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
202530354045505560

Find the sample standard deviation s of the data: 32, 45, 47, 17, 41, 60.

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

Exam scores are normally distributed with mean 170 and standard deviation 10. Find the z-score z of a score of 185.

μ = 170σ = 10x = 185
z =
3.ProbabilityMedium
xp(x)00.110.120.3530.140.35μ = 2.5

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

x: 0, 1, 2, 3, 4p(x): 0.1, 0.1, 0.35, 0.1, 0.35
Var(X) =
4.ProbabilityHard
xp(x)10.2520.130.140.3550.2μ

X has the distribution shown, and Y = 0.5X + 1. Find the standard deviation σY of Y.

x: 1, 2, 3, 4, 5p(x): 0.25, 0.1, 0.1, 0.35, 0.2Y = 0.5X + 1
σY =
5.Normal distributionMedium
−3−2−10123zshaded area to the right of z = 0.0202

Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.0202. Find z to two decimals.

A = 0.0202
z =
6.Normal distributionHard
380420460500540580620htop 95%

The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. Drag the cut-off so the shaded top 95% is to the right of it. Find the cut-off x.

μ = 500 hσ = 40 h95%
x = h
7.Confidence intervalsMedium
n = 20, sd​ = 5.90LU = ?d̄ = 0.6E90%

The blood pressure of 20 patients is measured before and after a treatment. The differences d = before − after (mmHg) have mean 0.6 and standard deviation 5.9. Find the upper limit of the 90% confidence interval for the mean difference.

n = 20d̄ = 0.6sd = 5.990%t(19) = 1.729
upper limit = mmHg
8.Confidence intervalsHard
σ = 30, n = ?x̄ − Ex̄ + Ex̄E = 1.599%

The population standard deviation is 30. How large a sample n do you need so that a 99% confidence interval for the mean has a margin of error of at most 1.5?

σ = 30E = 1.599%
n =
9.Hypothesis tests & confidence intervalsMedium
t −2.13102.131

A sample of 16 students has a mean score of 53.4 and a sample standard deviation of 8. Find the test statistic t for H₀: μ = 50.

μ₀ = 50x̄ = 53.4s = 8n = 16
t =
10.Hypothesis tests & confidence intervalsHard
−1.961.96z = ?

In a trial, 125 of 250 patients on treatment A recover, and 78 of 120 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.

x₁ = 125n₁ = 250x₂ = 78n₂ = 120
z =
11.Linear regression & correlationMedium
xy144357569781989xy56080

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

n = 5Σx = 25Σy = 340Σx² = 165Σy² = 24428Σxy = 1928
r =
12.Linear regression & correlationHard
xy254961582110271232xy510102030ŷ = 24

The table gives 6 paired observations of x = load in kN and y = extension in mm. Use the least-squares line ŷ = a + b·x. For which value x does the line predict ŷ = 24?

n = 6ŷ = 24
x =

3Solutions

1.Descriptive statistics

Answer: s = 14.61

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 = 242/6 = 40.33

Squared deviations

Σ(x − x̄)² = (−8.333)² + 4.667² + 6.667² + (−23.33)² + 0.6667² + 19.67² = 1067

Standard deviation

s = √(1067/5) = 14.61

Answer

s = 14.61
2.Descriptive statistics

Answer: z = 1.5

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 = (185 − 170)/10 = 1.5

Answer

z = 1.5
3.Probability

Answer: Var(X) = 1.75

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 = 0·0.1 + 1·0.1 + 2·0.35 + 3·0.1 + 4·0.35 = 2.5

Mean of the squares

E(X²) = 0·0.1 + 1·0.1 + 4·0.35 + 9·0.1 + 16·0.35 = 8

Variance

Var(X) = 8 − 2.5² = 1.75

Answer

Var(X) = 1.75
4.Probability

Answer: σY = 0.7462

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 = 3.15, Var(X) = 2.227, σX = 1.492

Linear change

σY = |0.5|·1.492 = 0.7462

Answer

σY = 0.7462

The mean does move: E(Y) = 0.5·3.15 + 1 = 2.575.

5.Normal distribution

Answer: z = 2.05

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 − 0.0202 = 0.9798

Read the z-table backwards

Find 0.9798 inside the table and read off z.

z = 2.05

Answer

z = 2.05
6.Normal distribution

Answer: x = 434 h

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

Φ(z) = 0.05

z from the table

z ≈ −1.64

Back to the original units

x = 500 − 1.64·40 = 434 h

Answer

x = 434 h
7.Confidence intervals

Answer: upper limit = 2.881 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 = 19, t = 1.729

Margin of error

E = 1.729·5.9/√20 = 2.281

Interval

0.6 ± 2.281 → [−1.681, 2.881]

Answer

upper limit = 2.881 mmHg
8.Confidence intervals

Answer: n = 2655

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.

Solve for n

n = (2.576·30/1.5)² = 2654

Round up

A sample size is a whole number, and rounding down would make the margin too large.

n = 2655

Answer

n = 2655
9.Hypothesis tests & confidence intervals

Answer: t = 1.7

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 = 8/√16 = 8/4 = 2

Test statistic

t = (53.4 − 50)/2 = 1.7

Answer

t = 1.7
t0.025; df = 15 = 2.131

|t| < 2.131, so H₀ is not rejected at the 5% level (two-sided).

10.Hypothesis tests & confidence intervals

Answer: z = −2.714

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 = 125/250 = 0.5
p̂₂ = 78/120 = 0.65
p̂ = 203/370 = 0.5486

Standard error under H₀

SE = √(0.5486·0.4514·(1/250 + 1/120)) = 0.05526

Test statistic

z = (0.5 − 0.65)/0.05526 = −2.714

Answer

z = −2.714

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

11.Linear regression & correlation

Answer: r = 0.9968

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 = 1928 − 25·340/5 = 228
Sxx = 165 − 25²/5 = 40
Syy = 24428 − 340²/5 = 1308

Correlation coefficient

r = 228/√(40·1308) = 0.9968

Answer

r = 0.9968
12.Linear regression & correlation

Answer: x = 9.094

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 = 42/6 = 7
ȳ = 109/6 = 18.17

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 195

Regression line

b = 195/70 = 2.786
a = 18.17 − 2.786·7 = −1.333
ŷ = −1.333 + 2.786·x

Solve for x

x = (24 − (−1.333))/2.786 = 9.094

Answer

x = 9.094

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