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
152025303540455055

Find the sample standard deviation s of the data: 28, 41, 13, 58, 40, 40.

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

A quantity is normally distributed with mean 170 and standard deviation 8. What proportion P of values is below 166?

μ = 170σ = 8x = 166
P =
3.ProbabilityMedium
xp(x)10.1520.3530.2540.25μ = 2.6

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

x: 1, 2, 3, 4p(x): 0.15, 0.35, 0.25, 0.25
σ =
4.ProbabilityHard
€p(x)−50.85150.1450.05

A game costs €5 to play. You win €20 with probability 0.1 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 = €5€20: p = 0.1€50: p = 0.05
E(G) = €
5.Normal distributionMedium
88115.75P = ?

The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. What fraction lies between 88 and 115.75 points? Give P as a decimal.

μ = 100σ = 15x1 = 88x2 = 115.75
P =
6.Normal distributionHard
4 s0X − Y?

Runner A's lap time X and runner B's lap time Y are independent and normal: X has mean 52 s and standard deviation 4 s, Y has mean 48 s and standard deviation 3 s. Find the probability P that X is greater than Y. Round z to two decimals.

μX = 52 s, σX = 4 sμY = 48 s, σY = 3 s
P =
7.Confidence intervalsMedium
n = 15, sd​ = 4.60LU = ?d̄ = −0.9E90%

The blood pressure of 15 patients is measured before and after a treatment. The differences d = before − after (mmHg) have mean −0.9 and standard deviation 4.6. Find the upper limit of the 90% confidence interval for the mean difference.

n = 15d̄ = −0.9sd = 4.690%t(14) = 1.761
upper limit = mmHg
8.Confidence intervalsHard
p = 0.5, n = ?p̂ − Ep̂ + Ep̂E = 0.0590%

Nothing is known in advance about the proportion of defective products. How many products n must be inspected so that a 90% confidence interval for the proportion has a margin of error of at most 0.05?

p = 0.5E = 0.0590%
n =
9.Hypothesis tests & confidence intervalsMedium
−2.0932.093t = ?df = 19

Two teaching methods are compared. Group 1: 16 students, mean score 72.5, standard deviation 8. Group 2: 5 students, mean score 64.9, standard deviation 7. 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 = 16x̄₁ = 72.5s₁ = 8n₂ = 5x̄₂ = 64.9s₂ = 7
t =
10.Hypothesis tests & confidence intervalsHard
σ = 12, n = ?x̄ − Ex̄ + Ex̄E = 299%

How large must a sample be so that the 99% confidence interval for a mean has a margin of error of at most 2? The standard deviation is 12. Find the smallest sample size n.

σ = 12E = 299%
n =
11.Linear regression & correlationMedium
xy186821682484271043012033130xy20253080100120

The table gives 6 paired observations of x = temperature in °C and y = ice creams sold. Their sums are given. Find Pearson's correlation coefficient r from these sums.

n = 6Σx = 153Σy = 574Σx² = 4059Σy² = 58420Σxy = 15366
r =
12.Linear regression & correlationHard
xy16581860206522752481xy20607080ŷ = 77

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 = 77?

n = 5ŷ = 77
x =

3Solutions

1.Descriptive statistics

Answer: s = 15.04

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 = 220/6 = 36.67

Squared deviations

Σ(x − x̄)² = (−8.667)² + 4.333² + (−23.67)² + 21.33² + 3.333² + 3.333² = 1131

Standard deviation

s = √(1131/5) = 15.04

Answer

s = 15.04
2.Descriptive statistics

Answer: P = 0.3085

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 = −0.5

Standard normal table

Φ(−0.5) = 0.3085
P = 0.3085

Answer

P = 0.3085
3.Probability

Answer: σ = 1.02

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 = 1·0.15 + 2·0.35 + 3·0.25 + 4·0.25 = 2.6

Mean of the squares

E(X²) = 1·0.15 + 4·0.35 + 9·0.25 + 16·0.25 = 7.8

Variance

Var(X) = 7.8 − 2.6² = 1.04
σ = √1.04 = 1.02

Answer

σ = 1.02
4.Probability

Answer: E(G) = −0.5 €

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: 15 (p = 0.1), 45 (p = 0.05), −5 (p = 0.85)

Expected value

E(G) = 15·0.1 + 45·0.05 − 5·0.85 = −0.5 €

Answer

E(G) = −0.5 €

A negative value means you lose money on average.

5.Normal distribution

Answer: P = 0.6413

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-scores

z₁ = (88 − 100)/15 = −0.8
z₂ = (115.75 − 100)/15 = 1.05

Area between

P = Φ(1.05) − Φ(−0.8) = 0.8531 − 0.2119 = 0.6413

Answer

P = 0.6413
6.Normal distribution

Answer: P = 0.7881

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 = 52 − 48 = 4 s, σ = 5 s

Standardise

z = (0 − 4)/5 = −0.8

Probability

P = 1 − Φ(−0.8) = 0.7881

Answer

P = 0.7881
7.Confidence intervals

Answer: upper limit = 1.192 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 = 1.761

Margin of error

E = 1.761·4.6/√15 = 2.092

Interval

−0.9 ± 2.092 → [−2.992, 1.192]

Answer

upper limit = 1.192 mmHg
8.Confidence intervals

Answer: n = 271

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 = 1.645²·0.5·0.5/0.05² = 270.6

Round up

n = 271

Answer

n = 271
9.Hypothesis tests & confidence intervals

Answer: t = 1.902

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 = (15·8² + 4·7²)/19 = 60.84
sp = 7.8

Standard error

SE = 7.8·√(1/16 + 1/5) = 3.996

Test statistic

t = (72.5 − 64.9)/3.996 = 1.902

Answer

t = 1.902

With df = 19, the two-sided 5% critical value is 2.093; |t| is smaller, so H₀ is not rejected.

10.Hypothesis tests & confidence intervals

Answer: n = 239

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 = (2.576·12/2)² = 238.9

Round up

n = 239

Answer

n = 239
11.Linear regression & correlation

Answer: r = 0.9808

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 = 15366 − 153·574/6 = 729
Sxx = 4059 − 153²/6 = 157.5
Syy = 58420 − 574²/6 = 3507

Correlation coefficient

r = 729/√(157.5·3507) = 0.9808

Answer

r = 0.9808
12.Linear regression & correlation

Answer: x = 23.02

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 = 100/5 = 20
ȳ = 339/5 = 67.8

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 122

Regression line

b = 122/40 = 3.05
a = 67.8 − 3.05·20 = 6.8
ŷ = 6.8 + 3.05·x

Solve for x

x = (77 − 6.8)/3.05 = 23.02

Answer

x = 23.02

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