Introductory Statistics: cram pack

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

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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
152025303540455055

Find the sample standard deviation s of the data: 15, 23, 31, 33, 22, 56.

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
xp(x)100.2200.3300.25400.25μ = 25.5

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

x: 10, 20, 30, 40p(x): 0.2, 0.3, 0.25, 0.25
σ =
4.ProbabilityHard
n = 11, p = 0.2: P(X ≤ 1) = ?k01234567891011

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

n = 11p = 0.2
P =
5.Normal distributionMedium
750 g745.1 gX + Y?

X is normal with mean 250 g and standard deviation 4 g; Y is independent and normal with mean 500 g and standard deviation 9 g. Find the probability P that X + Y is more than 745.1 g. Round z to two decimals.

μX = 250, σX = 4μY = 500, σY = 9X + Y > 745.1 g
P =
6.Normal distributionHard
μ = 65x = ?99%

The exam scores are normally distributed with mean 65 points and standard deviation 10 points. Find the value x below which 99% of them lie (the 99th percentile).

μ = 65σ = 10P = 99%
x = points
7.Confidence intervalsMedium
n = 11, s = 7LU = ?x̄ = 14E99%

A sample of 11 items has mean 14 and standard deviation s = 7. The population is roughly normal. Find the upper limit of the 99% confidence interval for the mean.

n = 11x̄ = 14s = 799%t(10) = 3.169
upper limit =
8.Confidence intervalsHard
p ≈ 0.8, n = ?p̂ − Ep̂ + Ep̂E = 0.0599%

A pilot study suggests that about 0.8 of products are defective. How many products n must be inspected so that a 99% confidence interval for the proportion has a margin of error of at most 0.05?

p ≈ 0.8E = 0.0599%
n =
9.Hypothesis tests & confidence intervalsMedium
1.46p = ?

The mean of a population is claimed to be 500; the population standard deviation is known to be 6. A random sample of 36 observations has a mean of 501.46. Test H₀: μ = 500 against H₁: μ > 500. Find the p-value p.

μ₀ = 500σ = 6n = 36x̄ = 501.46H₁: μ > μ₀
p =
10.Hypothesis tests & confidence intervalsHard
H₀H₁μ₀x̄cμ₀ + 3α = 0.01βσ = 16, power 95%, n = ?

A one-sided z test at α = 0.01 must detect a shift of the mean of 3 units with a power of 95%. The standard deviation is 16. Find the minimum sample size n.

α = 0.01power = 95%δ = 3σ = 16zα = 2.326zβ = 1.645
n =
11.Linear regression & correlationMedium
xy143244350453557658xy246455055

The table gives 6 paired observations of x = advertising spend in k€ and y = sales in k€. Find the coefficient of determination r² of the least-squares line.

n = 6
r² =
12.Linear regression & correlationHard
xy122218317410511xy24101520ŷ = 20

The table gives 5 paired observations of x = car age in years and y = price in k€. Use the least-squares line ŷ = a + b·x. For which value x does the line predict ŷ = 20?

n = 5ŷ = 20
x =

3Solutions

1.Descriptive statistics

Answer: s = 14.31

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 = 180/6 = 30

Squared deviations

Σ(x − x̄)² = (−15)² + (−7)² + 1² + 3² + (−8)² + 26² = 1024

Standard deviation

s = √(1024/5) = 14.31

Answer

s = 14.31
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: σ = 10.71

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 + 20·0.3 + 30·0.25 + 40·0.25 = 25.5

Mean of the squares

E(X²) = 100·0.2 + 400·0.3 + 900·0.25 + 1600·0.25 = 765

Variance

Var(X) = 765 − 25.5² = 114.8
σ = √114.8 = 10.71

Answer

σ = 10.71
4.Probability

Answer: P = 0.3221

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) = 0.811 = 0.0859
P(X = 1) = 11·0.2·0.810 = 0.2362

Add them

P = 0.0859 + 0.2362 = 0.3221

Answer

P = 0.3221
5.Normal distribution

Answer: P = 0.6915

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 sum

μ = 750 g, σ = √(4² + 9²) = 9.849 g

Standardise

z = (745.1 − 750)/9.849 = −0.5

Upper tail

P = 1 − Φ(−0.5) = 1 − 0.3085 = 0.6915

Answer

P = 0.6915
6.Normal distribution

Answer: x = 88.26 points

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) = 0.99 → z = 2.326

Back to x

x = 65 + 2.326·10 = 88.26 points

Answer

x = 88.26 points
7.Confidence intervals

Answer: upper limit = 20.69

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 = 10, t = 3.169

Margin of error

E = 3.169·7/√11 = 6.688

Interval

14 ± 6.688 → [7.312, 20.69]

Answer

upper limit = 20.69
8.Confidence intervals

Answer: n = 425

Confidence interval for a proportion: p̂ ± z·√(p̂(1 − p̂)/n), where p̂ = x/n. To reach a margin E you need n ≥ z²·p(1 − p)/E², rounded up; without a prior estimate use p = 0.5, which gives the largest n.

Solve for n

n = 2.576²·0.8·0.2/0.05² = 424.7

Round up

n = 425

Answer

n = 425
9.Hypothesis tests & confidence intervals

Answer: p = 0.07215

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 = 6/√36 = 1
z = (501.46 − 500)/1 = 1.46

Tail area

Φ(1.46) = 0.9279
1 − Φ(1.46) = 0.0721

p-value

p = 0.07215

Answer

p = 0.07215

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

10.Hypothesis tests & confidence intervals

Answer: n = 449

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 = ((2.326 + 1.645)·16/3)² = 448.5

Round up

n = 449

Answer

n = 449
11.Linear regression & correlation

Answer: r² = 0.9641

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 = 305/6 = 50.83

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = 58.5
Syy = Σ(y − ȳ)² = 202.8

Coefficient of determination

r² = 58.5²/(17.5·202.8) = 0.9641

Answer

r² = 0.9641

About 96.41% of the variation in y is explained by the line.

12.Linear regression & correlation

Answer: x = 1.533

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 = 78/5 = 15.6

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = −30

Regression line

b = −30/10 = −3
a = 15.6 − (−3)·3 = 24.6
ŷ = 24.6 − 3·x

Solve for x

x = (20 − 24.6)/(−3) = 1.533

Answer

x = 1.533

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