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
10152025303540455055

Find the sample standard deviation s of the data: 43, 57, 43, 10, 22.

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

A quantity is normally distributed with mean 60 and standard deviation 15. What proportion P of values is below 78.75?

μ = 60σ = 15x = 78.75
P =
3.ProbabilityMedium
xp(x)00.2520.240.360.25μ = 3.1

For the distribution shown, find the standard deviation σ of X.

x: 0, 2, 4, 6p(x): 0.25, 0.2, 0.3, 0.25
σ =
4.ProbabilityHard
€p(x)−100.83100.154900.02

A game costs €10 to play. You win €20 with probability 0.15 and €500 with probability 0.02; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.

stake = €10€20: p = 0.15€500: p = 0.02
E(G) = €
5.Normal distributionMedium
−3−2−10123zshaded area to the right of z = 0.0322

Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.0322. Find z to two decimals.

A = 0.0322
z =
6.Normal distributionHard
μ = 32x = ?99%

The delivery times are normally distributed with mean 32 min and standard deviation 6 min. Find the value x below which 99% of them lie (the 99th percentile).

μ = 32σ = 6P = 99%
x = min
7.Confidence intervalsMedium
μ₁ − μ₂0LU = ?x̄₁ − x̄₂E95%

The battery life of two phone models is compared. Model 1: 80 phones, mean 145 h, known standard deviation 3 h. Model 2: 80 phones, mean 146.5 h, known standard deviation 14 h. Find the upper limit of the 95% confidence interval for μ₁ − μ₂.

n₁ = 80x̄₁ = 145 hσ₁ = 3 hn₂ = 80x̄₂ = 146.5 hσ₂ = 14 h95%
upper limit = h
8.Confidence intervalsHard
n = 21Lσ = ?s = 10.895%

The diameters of 21 machined shafts have a sample standard deviation of 10.8 µm. Assuming the diameters are normally distributed, find the upper limit σ of the 95% confidence interval for the population standard deviation. With df = 20, 2.5% of the χ² distribution lies below 9.591 and 2.5% lies above 34.17.

n = 21s = 10.8 µmχ² = 9.591, 34.17
σ = µm
9.Hypothesis tests & confidence intervalsMedium
−2.1452.145t = ?df = 14

Two teaching methods are compared. Group 1: 6 students, mean score 100.5, standard deviation 11.5. Group 2: 10 students, mean score 113.7, standard deviation 3. 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 = 6x̄₁ = 100.5s₁ = 11.5n₂ = 10x̄₂ = 113.7s₂ = 3
t =
10.Hypothesis tests & confidence intervalsHard
−1.961.96z = ?

In a trial, 72 of 120 patients on treatment A recover, and 128 of 200 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.

x₁ = 72n₁ = 120x₂ = 128n₂ = 200
z =
11.Linear regression & correlationMedium
xy252362461577676xy24650607080

The table gives 5 paired observations of x = hours studied and y = exam score. Their sums are given. Find Pearson's correlation coefficient r from these sums.

n = 5Σx = 20Σy = 328Σx² = 90Σy² = 21974Σxy = 1375
r =
12.Linear regression & correlationHard
xy264136248291037xy51010203040ŷ = 26

The table gives 5 paired observations of x = load in kN and y = extension in mm. Use the least-squares line ŷ = a + b·x. For which value x does the line predict ŷ = 26?

n = 5ŷ = 26
x =

3Solutions

1.Descriptive statistics

Answer: s = 18.75

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 = 175/5 = 35

Squared deviations

Σ(x − x̄)² = 8² + 22² + 8² + (−25)² + (−13)² = 1406

Standard deviation

s = √(1406/4) = 18.75

Answer

s = 18.75
2.Descriptive statistics

Answer: P = 0.8944

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 = 1.25

Standard normal table

Φ(1.25) = 0.8944
P = 0.8944

Answer

P = 0.8944
3.Probability

Answer: σ = 2.234

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 = 0·0.25 + 2·0.2 + 4·0.3 + 6·0.25 = 3.1

Mean of the squares

E(X²) = 0·0.25 + 4·0.2 + 16·0.3 + 36·0.25 = 14.6

Variance

Var(X) = 14.6 − 3.1² = 4.99
σ = √4.99 = 2.234

Answer

σ = 2.234
4.Probability

Answer: E(G) = 3 €

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: 10 (p = 0.15), 490 (p = 0.02), −10 (p = 0.83)

Expected value

E(G) = 10·0.15 + 490·0.02 − 10·0.83 = 3 €

Answer

E(G) = 3 €

A positive value means you win money on average.

5.Normal distribution

Answer: z = 1.85

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 of z

Φ(z) = 1 − 0.0322 = 0.9678

Read the z-table backwards

Find 0.9678 inside the table and read off z.

z = 1.85

Answer

z = 1.85
6.Normal distribution

Answer: x = 45.96 min

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 for the percentile

Φ(z) = 0.99 → z = 2.326

Back to x

x = 32 + 2.326·6 = 45.96 min

Answer

x = 45.96 min
7.Confidence intervals

Answer: upper limit = 1.638 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 = 145 − 146.5 = −1.5

Standard error

SE = √(3²/80 + 14²/80) = 1.601

Margin of error

E = 1.96·1.601 = 3.138

Interval

−1.5 ± 3.138 → [−4.638, 1.638]

Answer

upper limit = 1.638 h

The interval contains 0, so the data do not show a significant difference.

8.Confidence intervals

Answer: σ = 15.6 µm

Interval for a variance: (n − 1)s²/σ² has a χ² distribution with n − 1 degrees of freedom. The 95% interval for σ² runs from (n − 1)s² divided by the upper χ² value (2.5% above it) to (n − 1)s² divided by the lower χ² value (2.5% below it); the interval for σ is the square root of both limits. It is not symmetric around s.

Sum of squares

(n − 1)s² = 20·10.8² = 2333

Limits for σ²

2333/34.17 = 68.27
2333/9.591 = 243.2

Limit for σ

σ = √243.2 = 15.6 µm

Answer

σ = 15.6 µm

95% interval for σ: [8.263, 15.6] µm.

9.Hypothesis tests & confidence intervals

Answer: t = −3.511

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 = (5·11.5² + 9·3²)/14 = 53.02
sp = 7.281

Standard error

SE = 7.281·√(1/6 + 1/10) = 3.76

Test statistic

t = (100.5 − 113.7)/3.76 = −3.511

Answer

t = −3.511

With df = 14, the two-sided 5% critical value is 2.145; |t| is larger, so H₀ is rejected.

10.Hypothesis tests & confidence intervals

Answer: z = −0.7155

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 = 72/120 = 0.6
p̂₂ = 128/200 = 0.64
p̂ = 200/320 = 0.625

Standard error under H₀

SE = √(0.625·0.375·(1/120 + 1/200)) = 0.0559

Test statistic

z = (0.6 − 0.64)/0.0559 = −0.7155

Answer

z = −0.7155

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

11.Linear regression & correlation

Answer: r = 0.9317

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 = 1375 − 20·328/5 = 63
Sxx = 90 − 20²/5 = 10
Syy = 21974 − 328²/5 = 457.2

Correlation coefficient

r = 63/√(10·457.2) = 0.9317

Answer

r = 0.9317
12.Linear regression & correlation

Answer: x = 7.077

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 = 30/5 = 6
ȳ = 109/5 = 21.8

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 156

Regression line

b = 156/40 = 3.9
a = 21.8 − 3.9·6 = −1.6
ŷ = −1.6 + 3.9·x

Solve for x

x = (26 − (−1.6))/3.9 = 7.077

Answer

x = 7.077

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