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
15202530354045

Find the sample standard deviation s of the data: 40, 40, 22, 12, 28, 46, 19.

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

A quantity is normally distributed with mean 170 and standard deviation 5. What proportion P of values is below 172?

μ = 170σ = 5x = 172
P =
3.ProbabilityMedium
xp(x)10.2520.130.240.3550.1μ = 2.95

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

x: 1, 2, 3, 4, 5p(x): 0.25, 0.1, 0.2, 0.35, 0.1
σ =
4.ProbabilityHard
€p(x)−50.7850.22450.02

A game costs €5 to play. You win €10 with probability 0.2 and €250 with probability 0.02; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.

stake = €5€10: p = 0.2€250: p = 0.02
E(G) = €
5.Normal distributionMedium
−3−2−10123zshaded area to the right of z = 0.6517

Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.6517. Find z to two decimals.

A = 0.6517
z =
6.Normal distributionHard
μ = 250x = ?25%

The bag weights are normally distributed with mean 250 g and standard deviation 5 g. Find the value x below which 25% of them lie (the 25th percentile).

μ = 250σ = 5P = 25%
x = g
7.Confidence intervalsMedium
n = 12, sd​ = 7.80LU = ?d̄ = 3.8E95%

The blood pressure of 12 patients is measured before and after a treatment. The differences d = before − after (mmHg) have mean 3.8 and standard deviation 7.8. Find the upper limit of the 95% confidence interval for the mean difference.

n = 12d̄ = 3.8sd = 7.895%t(11) = 2.201
upper limit = mmHg
8.Confidence intervalsHard
p = 0.5, n = ?p̂ − Ep̂ + Ep̂E = 0.0190%

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 0.01?

p = 0.5E = 0.0190%
n =
9.Hypothesis tests & confidence intervalsMedium
t −2.01102.011

A sample of 49 students has a mean score of 100.4 and a sample standard deviation of 4. Find the test statistic t for H₀: μ = 100.

μ₀ = 100x̄ = 100.4s = 4n = 49
t =
10.Hypothesis tests & confidence intervalsHard
z −1.9601.96

A sample of 16 measurements has mean 85. The population standard deviation is 20. Find the upper limit U of the 95% confidence interval for the mean.

x̄ = 85σ = 20n = 1695%
U =
11.Linear regression & correlationMedium
xy124218319416511xy2410152025

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 = 88Σx² = 55Σy² = 1638Σxy = 236
r =
12.Linear regression & correlationHard
xy18862010322107241262613628147xy2025100120140x = 21

The table gives 6 paired observations of x = temperature in °C and y = ice creams sold. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 21.

n = 6x = 21
ŷ =

3Solutions

1.Descriptive statistics

Answer: s = 12.7

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 = 207/7 = 29.57

Squared deviations

Σ(x − x̄)² = 10.43² + 10.43² + (−7.571)² + (−17.57)² + (−1.571)² + 16.43² + (−10.57)² = 967.7

Standard deviation

s = √(967.7/6) = 12.7

Answer

s = 12.7
2.Descriptive statistics

Answer: P = 0.6554

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 = 0.4

Standard normal table

Φ(0.4) = 0.6554
P = 0.6554

Answer

P = 0.6554
3.Probability

Answer: σ = 1.359

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 = 1·0.25 + 2·0.1 + 3·0.2 + 4·0.35 + 5·0.1 = 2.95

Mean of the squares

E(X²) = 1·0.25 + 4·0.1 + 9·0.2 + 16·0.35 + 25·0.1 = 10.55

Variance

Var(X) = 10.55 − 2.95² = 1.848
σ = √1.848 = 1.359

Answer

σ = 1.359
4.Probability

Answer: E(G) = 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: 5 (p = 0.2), 245 (p = 0.02), −5 (p = 0.78)

Expected value

E(G) = 5·0.2 + 245·0.02 − 5·0.78 = 2 €

Answer

E(G) = 2 €

A positive value means you win money on average.

5.Normal distribution

Answer: z = −0.39

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 − 0.6517 = 0.3483

Read the z-table backwards

Find 0.3483 inside the table and read off z.

z = −0.39

Answer

z = −0.39
6.Normal distribution

Answer: x = 246.6 g

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) = 0.25 → z = −0.6745

Back to x

x = 250 + −0.6745·5 = 246.63 g

Answer

x = 246.63 g
7.Confidence intervals

Answer: upper limit = 8.756 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 = 11, t = 2.201

Margin of error

E = 2.201·7.8/√12 = 4.956

Interval

3.8 ± 4.956 → [−1.156, 8.756]

Answer

upper limit = 8.756 mmHg
8.Confidence intervals

Answer: n = 6766

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 = 1.645²·0.5·0.5/0.01² = 6765

Round up

n = 6766

Answer

n = 6766
9.Hypothesis tests & confidence intervals

Answer: t = 0.7

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 = 4/√49 = 4/7 = 0.5714

Test statistic

t = (100.4 − 100)/0.5714 = 0.7

Answer

t = 0.7
t0.025; df = 48 = 2.011

|t| < 2.011, so H₀ is not rejected at the 5% level (two-sided).

10.Hypothesis tests & confidence intervals

Answer: U = 94.8

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).

Margin of error

E = 1.96·20/√16 = 9.8

Interval limit

U = 85 + 9.8 = 94.8

Answer

U = 94.8
11.Linear regression & correlation

Answer: r = −0.9375

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 = 236 − 15·88/5 = −28
Sxx = 55 − 15²/5 = 10
Syy = 1638 − 88²/5 = 89.2

Correlation coefficient

r = −28/√(10·89.2) = −0.9375

Answer

r = −0.9375
12.Linear regression & correlation

Answer: ŷ = 105.4

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 = 138/6 = 23
ȳ = 705/6 = 117.5

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 423

Regression line

b = 423/70 = 6.043
a = 117.5 − 6.043·23 = −21.49
ŷ = −21.49 + 6.043·x

Prediction

ŷ = −21.49 + 6.043·21 = 105.4

Answer

ŷ = 105.4

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