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: 54, 58, 36, 20, 25, 22, 21.

n = 7
s =
2.Descriptive statisticsHard
z −2−1012

A quantity is normally distributed with mean 70 and standard deviation 5. What proportion P of values is below 76.25?

μ = 70σ = 5x = 76.25
P =
3.ProbabilityMedium
xp(x)00.2510.320.130.1540.2μ = 1.75

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

x: 0, 1, 2, 3, 4p(x): 0.25, 0.3, 0.1, 0.15, 0.2
σ =
4.ProbabilityHard
€p(x)−10.810.1590.05

A game costs €1 to play. You win €2 with probability 0.15 and €10 with probability 0.05; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.

stake = €1€2: p = 0.15€10: p = 0.05
E(G) = €
5.Normal distributionMedium
750 g754.9 gX + Y?

X is normal with mean 250 g and standard deviation 4 g; Y is independent and normal with mean 500 g and standard deviation 9 g. Find the probability P that X + Y is more than 754.9 g. Round z to two decimals.

μX = 250, σX = 4μY = 500, σY = 9X + Y > 754.9 g
P =
6.Normal distributionHard
380420460500540580620htop 95%

The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. Drag the cut-off so the shaded top 95% is to the right of it. Find the cut-off x.

μ = 500 hσ = 40 h95%
x = h
7.Confidence intervalsMedium
n = 15, sd​ = 1.20LU = ?d̄ = 7E95%

The blood pressure of 15 patients is measured before and after a treatment. The differences d = before − after (mmHg) have mean 7 and standard deviation 1.2. Find the upper limit of the 95% confidence interval for the mean difference.

n = 15d̄ = 7sd = 1.295%t(14) = 2.145
upper limit = mmHg
8.Confidence intervalsHard
n = 11σ = ?Us = 7.495%

The diameters of 11 machined shafts have a sample standard deviation of 7.4 µm. Assuming the diameters are normally distributed, find the lower limit σ of the 95% confidence interval for the population standard deviation. With df = 10, 2.5% of the χ² distribution lies below 3.247 and 2.5% lies above 20.483.

n = 11s = 7.4 µmχ² = 3.247, 20.483
σ = µm
9.Hypothesis tests & confidence intervalsMedium
t −2.0302.03

A sample of 36 students has a mean score of 69.5 and a sample standard deviation of 15. Find the test statistic t for H₀: μ = 75.

μ₀ = 75x̄ = 69.5s = 15n = 36
t =
10.Hypothesis tests & confidence intervalsHard
σ = 15, n = ?x̄ − Ex̄ + Ex̄E = 295%

How large must a sample be so that the 95% confidence interval for a mean has a margin of error of at most 2? The standard deviation is 15. Find the smallest sample size n.

σ = 15E = 295%
n =
11.Linear regression & correlationMedium
xy127224319415513xy24152025

The table gives 5 paired observations of x = car age in years and y = price in k€. Their sums are given. Find Pearson's correlation coefficient r from these sums.

n = 5Σx = 15Σy = 98Σx² = 55Σy² = 2060Σxy = 257
r =
12.Linear regression & correlationHard
xy141249343454552xy2440455055ŷ = 46

The table gives 5 paired observations of x = hours studied and y = exam score. Use the least-squares line ŷ = a + b·x. For which value x does the line predict ŷ = 46?

n = 5ŷ = 46
x =

3Solutions

1.Descriptive statistics

Answer: s = 16.17

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 = 236/7 = 33.71

Squared deviations

Σ(x − x̄)² = 20.29² + 24.29² + 2.286² + (−13.71)² + (−8.714)² + (−11.71)² + (−12.71)² = 1569

Standard deviation

s = √(1569/6) = 16.17

Answer

s = 16.17
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 = 1.479

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 + 1·0.3 + 2·0.1 + 3·0.15 + 4·0.2 = 1.75

Mean of the squares

E(X²) = 0·0.25 + 1·0.3 + 4·0.1 + 9·0.15 + 16·0.2 = 5.25

Variance

Var(X) = 5.25 − 1.75² = 2.188
σ = √2.188 = 1.479

Answer

σ = 1.479
4.Probability

Answer: E(G) = −0.2 €

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: 1 (p = 0.15), 9 (p = 0.05), −1 (p = 0.8)

Expected value

E(G) = 1·0.15 + 9·0.05 − 1·0.8 = −0.2 €

Answer

E(G) = −0.2 €

A negative value means you lose money on average.

5.Normal distribution

Answer: P = 0.3085

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 = 750 g, σ = √(4² + 9²) = 9.849 g

Standardise

z = (754.9 − 750)/9.849 = 0.5

Upper tail

P = 1 − Φ(0.5) = 1 − 0.6915 = 0.3085

Answer

P = 0.3085
6.Normal distribution

Answer: x = 434 h

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) = 0.05

z from the table

z ≈ −1.64

Back to the original units

x = 500 − 1.64·40 = 434 h

Answer

x = 434 h
7.Confidence intervals

Answer: upper limit = 7.665 mmHg

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.

Critical value

df = 14, t = 2.145

Margin of error

E = 2.145·1.2/√15 = 0.6646

Interval

7 ± 0.6646 → [6.335, 7.665]

Answer

upper limit = 7.665 mmHg
8.Confidence intervals

Answer: σ = 5.171 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 = 10·7.4² = 547.6

Limits for σ²

547.6/20.483 = 26.73
547.6/3.247 = 168.6

Limit for σ

σ = √26.73 = 5.171 µm

Answer

σ = 5.171 µm

95% interval for σ: [5.171, 12.99] µm.

9.Hypothesis tests & confidence intervals

Answer: t = −2.2

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

SE = 15/√36 = 15/6 = 2.5

Test statistic

t = (69.5 − 75)/2.5 = −2.2

Answer

t = −2.2
t0.025; df = 35 = 2.03

|t| > 2.03, so H₀ is rejected at the 5% level (two-sided).

10.Hypothesis tests & confidence intervals

Answer: n = 217

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 = (1.96·15/2)² = 216.1

Round up

n = 217

Answer

n = 217
11.Linear regression & correlation

Answer: r = −0.9917

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 = 257 − 15·98/5 = −37
Sxx = 55 − 15²/5 = 10
Syy = 2060 − 98²/5 = 139.2

Correlation coefficient

r = −37/√(10·139.2) = −0.9917

Answer

r = −0.9917
12.Linear regression & correlation

Answer: x = 2.333

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 = 15/5 = 3
ȳ = 239/5 = 47.8

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = 27

Regression line

b = 27/10 = 2.7
a = 47.8 − 2.7·3 = 39.7
ŷ = 39.7 + 2.7·x

Solve for x

x = (46 − 39.7)/2.7 = 2.333

Answer

x = 2.333

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