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
253035404550

Find the sample standard deviation s of the data: 35, 38, 50, 32, 25, 35.

n = 6
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 95.

μ = 100σ = 5x = 95
z =
3.ProbabilityMedium
red: 8blue: 5

A bag holds 8 red and 5 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?

8 / 5
P =
4.ProbabilityHard
€p(x)−10.6520.3490.05

A game costs €1 to play. You win €3 with probability 0.3 and €50 with probability 0.05; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.

stake = €1€3: p = 0.3€50: p = 0.05
E(G) = €
5.Normal distributionMedium
510 kg568.8 kgT = X₁ + … + X₆?

The masses of passengers with luggage are independent and normal with mean 85 kg and standard deviation 12 kg. Find the probability P that the total mass of 6 of them is below 568.8 kg. Round z to two decimals.

μ = 85 kgσ = 12 kgn = 6< 568.8 kg
P =
6.Normal distributionHard
10 g0X − Y?

The mass X of a product from line 1 and Y from line 2 are independent and normal: X has mean 500 g and standard deviation 8 g, Y has mean 490 g and standard deviation 6 g. Find the probability P that X is greater than Y. Round z to two decimals.

μX = 500 g, σX = 8 gμY = 490 g, σY = 6 g
P =
7.Confidence intervalsMedium
μ₁ − μ₂0LUx̄₁ − x̄₂E = ?95%

The battery life of two phone models is compared. Model 1: 30 phones, mean 76.5 h, known standard deviation 9 h. Model 2: 60 phones, mean 64.5 h, known standard deviation 6 h. Find the margin of error E of the 95% confidence interval for μ₁ − μ₂.

n₁ = 30x̄₁ = 76.5 hσ₁ = 9 hn₂ = 60x̄₂ = 64.5 hσ₂ = 6 h95%
E = h
8.Confidence intervalsHard
μ₁ − μ₂, t(29) = 2.7560LU = ?x̄₁ − x̄₂E99%

Two fertilisers are tested. Plot group 1: 10 plots, mean yield 20 kg, standard deviation 6.5 kg. Plot group 2: 21 plots, mean 21.5 kg, standard deviation 6 kg. Assume equal variances (pooled t, df = n₁ + n₂ − 2). Find the upper limit of the 99% confidence interval for μ₁ − μ₂.

n₁ = 10x̄₁ = 20s₁ = 6.5n₂ = 21x̄₂ = 21.5s₂ = 699%t(29) = 2.756
upper limit = kg
9.Hypothesis tests & confidence intervalsMedium
−1.84p = ?

The mean of a population is claimed to be 120; the population standard deviation is known to be 15. A random sample of 100 observations has a mean of 117.24. Test H₀: μ = 120 against H₁: μ < 120. Find the p-value p.

μ₀ = 120σ = 15n = 100x̄ = 117.24H₁: μ < μ₀
p =
10.Hypothesis tests & confidence intervalsHard
−1.961.96z = ?

In a trial, 105 of 150 patients on treatment A recover, and 78 of 120 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.

x₁ = 105n₁ = 150x₂ = 78n₂ = 120
z =
11.Linear regression & correlationMedium
xy15411739194921672364xy152040506070

The table gives 5 paired observations of x = temperature in °C and y = ice creams sold. Find the coefficient of determination r² of the least-squares line.

n = 5
r² =
12.Linear regression & correlationHard
xy18532059226124812689xy20256080ŷ = 54

The table gives 5 paired observations of x = temperature in °C and y = ice creams sold. Use the least-squares line ŷ = a + b·x. For which value x does the line predict ŷ = 54?

n = 5ŷ = 54
x =

3Solutions

1.Descriptive statistics

Answer: s = 8.232

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 = 215/6 = 35.83

Squared deviations

Σ(x − x̄)² = (−0.8333)² + 2.167² + 14.17² + (−3.833)² + (−10.83)² + (−0.8333)² = 338.8

Standard deviation

s = √(338.8/5) = 8.232

Answer

s = 8.232
2.Descriptive statistics

Answer: z = −1

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 = (95 − 100)/5 = −1

Answer

z = −1
3.Probability

Answer: P = 0.359

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 = 8/13 · 7/12 = 56/156 = 0.359

Answer

P = 0.359
4.Probability

Answer: E(G) = 2.4 €

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.

Net outcomes

G: 2 (p = 0.3), 49 (p = 0.05), −1 (p = 0.65)

Expected value

E(G) = 2·0.3 + 49·0.05 − 1·0.65 = 2.4 €

Answer

E(G) = 2.4 €

A positive value means you win money on average.

5.Normal distribution

Answer: P = 0.9772

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 = 510 kg, σT = 12·√6 = 29.39 kg

Standardise

z = (568.8 − 510)/29.39 = 2

Probability

P = Φ(2) = 0.9772

Answer

P = 0.9772
6.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 difference

μ = 500 − 490 = 10 g, σ = 10 g

Standardise

z = (0 − 10)/10 = −1

Probability

P = 1 − Φ(−1) = 0.8413

Answer

P = 0.8413
7.Confidence intervals

Answer: E = 3.561 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 = 76.5 − 64.5 = 12

Standard error

SE = √(9²/30 + 6²/60) = 1.817

Margin of error

E = 1.96·1.817 = 3.561

Answer

E = 3.561 h

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

8.Confidence intervals

Answer: upper limit = 5.022 kg

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.

Pooled standard deviation

sp² = (9·6.5² + 20·6²)/29 = 37.94
sp = 6.16

Margin of error

E = 2.756·6.16·√(1/10 + 1/21) = 6.522

Interval

−1.5 ± 6.522 → [−8.022, 5.022]

Answer

upper limit = 5.022 kg
9.Hypothesis tests & confidence intervals

Answer: p = 0.03288

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.

Test statistic

SE = σ/√n = 15/√100 = 1.5
z = (117.24 − 120)/1.5 = −1.84

Tail area

Φ(1.84) = 0.9671
1 − Φ(1.84) = 0.0329

p-value

p = 0.03288

Answer

p = 0.03288

p ≤ 0.05, so H₀ is rejected at the 5% level.

10.Hypothesis tests & confidence intervals

Answer: z = 0.8736

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 = 105/150 = 0.7
p̂₂ = 78/120 = 0.65
p̂ = 183/270 = 0.6778

Standard error under H₀

SE = √(0.6778·0.3222·(1/150 + 1/120)) = 0.05724

Test statistic

z = (0.7 − 0.65)/0.05724 = 0.8736

Answer

z = 0.8736

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

11.Linear regression & correlation

Answer: r² = 0.8198

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 = 95/5 = 19
ȳ = 260/5 = 52

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 148
Syy = Σ(y − ȳ)² = 668

Coefficient of determination

r² = 148²/(40·668) = 0.8198

Answer

r² = 0.8198

About 81.98% of the variation in y is explained by the line.

12.Linear regression & correlation

Answer: x = 18.89

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 = 110/5 = 22
ȳ = 343/5 = 68.6

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 188

Regression line

b = 188/40 = 4.7
a = 68.6 − 4.7·22 = −34.8
ŷ = −34.8 + 4.7·x

Solve for x

x = (54 − (−34.8))/4.7 = 18.89

Answer

x = 18.89

ExamProblems · examproblems.com/courses/statistics-1