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
20253035404550

Find the sample standard deviation s of the data: 47, 17, 53, 32, 53, 47.

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

A quantity is normally distributed with mean 170 and standard deviation 8. What proportion P of values is above 178?

μ = 170σ = 8x = 178
P =
3.ProbabilityMedium
xp(x)00.1510.2520.330.3μ = 1.75

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

x: 0, 1, 2, 3p(x): 0.15, 0.25, 0.3, 0.3
σ =
4.ProbabilityHard
n = 11, p = 0.2: P(X = 1) = ?k01234567891011

A test has 11 multiple-choice questions. A student who guesses gets each one right with probability 0.2, independently. What is the probability P of exactly 1 correct answers?

n = 11p = 0.2k = 1
P =
5.Normal distributionMedium
−3−2−10123zshaded area between −z and z = 0.5285

Drag z so that the shaded area between −z and z under the standard normal curve is 0.5285. Find z to two decimals.

A = 0.5285
z =
6.Normal distributionHard
380420460500540580620htop 85%

The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. Drag the cut-off so the shaded top 85% is to the right of it. Find the cut-off x.

μ = 500 hσ = 40 h85%
x = h
7.Confidence intervalsMedium
n = 9, sd​ = 6.50LU = ?d̄ = −2.6E95%

The blood pressure of 9 patients is measured before and after a treatment. The differences d = before − after (mmHg) have mean −2.6 and standard deviation 6.5. Find the upper limit of the 95% confidence interval for the mean difference.

n = 9d̄ = −2.6sd = 6.595%t(8) = 2.306
upper limit = mmHg
8.Confidence intervalsHard
μ₁ − μ₂, t(15) = 2.1310LU = ?x̄₁ − x̄₂E95%

Two fertilisers are tested. Plot group 1: 11 plots, mean yield 14 kg, standard deviation 2 kg. Plot group 2: 6 plots, mean 24 kg, standard deviation 8 kg. Assume equal variances (pooled t, df = n₁ + n₂ − 2). Find the upper limit of the 95% confidence interval for μ₁ − μ₂.

n₁ = 11x̄₁ = 14s₁ = 2n₂ = 6x̄₂ = 24s₂ = 895%t(15) = 2.131
upper limit = kg
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 108.5 mm, and the line's known standard deviation is 12 mm. Line 2: a sample of 80 parts has mean 111.6 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 = 80x̄₁ = 108.5 mmσ₁ = 12 mmn₂ = 80x̄₂ = 111.6 mmσ₂ = 8 mm
z =
10.Hypothesis tests & confidence intervalsHard
σ = 10, n = ?x̄ − Ex̄ + Ex̄E = 299%

How large must a sample be so that the 99% confidence interval for a mean has a margin of error of at most 2? The standard deviation is 10. Find the smallest sample size n.

σ = 10E = 299%
n =
11.Linear regression & correlationMedium
xy126225319416515xy24152025

The table gives 5 paired observations of x = car age in years and y = price in k€. Find the coefficient of determination r² of the least-squares line.

n = 5
r² =
12.Linear regression & correlationHard
xy3355407499541160xy510405060x = 6

The table gives 5 paired observations of x = advertising spend in k€ and y = sales in k€. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 6.

n = 5x = 6
ŷ =

3Solutions

1.Descriptive statistics

Answer: s = 14.25

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 = 249/6 = 41.5

Squared deviations

Σ(x − x̄)² = 5.5² + (−24.5)² + 11.5² + (−9.5)² + 11.5² + 5.5² = 1016

Standard deviation

s = √(1016/5) = 14.25

Answer

s = 14.25
2.Descriptive statistics

Answer: P = 0.1587

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

Standard normal table

Φ(1) = 0.8413
P = 1 − 0.8413 = 0.1587

Answer

P = 0.1587
3.Probability

Answer: σ = 1.043

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 = 0·0.15 + 1·0.25 + 2·0.3 + 3·0.3 = 1.75

Mean of the squares

E(X²) = 0·0.15 + 1·0.25 + 4·0.3 + 9·0.3 = 4.15

Variance

Var(X) = 4.15 − 1.75² = 1.087
σ = √1.087 = 1.043

Answer

σ = 1.043
4.Probability

Answer: P = 0.2362

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.

Binomial formula

P(X = 1) = 11·0.21·0.810 = 11·0.2·0.1074 = 0.2362

Answer

P = 0.2362
5.Normal distribution

Answer: z = 0.72

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 + 0.5285)/2 = 0.7642

Read the z-table backwards

Find 0.7642 inside the table and read off z.

z = 0.72

Answer

z = 0.72
6.Normal distribution

Answer: x = 458 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) = 0.15

z from the table

z ≈ −1.04

Back to the original units

x = 500 − 1.04·40 = 458 h

Answer

x = 458 h
7.Confidence intervals

Answer: upper limit = 2.396 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 = 8, t = 2.306

Margin of error

E = 2.306·6.5/√9 = 4.996

Interval

−2.6 ± 4.996 → [−7.596, 2.396]

Answer

upper limit = 2.396 mmHg
8.Confidence intervals

Answer: upper limit = −4.702 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 = (10·2² + 5·8²)/15 = 24
sp = 4.899

Margin of error

E = 2.131·4.899·√(1/11 + 1/6) = 5.298

Interval

−10 ± 5.298 → [−15.3, −4.702]

Answer

upper limit = −4.702 kg
9.Hypothesis tests & confidence intervals

Answer: z = −1.923

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 = √(12²/80 + 8²/80) = √2.6 = 1.612

Test statistic

z = (108.5 − 111.6)/1.612 = −1.923

Answer

z = −1.923

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

10.Hypothesis tests & confidence intervals

Answer: n = 166

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).

Solve for n

n = (2.576·10/2)² = 165.9

Round up

n = 166

Answer

n = 166
11.Linear regression & correlation

Answer: r² = 0.9348

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 = 101/5 = 20.2

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = −31
Syy = Σ(y − ȳ)² = 102.8

Coefficient of determination

r² = −31²/(10·102.8) = 0.9348

Answer

r² = 0.9348

About 93.48% of the variation in y is explained by the line.

12.Linear regression & correlation

Answer: ŷ = 44.4

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 = 35/5 = 7
ȳ = 238/5 = 47.6

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 128

Regression line

b = 128/40 = 3.2
a = 47.6 − 3.2·7 = 25.2
ŷ = 25.2 + 3.2·x

Prediction

ŷ = 25.2 + 3.2·6 = 44.4

Answer

ŷ = 44.4

ExamProblems · examproblems.com/courses/statistics-1