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
1015202530

Find the sample standard deviation s of the data: 15, 19, 17, 10, 32.

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

A quantity is normally distributed with mean 100 and standard deviation 5. What proportion P of values is above 96?

μ = 100σ = 5x = 96
P =
3.ProbabilityMedium
xp(x)00.310.3520.35μ = 1.05

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

x: 0, 1, 2p(x): 0.3, 0.35, 0.35
σ =
4.ProbabilityHard
n = 10, p = 0.3: P(X = 5) = ?k012345678910

A test has 10 multiple-choice questions. A student who guesses gets each one right with probability 0.3, independently. What is the probability P of exactly 5 correct answers?

n = 10p = 0.3k = 5
P =
5.Normal distributionMedium
−3−2−10123zshaded area to the right of z = 0.9554

Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.9554. Find z to two decimals.

A = 0.9554
z =
6.Normal distributionHard
μ = 32x = ?20%

The delivery times are normally distributed with mean 32 min and standard deviation 6 min. Find the value x below which 20% of them lie (the 20th percentile).

μ = 32σ = 6P = 20%
x = min
7.Confidence intervalsMedium
n = 15, s = 7LU = ?x̄ = 82.5E90%

A sample of 15 items has mean 82.5 and standard deviation s = 7. The population is roughly normal. Find the upper limit of the 90% confidence interval for the mean.

n = 15x̄ = 82.5s = 790%t(14) = 1.761
upper limit =
8.Confidence intervalsHard
σ = 40, n = ?x̄ − Ex̄ + Ex̄E = 195%

The population standard deviation is 40. 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 = 40E = 195%
n =
9.Hypothesis tests & confidence intervalsMedium
−2.0862.086t = ?df = 20

Two teaching methods are compared. Group 1: 17 students, mean score 57, standard deviation 9. Group 2: 5 students, mean score 41.7, standard deviation 12. 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 = 17x̄₁ = 57s₁ = 9n₂ = 5x̄₂ = 41.7s₂ = 12
t =
10.Hypothesis tests & confidence intervalsHard
σ = 12, n = ?x̄ − Ex̄ + Ex̄E = 299%

How large must a sample be so that the 99% confidence interval for a mean has a margin of error of at most 2? The standard deviation is 12. Find the smallest sample size n.

σ = 12E = 299%
n =
11.Linear regression & correlationMedium
xy1212203164115967xy246101520

The table gives 6 paired observations of x = car age in years and y = price in k€. Their sums are given. Find Pearson's correlation coefficient r from these sums.

n = 6Σx = 21Σy = 84Σx² = 91Σy² = 1348Σxy = 240
r =
12.Linear regression & correlationHard
xy241343454553654762xy246405060x = 4.5

The table gives 6 paired observations of x = advertising spend in k€ and y = sales in k€. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 4.5.

n = 6x = 4.5
ŷ =

3Solutions

1.Descriptive statistics

Answer: s = 8.204

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 = 93/5 = 18.6

Squared deviations

Σ(x − x̄)² = (−3.6)² + 0.4² + (−1.6)² + (−8.6)² + 13.4² = 269.2

Standard deviation

s = √(269.2/4) = 8.204

Answer

s = 8.204
2.Descriptive statistics

Answer: P = 0.7881

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 = −0.8

Standard normal table

Φ(−0.8) = 0.2119
P = 1 − 0.2119 = 0.7881

Answer

P = 0.7881
3.Probability

Answer: σ = 0.8047

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 = 0·0.3 + 1·0.35 + 2·0.35 = 1.05

Mean of the squares

E(X²) = 0·0.3 + 1·0.35 + 4·0.35 = 1.75

Variance

Var(X) = 1.75 − 1.05² = 0.6475
σ = √0.6475 = 0.8047

Answer

σ = 0.8047
4.Probability

Answer: P = 0.1029

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) = 252·0.35·0.75 = 252·0.00243·0.1681 = 0.1029

Answer

P = 0.1029
5.Normal distribution

Answer: z = −1.7

From an area back to z: Φ(z) is the area to the left of z. For a left-tail area A, find z with Φ(z) = A in the table. A right-tail area A means Φ(z) = 1 − A. A central area A between −z and z leaves (1 − A)/2 in each tail, so Φ(z) = (1 + A)/2. For a value x, use x = μ + zσ.

Area to the left of z

Φ(z) = 1 − 0.9554 = 0.0446

Read the z-table backwards

Find 0.0446 inside the table and read off z.

z = −1.70

Answer

z = −1.70
6.Normal distribution

Answer: x = 26.95 min

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) = 0.2 → z = −0.8416

Back to x

x = 32 + −0.8416·6 = 26.95 min

Answer

x = 26.95 min
7.Confidence intervals

Answer: upper limit = 85.68

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 = 14, t = 1.761

Margin of error

E = 1.761·7/√15 = 3.183

Interval

82.5 ± 3.183 → [79.32, 85.68]

Answer

upper limit = 85.68
8.Confidence intervals

Answer: n = 6147

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 = (1.96·40/1)² = 6147

Round up

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

n = 6147

Answer

n = 6147
9.Hypothesis tests & confidence intervals

Answer: t = 3.109

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 = (16·9² + 4·12²)/20 = 93.6
sp = 9.675

Standard error

SE = 9.675·√(1/17 + 1/5) = 4.922

Test statistic

t = (57 − 41.7)/4.922 = 3.109

Answer

t = 3.109

With df = 20, the two-sided 5% critical value is 2.086; |t| is larger, so H₀ is rejected.

10.Hypothesis tests & confidence intervals

Answer: n = 239

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).

Solve for n

n = (2.576·12/2)² = 238.9

Round up

n = 239

Answer

n = 239
11.Linear regression & correlation

Answer: r = −0.9843

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 = 240 − 21·84/6 = −54
Sxx = 91 − 21²/6 = 17.5
Syy = 1348 − 84²/6 = 172

Correlation coefficient

r = −54/√(17.5·172) = −0.9843

Answer

r = −0.9843
12.Linear regression & correlation

Answer: ŷ = 51.17

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 = 307/6 = 51.17

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = 68.5

Regression line

b = 68.5/17.5 = 3.914
a = 51.17 − 3.914·4.5 = 33.55
ŷ = 33.55 + 3.914·x

Prediction

ŷ = 33.55 + 3.914·4.5 = 51.17

Answer

ŷ = 51.17

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