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
2025303540455055

Find the sample standard deviation s of the data: 20, 18, 36, 44, 20, 29, 55.

n = 7
s =
2.Descriptive statisticsHard
z −2−1012

A quantity is normally distributed with mean 60 and standard deviation 8. What proportion P of values is above 56?

μ = 60σ = 8x = 56
P =
3.ProbabilityMedium
xp(x)100.2200.35300.35400.1μ = 23.5

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

x: 10, 20, 30, 40p(x): 0.2, 0.35, 0.35, 0.1
σ =
4.ProbabilityHard
n = 8, p = 0.2: P(X ≤ 1) = ?k012345678

8 components are tested, and each one fails with probability 0.2, independently. What is the probability P that at most one fails?

n = 8p = 0.2
P =
5.Normal distributionMedium
167.15188.5P = ?

The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. What fraction lies between 167.15 and 188.5 cm? Give P as a decimal.

μ = 178σ = 7x1 = 167.15x2 = 188.5
P =
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
n = 12, sd​ = 7.40LUd̄ = −7.5E = ?90%

The blood pressure of 12 patients is measured before and after a treatment. The differences d = before − after (mmHg) have mean −7.5 and standard deviation 7.4. Find the margin of error E of the 90% confidence interval for the mean difference.

n = 12d̄ = −7.5sd = 7.490%t(11) = 1.796
E = mmHg
8.Confidence intervalsHard
σ = 24, n = ?x̄ − Ex̄ + Ex̄E = 499%

The population standard deviation is 24. 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 4?

σ = 24E = 499%
n =
9.Hypothesis tests & confidence intervalsMedium
t −1.9901.99

A sample of 81 students has a mean score of 117.6 and a sample standard deviation of 12. Find the test statistic t for H₀: μ = 120.

μ₀ = 120x̄ = 117.6s = 12n = 81
t =
10.Hypothesis tests & confidence intervalsHard
z −1.9601.96

A sample of 16 measurements has mean 117. The population standard deviation is 5. Find the lower limit L of the 95% confidence interval for the mean.

x̄ = 117σ = 5n = 1695%
L =
11.Linear regression & correlationMedium
xy260364470573681xy246607080

The table gives 5 paired observations of x = hours studied and y = exam score. Find Pearson's correlation coefficient r.

n = 5
r =
12.Linear regression & correlationHard
xy127224322421522616xy2462025x = 1.5

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. A new observation (1.5, 31) is made. Find its residual e = y − ŷ.

n = 6(1.5, 31)
e =

3Solutions

1.Descriptive statistics

Answer: s = 14.03

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 = 222/7 = 31.71

Squared deviations

Σ(x − x̄)² = (−11.71)² + (−13.71)² + 4.286² + 12.29² + (−11.71)² + (−2.714)² + 23.29² = 1181

Standard deviation

s = √(1181/6) = 14.03

Answer

s = 14.03
2.Descriptive statistics

Answer: P = 0.6915

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 = −0.5

Standard normal table

Φ(−0.5) = 0.3085
P = 1 − 0.3085 = 0.6915

Answer

P = 0.6915
3.Probability

Answer: σ = 9.097

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 = 10·0.2 + 20·0.35 + 30·0.35 + 40·0.1 = 23.5

Mean of the squares

E(X²) = 100·0.2 + 400·0.35 + 900·0.35 + 1600·0.1 = 635

Variance

Var(X) = 635 − 23.5² = 82.75
σ = √82.75 = 9.097

Answer

σ = 9.097
4.Probability

Answer: P = 0.5033

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) = 0.88 = 0.1678
P(X = 1) = 8·0.2·0.87 = 0.3355

Add them

P = 0.1678 + 0.3355 = 0.5033

Answer

P = 0.5033
5.Normal distribution

Answer: P = 0.8726

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 = (167.15 − 178)/7 = −1.55
z₂ = (188.5 − 178)/7 = 1.5

Area between

P = Φ(1.5) − Φ(−1.55) = 0.9332 − 0.06057 = 0.8726

Answer

P = 0.8726
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: E = 3.837 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 = 1.796

Margin of error

E = 1.796·7.4/√12 = 3.837

Answer

E = 3.837 mmHg
8.Confidence intervals

Answer: n = 239

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 = (2.576·24/4)² = 238.9

Round up

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

n = 239

Answer

n = 239
9.Hypothesis tests & confidence intervals

Answer: t = −1.8

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 = 12/√81 = 12/9 = 1.333

Test statistic

t = (117.6 − 120)/1.333 = −1.8

Answer

t = −1.8
t0.025; df = 80 = 1.99

|t| < 1.99, so H₀ is not rejected at the 5% level (two-sided).

10.Hypothesis tests & confidence intervals

Answer: L = 114.5

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).

Margin of error

E = 1.96·5/√16 = 2.45

Interval limit

L = 117 − 2.45 = 114.5

Answer

L = 114.5
11.Linear regression & correlation

Answer: r = 0.9903

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 = 20/5 = 4
ȳ = 348/5 = 69.6

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = 51
Syy = Σ(y − ȳ)² = 265.2

Correlation coefficient

r = 51/√(10·265.2) = 0.9903

Answer

r = 0.9903
12.Linear regression & correlation

Answer: e = 5.457

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 = 132/6 = 22

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −31

Regression line

b = −31/17.5 = −1.771
a = 22 − (−1.771)·3.5 = 28.2
ŷ = 28.2 − 1.771·x

Predicted value

ŷ = 28.2 + (−1.771)·1.5 = 25.54

Residual

e = 31 − 25.54 = 5.457

Answer

e = 5.457

The point lies above the line.

ExamProblems · examproblems.com/courses/statistics-1