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
25303540

Find the sample standard deviation s of the data: 21, 28, 39, 35, 44.

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

Exam scores are normally distributed with mean 60 and standard deviation 12. Find the z-score z of a score of 72.

μ = 60σ = 12x = 72
z =
3.ProbabilityMedium
xp(x)00.2510.120.2530.140.3μ = 2.1

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

x: 0, 1, 2, 3, 4p(x): 0.25, 0.1, 0.25, 0.1, 0.3
σ =
4.ProbabilityHard
n = 9, p = 0.3: P(X ≤ 1) = ?k0123456789

9 components are tested, and each one fails with probability 0.3, independently. What is the probability P that at most one fails?

n = 9p = 0.3
P =
5.Normal distributionMedium
−3−2−10123zshaded area to the right of z = 0.0113

Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.0113. Find z to two decimals.

A = 0.0113
z =
6.Normal distributionHard
35455565758595pointsbottom 5%

The exam scores are normally distributed with mean 65 points and standard deviation 10 points. Drag the cut-off so the shaded bottom 5% is to the left of it. Find the cut-off x.

μ = 65 pointsσ = 10 points5%
x = points
7.Confidence intervalsMedium
n = 10, s = 15LU = ?x̄ = 75.5E90%

A sample of 10 items has mean 75.5 and standard deviation s = 15. The population is roughly normal. Find the upper limit of the 90% confidence interval for the mean.

n = 10x̄ = 75.5s = 1590%t(9) = 1.833
upper limit =
8.Confidence intervalsHard
σ = 27, n = ?x̄ − Ex̄ + Ex̄E = 1.599%

The population standard deviation is 27. 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 = 27E = 1.599%
n =
9.Hypothesis tests & confidence intervalsMedium
t −2.13102.131

A sample of 16 students has a mean score of 43.8 and a sample standard deviation of 8. Find the test statistic t for H₀: μ = 50.

μ₀ = 50x̄ = 43.8s = 8n = 16
t =
10.Hypothesis tests & confidence intervalsHard
μ₀x̄c

A one-sided z test of H₀: μ = 200 against H₁: μ > 200 is done at α = 0.01 with a sample of 64; the standard deviation is known to be 4. In reality the mean is 201.65. Find the probability of a Type II error β in percent.

μ₀ = 200μ = 201.65σ = 4n = 64α = 0.01
β = %
11.Linear regression & correlationMedium
xy27476128131015xy5101015

The table gives 5 paired observations of x = load in kN and y = extension in mm. Their sums are given. Find Pearson's correlation coefficient r from these sums.

n = 5Σx = 30Σy = 54Σx² = 220Σy² = 636Σxy = 368
r =
12.Linear regression & correlationHard
xy125224319417517613xy246152025ŷ = 24

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. For which value x does the line predict ŷ = 24?

n = 6ŷ = 24
x =

3Solutions

1.Descriptive statistics

Answer: s = 9.072

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 = 167/5 = 33.4

Squared deviations

Σ(x − x̄)² = (−12.4)² + (−5.4)² + 5.6² + 1.6² + 10.6² = 329.2

Standard deviation

s = √(329.2/4) = 9.072

Answer

s = 9.072
2.Descriptive statistics

Answer: z = 1

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 = (72 − 60)/12 = 1

Answer

z = 1
3.Probability

Answer: σ = 1.546

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 = 0·0.25 + 1·0.1 + 2·0.25 + 3·0.1 + 4·0.3 = 2.1

Mean of the squares

E(X²) = 0·0.25 + 1·0.1 + 4·0.25 + 9·0.1 + 16·0.3 = 6.8

Variance

Var(X) = 6.8 − 2.1² = 2.39
σ = √2.39 = 1.546

Answer

σ = 1.546
4.Probability

Answer: P = 0.196

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) = 0.79 = 0.04035
P(X = 1) = 9·0.3·0.78 = 0.1556

Add them

P = 0.04035 + 0.1556 = 0.196

Answer

P = 0.196
5.Normal distribution

Answer: z = 2.28

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 − 0.0113 = 0.9887

Read the z-table backwards

Find 0.9887 inside the table and read off z.

z = 2.28

Answer

z = 2.28
6.Normal distribution

Answer: x = 48.6 points

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

Φ(z) = 0.05

z from the table

z ≈ −1.64

Back to the original units

x = 65 − 1.64·10 = 48.6 points

Answer

x = 48.6 points
7.Confidence intervals

Answer: upper limit = 84.19

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 = 9, t = 1.833

Margin of error

E = 1.833·15/√10 = 8.695

Interval

75.5 ± 8.695 → [66.81, 84.19]

Answer

upper limit = 84.19
8.Confidence intervals

Answer: n = 2150

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 = (2.576·27/1.5)² = 2150

Round up

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

n = 2150

Answer

n = 2150
9.Hypothesis tests & confidence intervals

Answer: t = −3.1

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).

Standard error

SE = 8/√16 = 8/4 = 2

Test statistic

t = (43.8 − 50)/2 = −3.1

Answer

t = −3.1
t0.025; df = 15 = 2.131

|t| > 2.131, so H₀ is rejected at the 5% level (two-sided).

10.Hypothesis tests & confidence intervals

Answer: β = 16.5 %

Power of a test: power = 1 − β is the probability of rejecting H₀ when it is false. Find the cutoff x̄c from μ₀ and z_α, then compute the probability of landing beyond x̄c using the true mean: z = (x̄c − μ)/(σ/√n).

Critical sample mean

SE = 4/√64 = 0.5
x̄c = 200 + 2.326·0.5 = 201.2

z under the true mean

z = (201.2 − 201.65)/0.5 = −0.974

Type II error

power = P(Z > −0.974) = 83.5%
β = 100% − 83.5% = 16.5%

Answer

β = 16.5%
11.Linear regression & correlation

Answer: r = 0.9574

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 = 368 − 30·54/5 = 44
Sxx = 220 − 30²/5 = 40
Syy = 636 − 54²/5 = 52.8

Correlation coefficient

r = 44/√(40·52.8) = 0.9574

Answer

r = 0.9574
12.Linear regression & correlation

Answer: x = 1.462

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 = 115/6 = 19.17

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −41.5

Regression line

b = −41.5/17.5 = −2.371
a = 19.17 − (−2.371)·3.5 = 27.47
ŷ = 27.47 − 2.371·x

Solve for x

x = (24 − 27.47)/(−2.371) = 1.462

Answer

x = 1.462

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