data: 34, 17, 20, 56, 26, 38.

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

A quantity is normally distributed with mean 70 and standard deviation 8. What proportion P of values is above 86?

μ = 70σ = 8x = 86
P =
3.ProbabilityMedium
n = 8k = 4N = ?

In how many ways N can a committee of 4 people be chosen from a group of 8? The order does not matter.

n = 8k = 4
N =
4.ProbabilityHard
n = 6, p = 0.2: P(X ≤ 1) = ?k0123456

6 components are tested, and each one fails with probability 0.2, independently. What is the probability P that at most one fails?

n = 6p = 0.2
P =
5.Normal distributionMedium
192 kg196.8 kgT = X₁ + … + X₁₆?

The masses of boxes are independent and normal with mean 12 kg and standard deviation 0.8 kg. Find the probability P that the total mass of 16 of them exceeds 196.8 kg. Round z to two decimals.

μ = 12 kgσ = 0.8 kgn = 16> 196.8 kg
P =
6.Normal distributionHard
0.03 mm0D − d?

Hole diameters D are normal with mean 25.03 mm and standard deviation 0.02 mm. Shaft diameters d are independent and normal with mean 25 mm and standard deviation 0.02 mm. A randomly chosen shaft fits when D − d > 0. Find the probability P that it fits. Round z to two decimals.

hole: μ = 25.03 mm, σ = 0.02 mmshaft: μ = 25 mm, σ = 0.02 mm
P =
7.Confidence intervalsMedium
n = 30, s = 14L = ?Ux̄ = 65.5E90%

A sample of 30 items has mean 65.5 and standard deviation s = 14. The population is roughly normal. Find the lower limit of the 90% confidence interval for the mean.

n = 30x̄ = 65.5s = 1490%t(29) = 1.699
lower limit =
8.Confidence intervalsHard
p ≈ 0.6, n = ?p̂ − Ep̂ + Ep̂E = 0.0190%

A pilot study suggests that about 0.6 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 0.01?

p ≈ 0.6E = 0.0190%
n =
9.Hypothesis tests & confidence intervalsMedium
−2.2282.228t = ?df = 10

Two teaching methods are compared. Group 1: 7 students, mean score 98, standard deviation 2. Group 2: 5 students, mean score 103.2, standard deviation 10.5. 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 = 7x̄₁ = 98s₁ = 2n₂ = 5x̄₂ = 103.2s₂ = 10.5
t =
10.Hypothesis tests & confidence intervalsHard
H₀H₁μ₀x̄cμ₀ + 3α = 0.01βσ = 18, power 80%, n = ?

A one-sided z test at α = 0.01 must detect a shift of the mean of 3 units with a power of 80%. The standard deviation is 18. Find the minimum sample size n.

α = 0.01power = 80%δ = 3σ = 18zα = 2.326zβ = 0.842
n =
11.Linear regression & correlationMedium
xy37347658669171028102xy4688090100

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
xy256372467575679793xy2466080x = 2.5

The table gives 6 paired observations of x = hours studied and y = exam score. 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 = 14.29

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 = 191/6 = 31.83

Squared deviations

Σ(x − x̄)² = 2.167² + (−14.83)² + (−11.83)² + 24.17² + (−5.833)² + 6.167² = 1021

Standard deviation

s = √(1021/5) = 14.29

Answer

s = 14.29
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) = 0.9772
P = 1 − 0.9772 = 0.02275

Answer

P = 0.02275
3.Probability

Answer: N = 70

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, 4) = (8·7·6·5)/(4·3·2·1) = 70

Answer

N = 70
4.Probability

Answer: P = 0.6554

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) = 0.86 = 0.2621
P(X = 1) = 6·0.2·0.85 = 0.3932

Add them

P = 0.2621 + 0.3932 = 0.6554

Answer

P = 0.6554
5.Normal distribution

Answer: P = 0.06681

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 = 192 kg, σT = 0.8·√16 = 3.2 kg

Standardise

z = (196.8 − 192)/3.2 = 1.5

Probability

P = 1 − Φ(1.5) = 0.06681

Answer

P = 0.06681
6.Normal distribution

Answer: P = 0.8554

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 difference

μ = 25.03 − 25 = 0.03 mm, σ = 0.02828 mm

Standardise

z = (0 − 0.03)/0.02828 = −1.06

Probability

P = 1 − Φ(−1.06) = 0.8554

Answer

P = 0.8554
7.Confidence intervals

Answer: lower limit = 61.16

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 = 29, t = 1.699

Margin of error

E = 1.699·14/√30 = 4.343

Interval

65.5 ± 4.343 → [61.16, 69.84]

Answer

lower limit = 61.16
8.Confidence intervals

Answer: n = 6495

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 = 1.645²·0.6·0.4/0.01² = 6494

Round up

n = 6495

Answer

n = 6495
9.Hypothesis tests & confidence intervals

Answer: t = −1.302

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 = (6·2² + 4·10.5²)/10 = 46.5
sp = 6.819

Standard error

SE = 6.819·√(1/7 + 1/5) = 3.993

Test statistic

t = (98 − 103.2)/3.993 = −1.302

Answer

t = −1.302

With df = 10, the two-sided 5% critical value is 2.228; |t| is smaller, so H₀ is not rejected.

10.Hypothesis tests & confidence intervals

Answer: n = 362

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 = ((2.326 + 0.842)·18/3)² = 361.3

Round up

n = 362

Answer

n = 362
11.Linear regression & correlation

Answer: r = 0.9799

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 = 530/6 = 88.33

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = 114
Syy = Σ(y − ȳ)² = 773.3

Correlation coefficient

r = 114/√(17.5·773.3) = 0.9799

Answer

r = 0.9799
12.Linear regression & correlation

Answer: ŷ = 61.44

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 = 27/6 = 4.5
ȳ = 442/6 = 73.67

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = 107

Regression line

b = 107/17.5 = 6.114
a = 73.67 − 6.114·4.5 = 46.15
ŷ = 46.15 + 6.114·x

Prediction

ŷ = 46.15 + 6.114·2.5 = 61.44

Answer

ŷ = 61.44

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