Introductory Statistics: cram pack

Formula sheets, the key method steps for every topic and 12 practice problems, with full solutions at the back.

New problemsBack to the course

In the print window, choose “Save as PDF” to keep a copy on your device.

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
30354045505560

Find the sample standard deviation s of the data: 60, 55, 27, 35, 49, 50.

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

Exam scores are normally distributed with mean 100 and standard deviation 10. Find the z-score z of a score of 120.

μ = 100σ = 10x = 120
z =
3.ProbabilityMedium
red: 4blue: 6

A bag holds 4 red and 6 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?

4 / 6
P =
4.ProbabilityHard
xp(x)00.220.3540.2560.2μ

X has the distribution shown, and Y = 0.5X + 10. Find the standard deviation σY of Y.

x: 0, 2, 4, 6p(x): 0.2, 0.35, 0.25, 0.2Y = 0.5X + 10
σY =
5.Normal distributionMedium
470548P = ?

The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. What fraction lies between 470 and 548 h? Give P as a decimal.

μ = 500σ = 40x1 = 470x2 = 548
P =
6.Normal distributionHard
157164171178185192199cmbottom 40%

The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. Drag the cut-off so the shaded bottom 40% is to the left of it. Find the cut-off x.

μ = 178 cmσ = 7 cm40%
x = cm
7.Confidence intervalsMedium
x̄E₁ = 1.795%, n = 50x̄E₂ = ?95%, n = 200

A 95% confidence interval for a mean, based on 50 observations and a known σ, has a margin of error of 1.7. What is the margin of error E₂ if the same study uses 200 observations instead?

95%n₁ = 50E₁ = 1.7n₂ = 200
E₂ =
8.Confidence intervalsHard
μ₁ − μ₂, t(29) = 2.0450LUx̄₁ − x̄₂E = ?95%

Two fertilisers are tested. Plot group 1: 26 plots, mean yield 51 kg, standard deviation 2.5 kg. Plot group 2: 5 plots, mean 48 kg, standard deviation 3 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 = 26x̄₁ = 51s₁ = 2.5n₂ = 5x̄₂ = 48s₂ = 395%t(29) = 2.045
E = kg
9.Hypothesis tests & confidence intervalsMedium
−2.0932.093t = ?df = 19

Two teaching methods are compared. Group 1: 13 students, mean score 107, standard deviation 7.5. Group 2: 8 students, mean score 101.5, standard deviation 8. 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 = 13x̄₁ = 107s₁ = 7.5n₂ = 8x̄₂ = 101.5s₂ = 8
t =
10.Hypothesis tests & confidence intervalsHard
H₀H₁μ₀x̄cμ₀ + 5α = 0.05βσ = 26, power 90%, n = ?

A one-sided z test at α = 0.05 must detect a shift of the mean of 5 units with a power of 90%. The standard deviation is 26. Find the minimum sample size n.

α = 0.05power = 90%δ = 5σ = 26zα = 1.645zβ = 1.282
n =
11.Linear regression & correlationMedium
xy25946968789510111xy5106080100

The table gives 5 paired observations of x = hours studied and y = exam score. Find the coefficient of determination r² of the least-squares line.

n = 5
r² =
12.Linear regression & correlationHard
xy1550186921832498279930121xy1520253050100ŷ = 88

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. For which value x does the line predict ŷ = 88?

n = 6ŷ = 88
x =

3Solutions

1.Descriptive statistics

Answer: s = 12.52

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 = 276/6 = 46

Squared deviations

Σ(x − x̄)² = 14² + 9² + (−19)² + (−11)² + 3² + 4² = 784

Standard deviation

s = √(784/5) = 12.52

Answer

s = 12.52
2.Descriptive statistics

Answer: z = 2

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 = (120 − 100)/10 = 2

Answer

z = 2
3.Probability

Answer: P = 0.1333

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 = 4/10 · 3/9 = 12/90 = 0.1333

Answer

P = 0.1333
4.Probability

Answer: σY = 1.023

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) = 4.19, σX = 2.047

Linear change

σY = |0.5|·2.047 = 1.023

Answer

σY = 1.023

The mean does move: E(Y) = 0.5·2.9 + 10 = 11.45.

5.Normal distribution

Answer: P = 0.6583

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 = (470 − 500)/40 = −0.75
z₂ = (548 − 500)/40 = 1.2

Area between

P = Φ(1.2) − Φ(−0.75) = 0.8849 − 0.2266 = 0.6583

Answer

P = 0.6583
6.Normal distribution

Answer: x = 176.3 cm

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) = 0.4

z from the table

z ≈ −0.25

Back to the original units

x = 178 − 0.25·7 = 176.3 cm

Answer

x = 176.3 cm
7.Confidence intervals

Answer: E₂ = 0.85

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 = (1.96/1.96)·√(50/200) = 0.5

New margin

E₂ = 1.7·0.5 = 0.85

Answer

E₂ = 0.85
8.Confidence intervals

Answer: E = 2.571 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 = (25·2.5² + 4·3²)/29 = 6.629
sp = 2.575

Margin of error

E = 2.045·2.575·√(1/26 + 1/5) = 2.571

Answer

E = 2.571 kg
9.Hypothesis tests & confidence intervals

Answer: t = 1.592

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 = (12·7.5² + 7·8²)/19 = 59.11
sp = 7.688

Standard error

SE = 7.688·√(1/13 + 1/8) = 3.455

Test statistic

t = (107 − 101.5)/3.455 = 1.592

Answer

t = 1.592

With df = 19, the two-sided 5% critical value is 2.093; |t| is smaller, so H₀ is not rejected.

10.Hypothesis tests & confidence intervals

Answer: n = 232

Sample size for a test: to detect a true shift δ with power 1 − β in a one-sided z test at level α, the standard error must satisfy δ = (z_α + z_β)·σ/√n, so n = ((z_α + z_β)·σ/δ)², rounded up.

Solve for n

n = ((1.645 + 1.282)·26/5)² = 231.7

Round up

n = 232

Answer

n = 232
11.Linear regression & correlation

Answer: r² = 0.989

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 = 421/5 = 84.2

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 260
Syy = Σ(y − ȳ)² = 1709

Coefficient of determination

r² = 260²/(40·1709) = 0.989

Answer

r² = 0.989

About 98.9% of the variation in y is explained by the line.

12.Linear regression & correlation

Answer: x = 22.8

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 = 520/6 = 86.67

Sums of squares and products

Sxx = Σ(x − x̄)² = 157.5
Sxy = Σ(x − x̄)(y − ȳ) = 690

Regression line

b = 690/157.5 = 4.381
a = 86.67 − 4.381·22.5 = −11.9
ŷ = −11.9 + 4.381·x

Solve for x

x = (88 − (−11.9))/4.381 = 22.8

Answer

x = 22.8

ExamProblems · examproblems.com/courses/statistics-1