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
2530354045505560

Find the sample standard deviation s of the data: 47, 24, 60, 42, 53.

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

A quantity is normally distributed with mean 170 and standard deviation 10. What proportion P of values is below 165?

μ = 170σ = 10x = 165
P =
3.ProbabilityMedium
xp(x)00.3510.220.230.25μ = 1.35

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

x: 0, 1, 2, 3p(x): 0.35, 0.2, 0.2, 0.25
σ =
4.ProbabilityHard
n = 11, p = 0.6: P(X ≤ 1) = ?k01234567891011

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

n = 11p = 0.6
P =
5.Normal distributionMedium
242256P = ?

The bag weights are normally distributed with mean 250 g and standard deviation 5 g. What fraction lies between 242 and 256 g? Give P as a decimal.

μ = 250σ = 5x1 = 242x2 = 256
P =
6.Normal distributionHard
μ = 500x = ?10%

The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. Find the value x below which 10% of them lie (the 10th percentile).

μ = 500σ = 40P = 10%
x = h
7.Confidence intervalsMedium
x̄E₁ = 5.399%, n = 100x̄E₂ = ?90%, n = 400

A 99% confidence interval for a mean, based on 100 observations and a known σ, has a margin of error of 5.3. What is the margin of error E₂ of a 90% interval based on 400 observations?

99%n₁ = 100E₁ = 5.390%n₂ = 400
E₂ =
8.Confidence intervalsHard
σ = 32, n = ?x̄ − Ex̄ + Ex̄E = 495%

The population standard deviation is 32. 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 4?

σ = 32E = 495%
n =
9.Hypothesis tests & confidence intervalsMedium
−1.36p = ?

The mean of a population is claimed to be 50; the population standard deviation is known to be 20. A random sample of 100 observations has a mean of 47.28. Test H₀: μ = 50 against H₁: μ < 50. Find the p-value p.

μ₀ = 50σ = 20n = 100x̄ = 47.28H₁: μ < μ₀
p =
10.Hypothesis tests & confidence intervalsHard
z −1.9601.96

A sample of 16 measurements has mean 77. The population standard deviation is 20. Find the lower limit L of the 95% confidence interval for the mean.

x̄ = 77σ = 20n = 1695%
L =
11.Linear regression & correlationMedium
xy128237341442553xy24304050

The table gives 5 paired observations of x = advertising spend in k€ and y = sales in k€. Their sums are given. Find Pearson's correlation coefficient r from these sums.

n = 5Σx = 15Σy = 201Σx² = 55Σy² = 8407Σxy = 658
r =
12.Linear regression & correlationHard
xy124222320417515611xy24610152025x = 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. Predict ŷ for x = 1.5.

n = 6x = 1.5
ŷ =

3Solutions

1.Descriptive statistics

Answer: s = 13.63

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 = 226/5 = 45.2

Squared deviations

Σ(x − x̄)² = 1.8² + (−21.2)² + 14.8² + (−3.2)² + 7.8² = 742.8

Standard deviation

s = √(742.8/4) = 13.63

Answer

s = 13.63
2.Descriptive statistics

Answer: P = 0.3085

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 = 0.3085

Answer

P = 0.3085
3.Probability

Answer: σ = 1.195

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 = 0·0.35 + 1·0.2 + 2·0.2 + 3·0.25 = 1.35

Mean of the squares

E(X²) = 0·0.35 + 1·0.2 + 4·0.2 + 9·0.25 = 3.25

Variance

Var(X) = 3.25 − 1.35² = 1.427
σ = √1.427 = 1.195

Answer

σ = 1.195
4.Probability

Answer: P = 0.000734

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) = 0.411 = 4.194 × 10⁻⁵
P(X = 1) = 11·0.6·0.410 = 0.0006921

Add them

P = 4.194 × 10⁻⁵ + 0.0006921 = 0.000734

Answer

P = 0.000734
5.Normal distribution

Answer: P = 0.8301

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 = (242 − 250)/5 = −1.6
z₂ = (256 − 250)/5 = 1.2

Area between

P = Φ(1.2) − Φ(−1.6) = 0.8849 − 0.0548 = 0.8301

Answer

P = 0.8301
6.Normal distribution

Answer: x = 448.7 h

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) = 0.1 → z = −1.282

Back to x

x = 500 + −1.282·40 = 448.74 h

Answer

x = 448.74 h
7.Confidence intervals

Answer: E₂ = 1.692

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 = (1.645/2.576)·√(100/400) = 0.3193

New margin

E₂ = 5.3·0.3193 = 1.692

Answer

E₂ = 1.692
8.Confidence intervals

Answer: n = 246

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 = (1.96·32/4)² = 245.9

Round up

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

n = 246

Answer

n = 246
9.Hypothesis tests & confidence intervals

Answer: p = 0.08692

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 = 20/√100 = 2
z = (47.28 − 50)/2 = −1.36

Tail area

Φ(1.36) = 0.9131
1 − Φ(1.36) = 0.0869

p-value

p = 0.08692

Answer

p = 0.08692

p > 0.05, so H₀ is not rejected at the 5% level.

10.Hypothesis tests & confidence intervals

Answer: L = 67.2

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 = 1.96·20/√16 = 9.8

Interval limit

L = 77 − 9.8 = 67.2

Answer

L = 67.2
11.Linear regression & correlation

Answer: r = 0.9621

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.

Centred sums

Sxy = 658 − 15·201/5 = 55
Sxx = 55 − 15²/5 = 10
Syy = 8407 − 201²/5 = 326.8

Correlation coefficient

r = 55/√(10·326.8) = 0.9621

Answer

r = 0.9621
12.Linear regression & correlation

Answer: ŷ = 23.25

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 = 109/6 = 18.17

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −44.5

Regression line

b = −44.5/17.5 = −2.543
a = 18.17 − (−2.543)·3.5 = 27.07
ŷ = 27.07 − 2.543·x

Prediction

ŷ = 27.07 + (−2.543)·1.5 = 23.25

Answer

ŷ = 23.25

ExamProblems · examproblems.com/courses/statistics-1