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: 12, 52, 34, 59, 48, 14, 11.

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

A quantity is normally distributed with mean 500 and standard deviation 50. What proportion P of values is above 530?

μ = 500σ = 50x = 530
P =
3.ProbabilityMedium
xp(x)10.320.330.2540.15μ = 2.25

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

x: 1, 2, 3, 4p(x): 0.3, 0.3, 0.25, 0.15
Var(X) =
4.ProbabilityHard
n = 10, p = 0.25: P(X = 5) = ?k012345678910

A test has 10 multiple-choice questions. A student who guesses gets each one right with probability 0.25, independently. What is the probability P of exactly 5 correct answers?

n = 10p = 0.25k = 5
P =
5.Normal distributionMedium
32 min37 minX + Y?

X is normal with mean 12 min and standard deviation 1.5 min; Y is independent and normal with mean 20 min and standard deviation 2 min. Find the probability P that X + Y is more than 37 min. Round z to two decimals.

μX = 12, σX = 1.5μY = 20, σY = 2X + Y > 37 min
P =
6.Normal distributionHard
μ = 65x = ?95%

The exam scores are normally distributed with mean 65 points and standard deviation 10 points. Find the value x below which 95% of them lie (the 95th percentile).

μ = 65σ = 10P = 95%
x = points
7.Confidence intervalsMedium
n = 13, s = 16LU = ?x̄ = 51E95%

A sample of 13 items has mean 51 and standard deviation s = 16. The population is roughly normal. Find the upper limit of the 95% confidence interval for the mean.

n = 13x̄ = 51s = 1695%t(12) = 2.179
upper limit =
8.Confidence intervalsHard
σ = 26, n = ?x̄ − Ex̄ + Ex̄E = 190%

The population standard deviation is 26. How large a sample n do you need so that a 90% confidence interval for the mean has a margin of error of at most 1?

σ = 26E = 190%
n =
9.Hypothesis tests & confidence intervalsMedium
−2.0642.064t = ?df = 24

Two teaching methods are compared. Group 1: 6 students, mean score 58, standard deviation 8.5. Group 2: 20 students, mean score 51.1, standard deviation 3.5. Assume equal population variances and use the pooled two-sample t test. Find the test statistic t for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (group 1 minus group 2).

n₁ = 6x̄₁ = 58s₁ = 8.5n₂ = 20x̄₂ = 51.1s₂ = 3.5
t =
10.Hypothesis tests & confidence intervalsHard
z −1.9601.96

A sample of 25 measurements has mean 55.8. The population standard deviation is 10. Find the upper limit U of the 95% confidence interval for the mean.

x̄ = 55.8σ = 10n = 2595%
U =
11.Linear regression & correlationMedium
xy1548185521632486279130102xy152025306080100

The table gives 6 paired observations of x = temperature in °C and y = ice creams sold. Their sums are given. Find Pearson's correlation coefficient r from these sums.

n = 6Σx = 135Σy = 445Σx² = 3195Σy² = 35379Σxy = 10614
r =
12.Linear regression & correlationHard
xy135346550756965xy5405060x = 6

The table gives 5 paired observations of x = advertising spend in k€ and y = sales in k€. Use the least-squares line ŷ = a + b·x. A new observation (6, 57) is made. Find its residual e = y − ŷ.

n = 5(6, 57)
e =

3Solutions

1.Descriptive statistics

Answer: s = 20.61

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 = 230/7 = 32.86

Squared deviations

Σ(x − x̄)² = (−20.86)² + 19.14² + 1.143² + 26.14² + 15.14² + (−18.86)² + (−21.86)² = 2549

Standard deviation

s = √(2549/6) = 20.61

Answer

s = 20.61
2.Descriptive statistics

Answer: P = 0.2743

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 = 0.6

Standard normal table

Φ(0.6) = 0.7257
P = 1 − 0.7257 = 0.2743

Answer

P = 0.2743
3.Probability

Answer: Var(X) = 1.088

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 = 1·0.3 + 2·0.3 + 3·0.25 + 4·0.15 = 2.25

Mean of the squares

E(X²) = 1·0.3 + 4·0.3 + 9·0.25 + 16·0.15 = 6.15

Variance

Var(X) = 6.15 − 2.25² = 1.088

Answer

Var(X) = 1.088
4.Probability

Answer: P = 0.0584

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 = 5) = 252·0.255·0.755 = 252·0.0009766·0.2373 = 0.0584

Answer

P = 0.0584
5.Normal distribution

Answer: P = 0.02275

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 = 32 min, σ = √(1.5² + 2²) = 2.5 min

Standardise

z = (37 − 32)/2.5 = 2

Upper tail

P = 1 − Φ(2) = 1 − 0.9772 = 0.02275

Answer

P = 0.02275
6.Normal distribution

Answer: x = 81.45 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) = 0.95 → z = 1.645

Back to x

x = 65 + 1.645·10 = 81.45 points

Answer

x = 81.45 points
7.Confidence intervals

Answer: upper limit = 60.67

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 = 12, t = 2.179

Margin of error

E = 2.179·16/√13 = 9.67

Interval

51 ± 9.67 → [41.33, 60.67]

Answer

upper limit = 60.67
8.Confidence intervals

Answer: n = 1830

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 = (1.645·26/1)² = 1829

Round up

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

n = 1830

Answer

n = 1830
9.Hypothesis tests & confidence intervals

Answer: t = 2.98

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.

Pooled variance

sp² = (5·8.5² + 19·3.5²)/24 = 24.75
sp = 4.975

Standard error

SE = 4.975·√(1/6 + 1/20) = 2.316

Test statistic

t = (58 − 51.1)/2.316 = 2.98

Answer

t = 2.98

With df = 24, the two-sided 5% critical value is 2.064; |t| is larger, so H₀ is rejected.

10.Hypothesis tests & confidence intervals

Answer: U = 59.72

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 = 1.96·10/√25 = 3.92

Interval limit

U = 55.8 + 3.92 = 59.72

Answer

U = 59.72
11.Linear regression & correlation

Answer: r = 0.9835

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 = 10614 − 135·445/6 = 601.5
Sxx = 3195 − 135²/6 = 157.5
Syy = 35379 − 445²/6 = 2375

Correlation coefficient

r = 601.5/√(157.5·2375) = 0.9835

Answer

r = 0.9835
12.Linear regression & correlation

Answer: e = 3.1

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 = 252/5 = 50.4

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 140

Regression line

b = 140/40 = 3.5
a = 50.4 − 3.5·5 = 32.9
ŷ = 32.9 + 3.5·x

Predicted value

ŷ = 32.9 + 3.5·6 = 53.9

Residual

e = 57 − 53.9 = 3.1

Answer

e = 3.1

The point lies above the line.

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