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
25303540455055

Find the sample standard deviation s of the data: 58, 22, 21, 21, 41.

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

A quantity is normally distributed with mean 500 and standard deviation 25. What proportion P of values is above 470?

μ = 500σ = 25x = 470
P =
3.ProbabilityMedium
xp(x)00.3510.0520.3530.25μ = 1.5

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

x: 0, 1, 2, 3p(x): 0.35, 0.05, 0.35, 0.25
σ =
4.ProbabilityHard
n = 11, p = 0.6: P(X ≤ 1) = ?k01234567891011

11 components are tested, and each one fails with probability 0.6, independently. What is the probability P that at most one fails?

n = 11p = 0.6
P =
5.Normal distributionMedium
−3−2−10123zshaded area between −z and z = 0.5285

Drag z so that the shaded area between −z and z under the standard normal curve is 0.5285. Find z to two decimals.

A = 0.5285
z =
6.Normal distributionHard
157164171178185192199cmtop 70%

The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. Drag the cut-off so the shaded top 70% is to the right of it. Find the cut-off x.

μ = 178 cmσ = 7 cm70%
x = cm
7.Confidence intervalsMedium
x̄E₁ = 6.399%, n = 100x̄E₂ = ?90%, n = 100

A 99% confidence interval for a mean, based on 100 observations and a known σ, has a margin of error of 6.3. What is the margin of error E₂ of a 90% interval from the same data?

99%n₁ = 100E₁ = 6.390%
E₂ =
8.Confidence intervalsHard
σ = 38, n = ?x̄ − Ex̄ + Ex̄E = 199%

The population standard deviation is 38. 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?

σ = 38E = 199%
n =
9.Hypothesis tests & confidence intervalsMedium
−1.441.44p = ?

The mean of a population is claimed to be 120; the population standard deviation is known to be 4. A random sample of 64 observations has a mean of 119.28. Test H₀: μ = 120 against H₁: μ ≠ 120. Find the p-value p.

μ₀ = 120σ = 4n = 64x̄ = 119.28H₁: μ ≠ μ₀
p =
10.Hypothesis tests & confidence intervalsHard
z −2.57602.576

A sample of 16 measurements has mean 122.5. The population standard deviation is 20. Find the upper limit U of the 99% confidence interval for the mean.

x̄ = 122.5σ = 20n = 1699%
U =
11.Linear regression & correlationMedium
xy144255354467566671xy246506070

The table gives 6 paired observations of x = hours studied and y = exam score. Find the coefficient of determination r² of the least-squares line.

n = 6
r² =
12.Linear regression & correlationHard
xy18862110424109271323014533161xy202530100150x = 19.5

The table gives 6 paired observations of x = temperature in °C and y = ice creams sold. Use the least-squares line ŷ = a + b·x. A new observation (19.5, 97) is made. Find its residual e = y − ŷ.

n = 6(19.5, 97)
e =

3Solutions

1.Descriptive statistics

Answer: s = 16.56

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 = 163/5 = 32.6

Squared deviations

Σ(x − x̄)² = 25.4² + (−10.6)² + (−11.6)² + (−11.6)² + 8.4² = 1097

Standard deviation

s = √(1097/4) = 16.56

Answer

s = 16.56
2.Descriptive statistics

Answer: P = 0.8849

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 = −1.2

Standard normal table

Φ(−1.2) = 0.1151
P = 1 − 0.1151 = 0.8849

Answer

P = 0.8849
3.Probability

Answer: σ = 1.204

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 = 0·0.35 + 1·0.05 + 2·0.35 + 3·0.25 = 1.5

Mean of the squares

E(X²) = 0·0.35 + 1·0.05 + 4·0.35 + 9·0.25 = 3.7

Variance

Var(X) = 3.7 − 1.5² = 1.45
σ = √1.45 = 1.204

Answer

σ = 1.204
4.Probability

Answer: P = 0.000734

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) = 0.411 = 4.194 × 10⁻⁵
P(X = 1) = 11·0.6·0.410 = 0.0006921

Add them

P = 4.194 × 10⁻⁵ + 0.0006921 = 0.000734

Answer

P = 0.000734
5.Normal distribution

Answer: z = 0.72

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 + 0.5285)/2 = 0.7642

Read the z-table backwards

Find 0.7642 inside the table and read off z.

z = 0.72

Answer

z = 0.72
6.Normal distribution

Answer: x = 174.4 cm

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) = 0.3

z from the table

z ≈ −0.52

Back to the original units

x = 178 − 0.52·7 = 174.4 cm

Answer

x = 174.4 cm
7.Confidence intervals

Answer: E₂ = 4.023

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.

Ratio of margins

E₂/E₁ = (1.645/2.576)·√(100/100) = 0.6386

New margin

E₂ = 6.3·0.6386 = 4.023

Answer

E₂ = 4.023
8.Confidence intervals

Answer: n = 9583

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 = (2.576·38/1)² = 9582

Round up

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

n = 9583

Answer

n = 9583
9.Hypothesis tests & confidence intervals

Answer: p = 0.1499

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 = 4/√64 = 0.5
z = (119.28 − 120)/0.5 = −1.44

Tail area

Φ(1.44) = 0.9251
1 − Φ(1.44) = 0.0749

p-value

p = 2·0.0749 = 0.1499

Answer

p = 0.1499

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

10.Hypothesis tests & confidence intervals

Answer: U = 135.4

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 = 2.576·20/√16 = 12.88

Interval limit

U = 122.5 + 12.88 = 135.4

Answer

U = 135.4
11.Linear regression & correlation

Answer: r² = 0.8974

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 = 357/6 = 59.5

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = 90.5
Syy = Σ(y − ȳ)² = 521.5

Coefficient of determination

r² = 90.5²/(17.5·521.5) = 0.8974

Answer

r² = 0.8974

About 89.74% of the variation in y is explained by the line.

12.Linear regression & correlation

Answer: e = 3.938

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 = 153/6 = 25.5
ȳ = 737/6 = 122.8

Sums of squares and products

Sxx = Σ(x − x̄)² = 157.5
Sxy = Σ(x − x̄)(y − ȳ) = 781.5

Regression line

b = 781.5/157.5 = 4.962
a = 122.8 − 4.962·25.5 = −3.695
ŷ = −3.695 + 4.962·x

Predicted value

ŷ = −3.695 + 4.962·19.5 = 93.06

Residual

e = 97 − 93.06 = 3.938

Answer

e = 3.938

The point lies above the line.

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