Introductory Statistics: cram pack

Formula sheets, the key method steps for every topic and 12 practice problems, with full solutions at the back.

New problemsBack to the course

In the print window, choose “Save as PDF” to keep a copy on your device.

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
101520253035404550

Find the sample standard deviation s of the data: 50, 41, 10, 28, 36.

n = 5
s =
2.Descriptive statisticsHard
z −2−1012

Exam scores are normally distributed with mean 100 and standard deviation 5. Find the z-score z of a score of 107.5.

μ = 100σ = 5x = 107.5
z =
3.ProbabilityMedium
xp(x)10.320.2530.3540.1μ = 2.25

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

x: 1, 2, 3, 4p(x): 0.3, 0.25, 0.35, 0.1
Var(X) =
4.ProbabilityHard
n = 10, p = 0.25: P(X = 2) = ?k012345678910

A test has 10 multiple-choice questions. A student who guesses gets each one right with probability 0.25, independently. What is the probability P of exactly 2 correct answers?

n = 10p = 0.25k = 2
P =
5.Normal distributionMedium
48 kg45.6 kgT = X₁ + … + X₄?

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 4 of them is below 45.6 kg. Round z to two decimals.

μ = 12 kgσ = 0.8 kgn = 4< 45.6 kg
P =
6.Normal distributionHard
10 cm0X − Y?

The heights X of men and Y of women in a group are independent and normal: X has mean 180 cm and standard deviation 10 cm, Y has mean 170 cm and standard deviation 12 cm. Find the probability P that X is greater than Y. Round z to two decimals.

μX = 180 cm, σX = 10 cmμY = 170 cm, σY = 12 cm
P =
7.Confidence intervalsMedium
μ₁ − μ₂0LU = ?x̄₁ − x̄₂E90%

The battery life of two phone models is compared. Model 1: 60 phones, mean 137.5 h, known standard deviation 18 h. Model 2: 60 phones, mean 146 h, known standard deviation 11 h. Find the upper limit of the 90% confidence interval for μ₁ − μ₂.

n₁ = 60x̄₁ = 137.5 hσ₁ = 18 hn₂ = 60x̄₂ = 146 hσ₂ = 11 h90%
upper limit = h
8.Confidence intervalsHard
p = 0.5, n = ?p̂ − Ep̂ + Ep̂E = 0.0299%

Nothing is known in advance about the proportion of defective products. How many products n must be inspected so that a 99% confidence interval for the proportion has a margin of error of at most 0.02?

p = 0.5E = 0.0299%
n =
9.Hypothesis tests & confidence intervalsMedium
−2.0862.086t = ?df = 20

Two teaching methods are compared. Group 1: 13 students, mean score 74, standard deviation 9.5. Group 2: 9 students, mean score 84.1, standard deviation 11. 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 = 13x̄₁ = 74s₁ = 9.5n₂ = 9x̄₂ = 84.1s₂ = 11
t =
10.Hypothesis tests & confidence intervalsHard
σ = 4, n = ?x̄ − Ex̄ + Ex̄E = 2.590%

How large must a sample be so that the 90% confidence interval for a mean has a margin of error of at most 2.5? The standard deviation is 4. Find the smallest sample size n.

σ = 4E = 2.590%
n =
11.Linear regression & correlationMedium
xy2541161581910201224xy5101020

The table gives 6 paired observations of x = load in kN and y = extension in mm. Find Pearson's correlation coefficient r.

n = 6
r =
12.Linear regression & correlationHard
xy3445527629771191xy5106080x = 6

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. Predict ŷ for x = 6.

n = 5x = 6
ŷ =

3Solutions

1.Descriptive statistics

Answer: s = 15.13

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 = 165/5 = 33

Squared deviations

Σ(x − x̄)² = 17² + 8² + (−23)² + (−5)² + 3² = 916

Standard deviation

s = √(916/4) = 15.13

Answer

s = 15.13
2.Descriptive statistics

Answer: z = 1.5

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 = (107.5 − 100)/5 = 1.5

Answer

z = 1.5
3.Probability

Answer: Var(X) = 0.9875

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 = 1·0.3 + 2·0.25 + 3·0.35 + 4·0.1 = 2.25

Mean of the squares

E(X²) = 1·0.3 + 4·0.25 + 9·0.35 + 16·0.1 = 6.05

Variance

Var(X) = 6.05 − 2.25² = 0.9875

Answer

Var(X) = 0.9875
4.Probability

Answer: P = 0.2816

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 = 2) = 45·0.252·0.758 = 45·0.0625·0.1001 = 0.2816

Answer

P = 0.2816
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 = 48 kg, σT = 0.8·√4 = 1.6 kg

Standardise

z = (45.6 − 48)/1.6 = −1.5

Probability

P = Φ(−1.5) = 0.06681

Answer

P = 0.06681
6.Normal distribution

Answer: P = 0.7389

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 = 180 − 170 = 10 cm, σ = 15.62 cm

Standardise

z = (0 − 10)/15.62 = −0.64

Probability

P = 1 − Φ(−0.64) = 0.7389

Answer

P = 0.7389
7.Confidence intervals

Answer: upper limit = −4.02 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 = 137.5 − 146 = −8.5

Standard error

SE = √(18²/60 + 11²/60) = 2.723

Margin of error

E = 1.645·2.723 = 4.48

Interval

−8.5 ± 4.48 → [−12.98, −4.02]

Answer

upper limit = −4.02 h

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

8.Confidence intervals

Answer: n = 4148

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 = 2.576²·0.5·0.5/0.02² = 4147

Round up

n = 4148

Answer

n = 4148
9.Hypothesis tests & confidence intervals

Answer: t = −2.3

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 = (12·9.5² + 8·11²)/20 = 102.5
sp = 10.13

Standard error

SE = 10.13·√(1/13 + 1/9) = 4.391

Test statistic

t = (74 − 84.1)/4.391 = −2.3

Answer

t = −2.3

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: n = 7

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).

Solve for n

n = (1.645·4/2.5)² = 6.927

Round up

n = 7

Answer

n = 7
11.Linear regression & correlation

Answer: r = 0.9817

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 = 94/6 = 15.67

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 126
Syy = Σ(y − ȳ)² = 235.3

Correlation coefficient

r = 126/√(70·235.3) = 0.9817

Answer

r = 0.9817
12.Linear regression & correlation

Answer: ŷ = 59.25

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 = 35/5 = 7
ȳ = 326/5 = 65.2

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 238

Regression line

b = 238/40 = 5.95
a = 65.2 − 5.95·7 = 23.55
ŷ = 23.55 + 5.95·x

Prediction

ŷ = 23.55 + 5.95·6 = 59.25

Answer

ŷ = 59.25

ExamProblems · examproblems.com/courses/statistics-1