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
15202530354045505560

Find the sample standard deviation s of the data: 60, 12, 38, 52, 50, 53.

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

Exam scores are normally distributed with mean 70 and standard deviation 20. Find the z-score z of a score of 54.

μ = 70σ = 20x = 54
z =
3.ProbabilityMedium
xp(x)10.2520.230.2540.2550.05μ = 2.65

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

x: 1, 2, 3, 4, 5p(x): 0.25, 0.2, 0.25, 0.25, 0.05
σ =
4.ProbabilityHard
n = 8, p = 0.6: P(X ≤ 1) = ?k012345678

8 components are tested, and each one fails with probability 0.6, independently. What is the probability P that at most one fails?

n = 8p = 0.6
P =
5.Normal distributionMedium
172.4183.95P = ?

The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. What fraction lies between 172.4 and 183.95 cm? Give P as a decimal.

μ = 178σ = 7x1 = 172.4x2 = 183.95
P =
6.Normal distributionHard
μ = 250x = ?80%

The bag weights are normally distributed with mean 250 g and standard deviation 5 g. Find the value x below which 80% of them lie (the 80th percentile).

μ = 250σ = 5P = 80%
x = g
7.Confidence intervalsMedium
n = 12, sd​ = 7.60LU = ?d̄ = −4.2E95%

The blood pressure of 12 patients is measured before and after a treatment. The differences d = before − after (mmHg) have mean −4.2 and standard deviation 7.6. Find the upper limit of the 95% confidence interval for the mean difference.

n = 12d̄ = −4.2sd = 7.695%t(11) = 2.201
upper limit = mmHg
8.Confidence intervalsHard
σ = 18, n = ?x̄ − Ex̄ + Ex̄E = 590%

The population standard deviation is 18. How large a sample n do you need so that a 90% confidence interval for the mean has a margin of error of at most 5?

σ = 18E = 590%
n =
9.Hypothesis tests & confidence intervalsMedium
−2.0452.045t = ?df = 29

Two teaching methods are compared. Group 1: 20 students, mean score 98.5, standard deviation 2. Group 2: 11 students, mean score 93, standard deviation 10. 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 = 20x̄₁ = 98.5s₁ = 2n₂ = 11x̄₂ = 93s₂ = 10
t =
10.Hypothesis tests & confidence intervalsHard
−1.961.96z = ?

In a trial, 135 of 300 patients on treatment A recover, and 42 of 80 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.

x₁ = 135n₁ = 300x₂ = 42n₂ = 80
z =
11.Linear regression & correlationMedium
xy235339450553653760xy246405060

The table gives 6 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 = 6
r² =
12.Linear regression & correlationHard
xy123221321417516613xy2461520ŷ = 18

The table gives 6 paired observations of x = car age in years and y = price in k€. Use the least-squares line ŷ = a + b·x. For which value x does the line predict ŷ = 18?

n = 6ŷ = 18
x =

3Solutions

1.Descriptive statistics

Answer: s = 17.3

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 = 265/6 = 44.17

Squared deviations

Σ(x − x̄)² = 15.83² + (−32.17)² + (−6.167)² + 7.833² + 5.833² + 8.833² = 1497

Standard deviation

s = √(1497/5) = 17.3

Answer

s = 17.3
2.Descriptive statistics

Answer: z = −0.8

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 = (54 − 70)/20 = −0.8

Answer

z = −0.8
3.Probability

Answer: σ = 1.236

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 + 2·0.2 + 3·0.25 + 4·0.25 + 5·0.05 = 2.65

Mean of the squares

E(X²) = 1·0.25 + 4·0.2 + 9·0.25 + 16·0.25 + 25·0.05 = 8.55

Variance

Var(X) = 8.55 − 2.65² = 1.528
σ = √1.528 = 1.236

Answer

σ = 1.236
4.Probability

Answer: P = 0.00852

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.

Zero and one failure

P(X = 0) = 0.48 = 0.0006554
P(X = 1) = 8·0.6·0.47 = 0.007864

Add them

P = 0.0006554 + 0.007864 = 0.00852

Answer

P = 0.00852
5.Normal distribution

Answer: P = 0.5905

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 = (172.4 − 178)/7 = −0.8
z₂ = (183.95 − 178)/7 = 0.85

Area between

P = Φ(0.85) − Φ(−0.8) = 0.8023 − 0.2119 = 0.5905

Answer

P = 0.5905
6.Normal distribution

Answer: x = 254.2 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) = 0.8 → z = 0.8416

Back to x

x = 250 + 0.8416·5 = 254.21 g

Answer

x = 254.21 g
7.Confidence intervals

Answer: upper limit = 0.6288 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 = 2.201·7.6/√12 = 4.829

Interval

−4.2 ± 4.829 → [−9.029, 0.6288]

Answer

upper limit = 0.6288 mmHg
8.Confidence intervals

Answer: n = 36

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 = (1.645·18/5)² = 35.07

Round up

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

n = 36

Answer

n = 36
9.Hypothesis tests & confidence intervals

Answer: t = 2.405

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 = (19·2² + 10·10²)/29 = 37.1
sp = 6.091

Standard error

SE = 6.091·√(1/20 + 1/11) = 2.287

Test statistic

t = (98.5 − 93)/2.287 = 2.405

Answer

t = 2.405

With df = 29, the two-sided 5% critical value is 2.045; |t| is larger, so H₀ is rejected.

10.Hypothesis tests & confidence intervals

Answer: z = −1.195

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 = 135/300 = 0.45
p̂₂ = 42/80 = 0.525
p̂ = 177/380 = 0.4658

Standard error under H₀

SE = √(0.4658·0.5342·(1/300 + 1/80)) = 0.06277

Test statistic

z = (0.45 − 0.525)/0.06277 = −1.195

Answer

z = −1.195

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

11.Linear regression & correlation

Answer: r² = 0.9229

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 = 27/6 = 4.5
ȳ = 290/6 = 48.33

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = 85
Syy = Σ(y − ȳ)² = 447.3

Coefficient of determination

r² = 85²/(17.5·447.3) = 0.9229

Answer

r² = 0.9229

About 92.29% of the variation in y is explained by the line.

12.Linear regression & correlation

Answer: x = 3.754

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 = 21/6 = 3.5
ȳ = 111/6 = 18.5

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −34.5

Regression line

b = −34.5/17.5 = −1.971
a = 18.5 − (−1.971)·3.5 = 25.4
ŷ = 25.4 − 1.971·x

Solve for x

x = (18 − 25.4)/(−1.971) = 3.754

Answer

x = 3.754

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