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
15202530354045

Find the sample standard deviation s of the data: 12, 46, 38, 38, 24, 43.

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

A quantity is normally distributed with mean 500 and standard deviation 75. What proportion P of values is below 612.5?

μ = 500σ = 75x = 612.5
P =
3.ProbabilityMedium
xp(x)100.25200.35300.1400.3μ = 24.5

For the distribution shown, find the variance Var(X) of X.

x: 10, 20, 30, 40p(x): 0.25, 0.35, 0.1, 0.3
Var(X) =
4.ProbabilityHard
xp(x)100.3200.2300.15400.35μ

X has the distribution shown, and Y = 10X + 4. Find the standard deviation σY of Y.

x: 10, 20, 30, 40p(x): 0.3, 0.2, 0.15, 0.35Y = 10X + 4
σY =
5.Normal distributionMedium
105 s113 sX + Y?

X is normal with mean 60 s and standard deviation 5 s; Y is independent and normal with mean 45 s and standard deviation 4 s. Find the probability P that X + Y is more than 113 s. Round z to two decimals.

μX = 60, σX = 5μY = 45, σY = 4X + Y > 113 s
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
n = 21, s = 18.5L = ?Ux̄ = 71E90%

A sample of 21 items has mean 71 and standard deviation s = 18.5. The population is roughly normal. Find the lower limit of the 90% confidence interval for the mean.

n = 21x̄ = 71s = 18.590%t(20) = 1.725
lower limit =
8.Confidence intervalsHard
σ = 29, n = ?x̄ − Ex̄ + Ex̄E = 1.599%

The population standard deviation is 29. 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 1.5?

σ = 29E = 1.599%
n =
9.Hypothesis tests & confidence intervalsMedium
2.1p = ?

The mean of a population is claimed to be 120; the population standard deviation is known to be 10. A random sample of 100 observations has a mean of 122.1. Test H₀: μ = 120 against H₁: μ > 120. Find the p-value p.

μ₀ = 120σ = 10n = 100x̄ = 122.1H₁: μ > μ₀
p =
10.Hypothesis tests & confidence intervalsHard
H₀H₁μ₀x̄cμ₀ + 4α = 0.01βσ = 20, power 80%, n = ?

A one-sided z test at α = 0.01 must detect a shift of the mean of 4 units with a power of 80%. The standard deviation is 20. Find the minimum sample size n.

α = 0.01power = 80%δ = 4σ = 20zα = 2.326zβ = 0.842
n =
11.Linear regression & correlationMedium
xy127224323420523xy242025

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
xy249352454558670775xy246506070ŷ = 61

The table gives 6 paired observations of x = hours studied and y = exam score. Use the least-squares line ŷ = a + b·x. For which value x does the line predict ŷ = 61?

n = 6ŷ = 61
x =

3Solutions

1.Descriptive statistics

Answer: s = 12.96

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 = 201/6 = 33.5

Squared deviations

Σ(x − x̄)² = (−21.5)² + 12.5² + 4.5² + 4.5² + (−9.5)² + 9.5² = 839.5

Standard deviation

s = √(839.5/5) = 12.96

Answer

s = 12.96
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) = 0.9332
P = 0.9332

Answer

P = 0.9332
3.Probability

Answer: Var(X) = 134.8

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 = 10·0.25 + 20·0.35 + 30·0.1 + 40·0.3 = 24.5

Mean of the squares

E(X²) = 100·0.25 + 400·0.35 + 900·0.1 + 1600·0.3 = 735

Variance

Var(X) = 735 − 24.5² = 134.8

Answer

Var(X) = 134.8
4.Probability

Answer: σY = 124.4

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.

Spread of X

μ = 25.5, Var(X) = 154.8, σX = 12.44

Linear change

σY = |10|·12.44 = 124.4

Answer

σY = 124.4

The mean does move: E(Y) = 10·25.5 + 4 = 259.

5.Normal distribution

Answer: P = 0.1056

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 = 105 s, σ = √(5² + 4²) = 6.403 s

Standardise

z = (113 − 105)/6.403 = 1.25

Upper tail

P = 1 − Φ(1.25) = 1 − 0.8944 = 0.1056

Answer

P = 0.1056
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: lower limit = 64.04

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 = 20, t = 1.725

Margin of error

E = 1.725·18.5/√21 = 6.964

Interval

71 ± 6.964 → [64.04, 77.96]

Answer

lower limit = 64.04
8.Confidence intervals

Answer: n = 2481

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 = (2.576·29/1.5)² = 2480

Round up

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

n = 2481

Answer

n = 2481
9.Hypothesis tests & confidence intervals

Answer: p = 0.01786

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 = 10/√100 = 1
z = (122.1 − 120)/1 = 2.1

Tail area

Φ(2.1) = 0.9821
1 − Φ(2.1) = 0.0179

p-value

p = 0.01786

Answer

p = 0.01786

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

10.Hypothesis tests & confidence intervals

Answer: n = 251

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 + 0.842)·20/4)² = 250.9

Round up

n = 251

Answer

n = 251
11.Linear regression & correlation

Answer: r² = 0.5714

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 = 117/5 = 23.4

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = −12
Syy = Σ(y − ȳ)² = 25.2

Coefficient of determination

r² = −12²/(10·25.2) = 0.5714

Answer

r² = 0.5714

About 57.14% of the variation in y is explained by the line.

12.Linear regression & correlation

Answer: x = 4.748

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 = 27/6 = 4.5
ȳ = 358/6 = 59.67

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = 94

Regression line

b = 94/17.5 = 5.371
a = 59.67 − 5.371·4.5 = 35.5
ŷ = 35.5 + 5.371·x

Solve for x

x = (61 − 35.5)/5.371 = 4.748

Answer

x = 4.748

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