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: 18, 49, 27, 58, 15, 20.

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

Exam scores are normally distributed with mean 70 and standard deviation 15. Find the z-score z of a score of 52.

μ = 70σ = 15x = 52
z =
3.ProbabilityMedium
xp(x)00.110.320.2530.0540.3μ = 2.15

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

x: 0, 1, 2, 3, 4p(x): 0.1, 0.3, 0.25, 0.05, 0.3
Var(X) =
4.ProbabilityHard
n = 9, p = 0.1: P(X = 5) = ?k0123456789

A test has 9 multiple-choice questions. A student who guesses gets each one right with probability 0.1, independently. What is the probability P of exactly 5 correct answers?

n = 9p = 0.1k = 5
P =
5.Normal distributionMedium
200 mm199.5 mmX + Y?

X is normal with mean 120 mm and standard deviation 0.3 mm; Y is independent and normal with mean 80 mm and standard deviation 0.4 mm. Find the probability P that X + Y is more than 199.5 mm. Round z to two decimals.

μX = 120, σX = 0.3μY = 80, σY = 0.4X + Y > 199.5 mm
P =
6.Normal distributionHard
35455565758595pointstop 25%

The exam scores are normally distributed with mean 65 points and standard deviation 10 points. Drag the cut-off so the shaded top 25% is to the right of it. Find the cut-off x.

μ = 65 pointsσ = 10 points25%
x = points
7.Confidence intervalsMedium
μ₁ − μ₂0LU = ?x̄₁ − x̄₂E90%

The battery life of two phone models is compared. Model 1: 60 phones, mean 95 h, known standard deviation 17 h. Model 2: 60 phones, mean 81.5 h, known standard deviation 20 h. Find the upper limit of the 90% confidence interval for μ₁ − μ₂.

n₁ = 60x̄₁ = 95 hσ₁ = 17 hn₂ = 60x̄₂ = 81.5 hσ₂ = 20 h90%
upper limit = h
8.Confidence intervalsHard
p ≈ 0.1, n = ?p̂ − Ep̂ + Ep̂E = 0.0490%

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 0.04?

p ≈ 0.1E = 0.0490%
n =
9.Hypothesis tests & confidence intervalsMedium
−2.3062.306t = ?df = 8

9 employees take a typing test before and after a training course. For each person the difference d = after − before (words per minute) is recorded. The mean difference is −3.45 and the standard deviation of the differences is 7.4. Find the paired test statistic t for H₀: μd = 0.

n = 9d̄ = −3.45sd = 7.4
t =
10.Hypothesis tests & confidence intervalsHard
−1.961.96z = ?

In a trial, 113 of 250 patients on treatment A recover, and 60 of 150 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.

x₁ = 113n₁ = 250x₂ = 60n₂ = 150
z =
11.Linear regression & correlationMedium
xy146116701889201012298xy15206080100

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 = 90Σy = 419Σx² = 1660Σy² = 36347Σxy = 7752
r =
12.Linear regression & correlationHard
xy2841862583310421246xy5102040x = 5

The table gives 6 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 = 5.

n = 6x = 5
ŷ =

3Solutions

1.Descriptive statistics

Answer: s = 17.97

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 = 187/6 = 31.17

Squared deviations

Σ(x − x̄)² = (−13.17)² + 17.83² + (−4.167)² + 26.83² + (−16.17)² + (−11.17)² = 1615

Standard deviation

s = √(1615/5) = 17.97

Answer

s = 17.97
2.Descriptive statistics

Answer: z = −1.2

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 = (52 − 70)/15 = −1.2

Answer

z = −1.2
3.Probability

Answer: Var(X) = 1.928

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 = 0·0.1 + 1·0.3 + 2·0.25 + 3·0.05 + 4·0.3 = 2.15

Mean of the squares

E(X²) = 0·0.1 + 1·0.3 + 4·0.25 + 9·0.05 + 16·0.3 = 6.55

Variance

Var(X) = 6.55 − 2.15² = 1.928

Answer

Var(X) = 1.928
4.Probability

Answer: P = 0.0008267

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 = 5) = 126·0.15·0.94 = 126·1 × 10⁻⁵·0.6561 = 0.0008267

Answer

P = 0.0008267
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 sum

μ = 200 mm, σ = √(0.3² + 0.4²) = 0.5 mm

Standardise

z = (199.5 − 200)/0.5 = −1

Upper tail

P = 1 − Φ(−1) = 1 − 0.1587 = 0.8413

Answer

P = 0.8413
6.Normal distribution

Answer: x = 71.7 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) = 0.75

z from the table

z ≈ 0.67

Back to the original units

x = 65 + 0.67·10 = 71.7 points

Answer

x = 71.7 points
7.Confidence intervals

Answer: upper limit = 19.07 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 = 95 − 81.5 = 13.5

Standard error

SE = √(17²/60 + 20²/60) = 3.389

Margin of error

E = 1.645·3.389 = 5.574

Interval

13.5 ± 5.574 → [7.926, 19.07]

Answer

upper limit = 19.07 h

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

8.Confidence intervals

Answer: n = 153

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 = 1.645²·0.1·0.9/0.04² = 152.2

Round up

n = 153

Answer

n = 153
9.Hypothesis tests & confidence intervals

Answer: t = −1.399

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 of the mean difference

SE = 7.4/√9 = 2.467

Test statistic

t = −3.45/2.467 = −1.399

Answer

t = −1.399

With df = 8, the two-sided 5% critical value is 2.306; |t| is smaller, so H₀ is not rejected.

10.Hypothesis tests & confidence intervals

Answer: z = 1.016

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.

Sample proportions

p̂₁ = 113/250 = 0.452
p̂₂ = 60/150 = 0.4
p̂ = 173/400 = 0.4325

Standard error under H₀

SE = √(0.4325·0.5675·(1/250 + 1/150)) = 0.05117

Test statistic

z = (0.452 − 0.4)/0.05117 = 1.016

Answer

z = 1.016

|z| < 1.96, so H₀ is not rejected at the 5% level (two-sided).

11.Linear regression & correlation

Answer: r = 0.9449

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 = 7752 − 90·419/5 = 210
Sxx = 1660 − 90²/5 = 40
Syy = 36347 − 419²/5 = 1235

Correlation coefficient

r = 210/√(40·1235) = 0.9449

Answer

r = 0.9449
12.Linear regression & correlation

Answer: ŷ = 20.95

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 = 42/6 = 7
ȳ = 172/6 = 28.67

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 270

Regression line

b = 270/70 = 3.857
a = 28.67 − 3.857·7 = 1.667
ŷ = 1.667 + 3.857·x

Prediction

ŷ = 1.667 + 3.857·5 = 20.95

Answer

ŷ = 20.95

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