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
1520253035404550

Find the sample standard deviation s of the data: 52, 14, 13, 22, 54.

n = 5
s =
2.Descriptive statisticsHard
z −2−1012

A quantity is normally distributed with mean 50 and standard deviation 10. What proportion P of values is above 35?

μ = 50σ = 10x = 35
P =
3.ProbabilityMedium
P(at least one six) = ??1?2?3P(six) = 1/6 on each roll

A fair die is rolled 3 times. What is the probability P of getting at least one six?

n = 3
P =
4.ProbabilityHard
n = 5, p = 0.5: P(X = 1) = ?k012345

A test has 5 multiple-choice questions. A student who guesses gets each one right with probability 0.5, independently. What is the probability P of exactly 1 correct answers?

n = 5p = 0.5k = 1
P =
5.Normal distributionMedium
57.567P = ?

The exam scores are normally distributed with mean 65 points and standard deviation 10 points. What fraction lies between 57.5 and 67 points? Give P as a decimal.

μ = 65σ = 10x1 = 57.5x2 = 67
P =
6.Normal distributionHard
0.04 mm0D − d?

Hole diameters D are normal with mean 25.04 mm and standard deviation 0.02 mm. Shaft diameters d are independent and normal with mean 25 mm and standard deviation 0.01 mm. A randomly chosen shaft fits when D − d > 0. Find the probability P that it fits. Round z to two decimals.

hole: μ = 25.04 mm, σ = 0.02 mmshaft: μ = 25 mm, σ = 0.01 mm
P =
7.Confidence intervalsMedium
n = 16, s = 5.5LU = ?x̄ = 78E99%

A sample of 16 items has mean 78 and standard deviation s = 5.5. The population is roughly normal. Find the upper limit of the 99% confidence interval for the mean.

n = 16x̄ = 78s = 5.599%t(15) = 2.947
upper limit =
8.Confidence intervalsHard
σ = 38, n = ?x̄ − Ex̄ + Ex̄E = 195%

The population standard deviation is 38. How large a sample n do you need so that a 95% confidence interval for the mean has a margin of error of at most 1?

σ = 38E = 195%
n =
9.Hypothesis tests & confidence intervalsMedium
−2.242.24p = ?

The mean of a population is claimed to be 50; the population standard deviation is known to be 24. A random sample of 36 observations has a mean of 58.96. Test H₀: μ = 50 against H₁: μ ≠ 50. Find the p-value p.

μ₀ = 50σ = 24n = 36x̄ = 58.96H₁: μ ≠ μ₀
p =
10.Hypothesis tests & confidence intervalsHard
H₀H₁μ₀x̄cμ₀ + 4α = 0.05βσ = 27, power 80%, n = ?

A one-sided z test at α = 0.05 must detect a shift of the mean of 4 units with a power of 80%. The standard deviation is 27. Find the minimum sample size n.

α = 0.05power = 80%δ = 4σ = 27zα = 1.645zβ = 0.842
n =
11.Linear regression & correlationMedium
xy274136168241029xy510102030

The table gives 5 paired observations of x = load in kN and y = extension in mm. Find the coefficient of determination r² of the least-squares line.

n = 5
r² =
12.Linear regression & correlationHard
xy366471575688788xy46708090x = 3.5

The table gives 5 paired observations of x = hours studied and y = exam score. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 3.5.

n = 5x = 3.5
ŷ =

3Solutions

1.Descriptive statistics

Answer: s = 20.4

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 = 155/5 = 31

Squared deviations

Σ(x − x̄)² = 21² + (−17)² + (−18)² + (−9)² + 23² = 1664

Standard deviation

s = √(1664/4) = 20.4

Answer

s = 20.4
2.Descriptive statistics

Answer: P = 0.9332

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 = −1.5

Standard normal table

Φ(−1.5) = 0.06681
P = 1 − 0.06681 = 0.9332

Answer

P = 0.9332
3.Probability

Answer: P = 0.4213

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.

Complement

P(no six) = (5/6)3 = 0.5787
P = 1 − 0.5787 = 0.4213

Answer

P = 0.4213
4.Probability

Answer: P = 0.1563

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) = 5·0.51·0.54 = 5·0.5·0.0625 = 0.1563

Answer

P = 0.1563
5.Normal distribution

Answer: P = 0.3526

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 = (57.5 − 65)/10 = −0.75
z₂ = (67 − 65)/10 = 0.2

Area between

P = Φ(0.2) − Φ(−0.75) = 0.5793 − 0.2266 = 0.3526

Answer

P = 0.3526
6.Normal distribution

Answer: P = 0.9633

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 = 25.04 − 25 = 0.04 mm, σ = 0.02236 mm

Standardise

z = (0 − 0.04)/0.02236 = −1.79

Probability

P = 1 − Φ(−1.79) = 0.9633

Answer

P = 0.9633
7.Confidence intervals

Answer: upper limit = 82.05

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.

Critical value

df = n − 1 = 15, t = 2.947

Margin of error

E = 2.947·5.5/√16 = 4.052

Interval

78 ± 4.052 → [73.95, 82.05]

Answer

upper limit = 82.05
8.Confidence intervals

Answer: n = 5548

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 = (1.96·38/1)² = 5547

Round up

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

n = 5548

Answer

n = 5548
9.Hypothesis tests & confidence intervals

Answer: p = 0.02509

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.

Test statistic

SE = σ/√n = 24/√36 = 4
z = (58.96 − 50)/4 = 2.24

Tail area

Φ(2.24) = 0.9875
1 − Φ(2.24) = 0.0125

p-value

p = 2·0.0125 = 0.02509

Answer

p = 0.02509

p ≤ 0.05, so H₀ is rejected at the 5% level.

10.Hypothesis tests & confidence intervals

Answer: n = 282

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 + 0.842)·27/4)² = 281.8

Round up

n = 282

Answer

n = 282
11.Linear regression & correlation

Answer: r² = 0.986

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 = 89/5 = 17.8

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 110
Syy = Σ(y − ȳ)² = 306.8

Coefficient of determination

r² = 110²/(40·306.8) = 0.986

Answer

r² = 0.986

About 98.6% of the variation in y is explained by the line.

12.Linear regression & correlation

Answer: ŷ = 68.45

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 = 25/5 = 5
ȳ = 388/5 = 77.6

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = 61

Regression line

b = 61/10 = 6.1
a = 77.6 − 6.1·5 = 47.1
ŷ = 47.1 + 6.1·x

Prediction

ŷ = 47.1 + 6.1·3.5 = 68.45

Answer

ŷ = 68.45

ExamProblems · examproblems.com/courses/statistics-1