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: 55, 26, 16, 20, 58, 44.

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

Exam scores are normally distributed with mean 500 and standard deviation 25. Find the z-score z of a score of 550.

μ = 500σ = 25x = 550
z =
3.ProbabilityMedium
xp(x)10.220.3530.3540.1μ = 2.35

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

x: 1, 2, 3, 4p(x): 0.2, 0.35, 0.35, 0.1
Var(X) =
4.ProbabilityHard
€p(x)−30.8120.15570.05

A game costs €3 to play. You win €15 with probability 0.15 and €60 with probability 0.05; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.

stake = €3€15: p = 0.15€60: p = 0.05
E(G) = €
5.Normal distributionMedium
−3−2−10123zshaded area to the right of z = 0.1446

Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.1446. Find z to two decimals.

A = 0.1446
z =
6.Normal distributionHard
μ = 65x = ?99%

The exam scores are normally distributed with mean 65 points and standard deviation 10 points. Find the value x below which 99% of them lie (the 99th percentile).

μ = 65σ = 10P = 99%
x = points
7.Confidence intervalsMedium
n = 10, sd​ = 20LU = ?d̄ = −8.2E90%

The blood pressure of 10 patients is measured before and after a treatment. The differences d = before − after (mmHg) have mean −8.2 and standard deviation 2. Find the upper limit of the 90% confidence interval for the mean difference.

n = 10d̄ = −8.2sd = 290%t(9) = 1.833
upper limit = mmHg
8.Confidence intervalsHard
n = 30Lσ = ?s = 10.895%

The diameters of 30 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 = 29, 2.5% of the χ² distribution lies below 16.047 and 2.5% lies above 45.722.

n = 30s = 10.8 µmχ² = 16.047, 45.722
σ = µm
9.Hypothesis tests & confidence intervalsMedium
1.6p = ?

The mean of a population is claimed to be 500; the population standard deviation is known to be 5. A random sample of 25 observations has a mean of 501.6. Test H₀: μ = 500 against H₁: μ > 500. Find the p-value p.

μ₀ = 500σ = 5n = 25x̄ = 501.6H₁: μ > μ₀
p =
10.Hypothesis tests & confidence intervalsHard
σ = 15, n = ?x̄ − Ex̄ + Ex̄E = 1.595%

How large must a sample be so that the 95% confidence interval for a mean has a margin of error of at most 1.5? The standard deviation is 15. Find the smallest sample size n.

σ = 15E = 1.595%
n =
11.Linear regression & correlationMedium
xy264116158201022xy5105101520

The table gives 5 paired observations of x = load in kN and y = extension in mm. Their sums are given. Find Pearson's correlation coefficient r from these sums.

n = 5Σx = 30Σy = 74Σx² = 220Σy² = 1266Σxy = 526
r =
12.Linear regression & correlationHard
xy33654275496511751382xy510406080x = 12

The table gives 6 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 = 12.

n = 6x = 12
ŷ =

3Solutions

1.Descriptive statistics

Answer: s = 18.24

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 = 219/6 = 36.5

Squared deviations

Σ(x − x̄)² = 18.5² + (−10.5)² + (−20.5)² + (−16.5)² + 21.5² + 7.5² = 1664

Standard deviation

s = √(1664/5) = 18.24

Answer

s = 18.24
2.Descriptive statistics

Answer: z = 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 = (550 − 500)/25 = 2

Answer

z = 2
3.Probability

Answer: Var(X) = 0.8275

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 = 1·0.2 + 2·0.35 + 3·0.35 + 4·0.1 = 2.35

Mean of the squares

E(X²) = 1·0.2 + 4·0.35 + 9·0.35 + 16·0.1 = 6.35

Variance

Var(X) = 6.35 − 2.35² = 0.8275

Answer

Var(X) = 0.8275
4.Probability

Answer: E(G) = 2.25 €

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: 12 (p = 0.15), 57 (p = 0.05), −3 (p = 0.8)

Expected value

E(G) = 12·0.15 + 57·0.05 − 3·0.8 = 2.25 €

Answer

E(G) = 2.25 €

A positive value means you win money on average.

5.Normal distribution

Answer: z = 1.06

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 − 0.1446 = 0.8554

Read the z-table backwards

Find 0.8554 inside the table and read off z.

z = 1.06

Answer

z = 1.06
6.Normal distribution

Answer: x = 88.26 points

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 = 65 + 2.326·10 = 88.26 points

Answer

x = 88.26 points
7.Confidence intervals

Answer: upper limit = −7.041 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 = 9, t = 1.833

Margin of error

E = 1.833·2/√10 = 1.159

Interval

−8.2 ± 1.159 → [−9.359, −7.041]

Answer

upper limit = −7.041 mmHg
8.Confidence intervals

Answer: σ = 14.52 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 = 29·10.8² = 3383

Limits for σ²

3383/45.722 = 73.98
3383/16.047 = 210.8

Limit for σ

σ = √210.8 = 14.52 µm

Answer

σ = 14.52 µm

95% interval for σ: [8.601, 14.52] µm.

9.Hypothesis tests & confidence intervals

Answer: p = 0.0548

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 = 5/√25 = 1
z = (501.6 − 500)/1 = 1.6

Tail area

Φ(1.6) = 0.9452
1 − Φ(1.6) = 0.0548

p-value

p = 0.0548

Answer

p = 0.0548

p > 0.05, so H₀ is not rejected at the 5% level.

10.Hypothesis tests & confidence intervals

Answer: n = 385

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 = (1.96·15/1.5)² = 384.2

Round up

n = 385

Answer

n = 385
11.Linear regression & correlation

Answer: r = 0.9921

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 = 526 − 30·74/5 = 82
Sxx = 220 − 30²/5 = 40
Syy = 1266 − 74²/5 = 170.8

Correlation coefficient

r = 82/√(40·170.8) = 0.9921

Answer

r = 0.9921
12.Linear regression & correlation

Answer: ŷ = 78.43

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 = 48/6 = 8
ȳ = 354/6 = 59

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 340

Regression line

b = 340/70 = 4.857
a = 59 − 4.857·8 = 20.14
ŷ = 20.14 + 4.857·x

Prediction

ŷ = 20.14 + 4.857·12 = 78.43

Answer

ŷ = 78.43

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