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
253035404550

Find the sample standard deviation s of the data: 42, 23, 21, 37, 52, 36, 37.

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

A quantity is normally distributed with mean 60 and standard deviation 8. What proportion P of values is above 72?

μ = 60σ = 8x = 72
P =
3.ProbabilityMedium
xp(x)10.320.1530.340.25μ = 2.5

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

x: 1, 2, 3, 4p(x): 0.3, 0.15, 0.3, 0.25
σ =
4.ProbabilityHard
xp(x)100.25200.35300.3400.1μ

X has the distribution shown, and Y = 3X + 10. Find the standard deviation σY of Y.

x: 10, 20, 30, 40p(x): 0.25, 0.35, 0.3, 0.1Y = 3X + 10
σY =
5.Normal distributionMedium
480 g486.2 gT = X₁ + … + X₁₂?

The masses of bolts are independent and normal with mean 40 g and standard deviation 1.2 g. Find the probability P that the total mass of 12 of them is below 486.2 g. Round z to two decimals.

μ = 40 gσ = 1.2 gn = 12< 486.2 g
P =
6.Normal distributionHard
10 cm0X − Y?

The heights X of men and Y of women in a group are independent and normal: X has mean 180 cm and standard deviation 10 cm, Y has mean 170 cm and standard deviation 12 cm. Find the probability P that X is greater than Y. Round z to two decimals.

μX = 180 cm, σX = 10 cmμY = 170 cm, σY = 12 cm
P =
7.Confidence intervalsMedium
x̄E₁ = 10.899%, n = 50x̄E₂ = ?99%, n = 200

A 99% confidence interval for a mean, based on 50 observations and a known σ, has a margin of error of 10.8. What is the margin of error E₂ if the same study uses 200 observations instead?

99%n₁ = 50E₁ = 10.8n₂ = 200
E₂ =
8.Confidence intervalsHard
p ≈ 0.4, n = ?p̂ − Ep̂ + Ep̂E = 0.0399%

A pilot study suggests that about 0.4 of products are defective. How many products n must be inspected so that a 99% confidence interval for the proportion has a margin of error of at most 0.03?

p ≈ 0.4E = 0.0399%
n =
9.Hypothesis tests & confidence intervalsMedium
−1.961.96z = ?

Two production lines make the same part. Line 1: a sample of 80 parts has mean length 80 mm, and the line's known standard deviation is 5 mm. Line 2: a sample of 100 parts has mean 83.4 mm, known standard deviation 10 mm. Find the test statistic z for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (line 1 minus line 2).

n₁ = 80x̄₁ = 80 mmσ₁ = 5 mmn₂ = 100x̄₂ = 83.4 mmσ₂ = 10 mm
z =
10.Hypothesis tests & confidence intervalsHard
−1.961.96z = ?

In a trial, 72 of 120 patients on treatment A recover, and 64 of 120 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.

x₁ = 72n₁ = 120x₂ = 64n₂ = 120
z =
11.Linear regression & correlationMedium
xy139247352453555xy2440455055

The table gives 5 paired observations of x = advertising spend in k€ and y = sales in k€. Find the coefficient of determination r² of the least-squares line.

n = 5
r² =
12.Linear regression & correlationHard
xy1575189521107241252714330153xy15202530100150x = 28.5

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 = 28.5.

n = 6x = 28.5
ŷ =

3Solutions

1.Descriptive statistics

Answer: s = 10.69

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 = 248/7 = 35.43

Squared deviations

Σ(x − x̄)² = 6.571² + (−12.43)² + (−14.43)² + 1.571² + 16.57² + 0.5714² + 1.571² = 685.7

Standard deviation

s = √(685.7/6) = 10.69

Answer

s = 10.69
2.Descriptive statistics

Answer: P = 0.06681

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 = 1.5

Standard normal table

Φ(1.5) = 0.9332
P = 1 − 0.9332 = 0.06681

Answer

P = 0.06681
3.Probability

Answer: σ = 1.162

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 = 1·0.3 + 2·0.15 + 3·0.3 + 4·0.25 = 2.5

Mean of the squares

E(X²) = 1·0.3 + 4·0.15 + 9·0.3 + 16·0.25 = 7.6

Variance

Var(X) = 7.6 − 2.5² = 1.35
σ = √1.35 = 1.162

Answer

σ = 1.162
4.Probability

Answer: σY = 28.26

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.

Spread of X

μ = 22.5, Var(X) = 88.75, σX = 9.421

Linear change

σY = |3|·9.421 = 28.26

Answer

σY = 28.26

The mean does move: E(Y) = 3·22.5 + 10 = 77.5.

5.Normal distribution

Answer: P = 0.9319

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 total

μT = 480 g, σT = 1.2·√12 = 4.157 g

Standardise

z = (486.2 − 480)/4.157 = 1.49

Probability

P = Φ(1.49) = 0.9319

Answer

P = 0.9319
6.Normal distribution

Answer: P = 0.7389

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 = 180 − 170 = 10 cm, σ = 15.62 cm

Standardise

z = (0 − 10)/15.62 = −0.64

Probability

P = 1 − Φ(−0.64) = 0.7389

Answer

P = 0.7389
7.Confidence intervals

Answer: E₂ = 5.4

Confidence interval for a mean: x̄ ± margin of error. With a known population σ the margin is z·σ/√n; with only the sample standard deviation s, use t with n − 1 degrees of freedom: t·s/√n. Critical values: z = 1.645 (90%), 1.96 (95%), 2.576 (99%). To get a margin E or smaller, you need n ≥ (z·σ/E)², rounded up.

Ratio of margins

E₂/E₁ = (2.576/2.576)·√(50/200) = 0.5

New margin

E₂ = 10.8·0.5 = 5.4

Answer

E₂ = 5.4
8.Confidence intervals

Answer: n = 1770

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 = 2.576²·0.4·0.6/0.03² = 1770

Round up

n = 1770

Answer

n = 1770
9.Hypothesis tests & confidence intervals

Answer: z = −2.968

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.

Standard error of the difference

SE = √(5²/80 + 10²/100) = √1.313 = 1.146

Test statistic

z = (80 − 83.4)/1.146 = −2.968

Answer

z = −2.968

|z| > 1.96, so H₀ is rejected at the 5% level (two-sided).

10.Hypothesis tests & confidence intervals

Answer: z = 1.042

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 = 64/120 = 0.5333
p̂ = 136/240 = 0.5667

Standard error under H₀

SE = √(0.5667·0.4333·(1/120 + 1/120)) = 0.06397

Test statistic

z = (0.6 − 0.5333)/0.06397 = 1.042

Answer

z = 1.042

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

11.Linear regression & correlation

Answer: r² = 0.8762

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 = 246/5 = 49.2

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = 38
Syy = Σ(y − ȳ)² = 164.8

Coefficient of determination

r² = 38²/(10·164.8) = 0.8762

Answer

r² = 0.8762

About 87.62% of the variation in y is explained by the line.

12.Linear regression & correlation

Answer: ŷ = 147.9

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 = 135/6 = 22.5
ȳ = 698/6 = 116.3

Sums of squares and products

Sxx = Σ(x − x̄)² = 157.5
Sxy = Σ(x − x̄)(y − ȳ) = 828

Regression line

b = 828/157.5 = 5.257
a = 116.3 − 5.257·22.5 = −1.952
ŷ = −1.952 + 5.257·x

Prediction

ŷ = −1.952 + 5.257·28.5 = 147.9

Answer

ŷ = 147.9

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