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: 33, 49, 13, 21, 33.

n = 5
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 above 177.5?

μ = 170σ = 5x = 177.5
P =
3.ProbabilityMedium
red: 4blue: 8

A bag holds 4 red and 8 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?

4 / 8
P =
4.ProbabilityHard
xp(x)10.320.0530.3540.0550.25μ

X has the distribution shown, and Y = 2X + 4. Find the standard deviation σY of Y.

x: 1, 2, 3, 4, 5p(x): 0.3, 0.05, 0.35, 0.05, 0.25Y = 2X + 4
σY =
5.Normal distributionMedium
−3−2−10123zshaded area to the right of z = 0.7157

Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.7157. Find z to two decimals.

A = 0.7157
z =
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
x̄E₁ = 8.999%, n = 36x̄E₂ = ?99%, n = 9

A 99% confidence interval for a mean, based on 36 observations and a known σ, has a margin of error of 8.9. What is the margin of error E₂ if the same study uses 9 observations instead?

99%n₁ = 36E₁ = 8.9n₂ = 9
E₂ =
8.Confidence intervalsHard
μ₁ − μ₂, t(19) = 2.0930LUx̄₁ − x̄₂E = ?95%

Two fertilisers are tested. Plot group 1: 15 plots, mean yield 16 kg, standard deviation 8.5 kg. Plot group 2: 6 plots, mean 19 kg, standard deviation 5 kg. Assume equal variances (pooled t, df = n₁ + n₂ − 2). Find the margin of error E of the 95% confidence interval for μ₁ − μ₂.

n₁ = 15x̄₁ = 16s₁ = 8.5n₂ = 6x̄₂ = 19s₂ = 595%t(19) = 2.093
E = kg
9.Hypothesis tests & confidence intervalsMedium
−1.961.96z = ?

Two production lines make the same part. Line 1: a sample of 50 parts has mean length 77 mm, and the line's known standard deviation is 17 mm. Line 2: a sample of 60 parts has mean 72 mm, known standard deviation 8 mm. Find the test statistic z for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (line 1 minus line 2).

n₁ = 50x̄₁ = 77 mmσ₁ = 17 mmn₂ = 60x̄₂ = 72 mmσ₂ = 8 mm
z =
10.Hypothesis tests & confidence intervalsHard
−1.961.96z = ?

In a trial, 36 of 120 patients on treatment A recover, and 49 of 120 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.

x₁ = 36n₁ = 120x₂ = 49n₂ = 120
z =
11.Linear regression & correlationMedium
xy2324316398481053xy510304050

The table gives 5 paired observations of x = advertising spend in k€ and y = sales in k€. Find Pearson's correlation coefficient r.

n = 5
r =
12.Linear regression & correlationHard
xy17761988211122311025126xy202580100120ŷ = 109

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 = 109?

n = 5ŷ = 109
x =

3Solutions

1.Descriptive statistics

Answer: s = 13.68

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 = 149/5 = 29.8

Squared deviations

Σ(x − x̄)² = 3.2² + 19.2² + (−16.8)² + (−8.8)² + 3.2² = 748.8

Standard deviation

s = √(748.8/4) = 13.68

Answer

s = 13.68
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: P = 0.09091

Probability: with equally likely outcomes, P(A) = favourable/total. For independent events multiply: P(A and B) = P(A)·P(B). The complement often helps: P(at least one) = 1 − P(none). Binomial: in n independent trials with success probability p, P(X = k) = C(n, k)·pᵏ·(1 − p)ⁿ⁻ᵏ.

First and second draw

P = 4/12 · 3/11 = 12/132 = 0.09091

Answer

P = 0.09091
4.Probability

Answer: σY = 3.027

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 = 2.9, Var(X) = 2.29, σX = 1.513

Linear change

σY = |2|·1.513 = 3.027

Answer

σY = 3.027

The mean does move: E(Y) = 2·2.9 + 4 = 9.8.

5.Normal distribution

Answer: z = −0.57

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 − 0.7157 = 0.2843

Read the z-table backwards

Find 0.2843 inside the table and read off z.

z = −0.57

Answer

z = −0.57
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: E₂ = 17.8

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)·√(36/9) = 2

New margin

E₂ = 8.9·2 = 17.8

Answer

E₂ = 17.8
8.Confidence intervals

Answer: E = 7.819 kg

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.

Pooled standard deviation

sp² = (14·8.5² + 5·5²)/19 = 59.82
sp = 7.734

Margin of error

E = 2.093·7.734·√(1/15 + 1/6) = 7.819

Answer

E = 7.819 kg
9.Hypothesis tests & confidence intervals

Answer: z = 1.911

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 = √(17²/50 + 8²/60) = √6.847 = 2.617

Test statistic

z = (77 − 72)/2.617 = 1.911

Answer

z = 1.911

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

10.Hypothesis tests & confidence intervals

Answer: z = −1.755

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 = 36/120 = 0.3
p̂₂ = 49/120 = 0.4083
p̂ = 85/240 = 0.3542

Standard error under H₀

SE = √(0.3542·0.6458·(1/120 + 1/120)) = 0.06174

Test statistic

z = (0.3 − 0.4083)/0.06174 = −1.755

Answer

z = −1.755

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

11.Linear regression & correlation

Answer: r = 0.9607

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 = 30/5 = 6
ȳ = 203/5 = 40.6

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 118
Syy = Σ(y − ȳ)² = 377.2

Correlation coefficient

r = 118/√(40·377.2) = 0.9607

Answer

r = 0.9607
12.Linear regression & correlation

Answer: x = 22.08

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 = 105/5 = 21
ȳ = 512/5 = 102.4

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 244

Regression line

b = 244/40 = 6.1
a = 102.4 − 6.1·21 = −25.7
ŷ = −25.7 + 6.1·x

Solve for x

x = (109 − (−25.7))/6.1 = 22.08

Answer

x = 22.08

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