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
1520253035404550

Find the sample standard deviation s of the data: 14, 49, 34, 41, 51, 54.

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

A quantity is normally distributed with mean 170 and standard deviation 20. What proportion P of values is below 146?

μ = 170σ = 20x = 146
P =
3.ProbabilityMedium
xp(x)00.2520.2540.3560.15μ = 2.8

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

x: 0, 2, 4, 6p(x): 0.25, 0.25, 0.35, 0.15
σ =
4.ProbabilityHard
n = 12, p = 0.2: P(X = 3) = ?k0123456789101112

A test has 12 multiple-choice questions. A student who guesses gets each one right with probability 0.2, independently. What is the probability P of exactly 3 correct answers?

n = 12p = 0.2k = 3
P =
5.Normal distributionMedium
−3−2−10123zshaded area between −z and z = 0.9523

Drag z so that the shaded area between −z and z under the standard normal curve is 0.9523. Find z to two decimals.

A = 0.9523
z =
6.Normal distributionHard
380420460500540580620hbottom 10%

The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. Drag the cut-off so the shaded bottom 10% is to the left of it. Find the cut-off x.

μ = 500 hσ = 40 h10%
x = h
7.Confidence intervalsMedium
x̄E₁ = 3.495%, n = 36x̄E₂ = ?99%, n = 144

A 95% confidence interval for a mean, based on 36 observations and a known σ, has a margin of error of 3.4. What is the margin of error E₂ of a 99% interval based on 144 observations?

95%n₁ = 36E₁ = 3.499%n₂ = 144
E₂ =
8.Confidence intervalsHard
μ₁ − μ₂, t(29) = 2.0450L = ?Ux̄₁ − x̄₂E95%

Two fertilisers are tested. Plot group 1: 17 plots, mean yield 83 kg, standard deviation 5 kg. Plot group 2: 14 plots, mean 81 kg, standard deviation 1.5 kg. Assume equal variances (pooled t, df = n₁ + n₂ − 2). Find the lower limit of the 95% confidence interval for μ₁ − μ₂.

n₁ = 17x̄₁ = 83s₁ = 5n₂ = 14x̄₂ = 81s₂ = 1.595%t(29) = 2.045
lower limit = kg
9.Hypothesis tests & confidence intervalsMedium
−1.78p = ?

The mean of a population is claimed to be 80; the population standard deviation is known to be 32. A random sample of 64 observations has a mean of 72.88. Test H₀: μ = 80 against H₁: μ < 80. Find the p-value p.

μ₀ = 80σ = 32n = 64x̄ = 72.88H₁: μ < μ₀
p =
10.Hypothesis tests & confidence intervalsHard
σ = 20, n = ?x̄ − Ex̄ + Ex̄E = 190%

How large must a sample be so that the 90% confidence interval for a mean has a margin of error of at most 1? The standard deviation is 20. Find the smallest sample size n.

σ = 20E = 190%
n =
11.Linear regression & correlationMedium
xy24245465686810741281xy510406080

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
xy1876218724112271173013733144xy202530100150ŷ = 113

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. For which value x does the line predict ŷ = 113?

n = 6ŷ = 113
x =

3Solutions

1.Descriptive statistics

Answer: s = 14.9

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 = 243/6 = 40.5

Squared deviations

Σ(x − x̄)² = (−26.5)² + 8.5² + (−6.5)² + 0.5² + 10.5² + 13.5² = 1110

Standard deviation

s = √(1110/5) = 14.9

Answer

s = 14.9
2.Descriptive statistics

Answer: P = 0.1151

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 = 0.1151

Answer

P = 0.1151
3.Probability

Answer: σ = 2.04

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 + 2·0.25 + 4·0.35 + 6·0.15 = 2.8

Mean of the squares

E(X²) = 0·0.25 + 4·0.25 + 16·0.35 + 36·0.15 = 12

Variance

Var(X) = 12 − 2.8² = 4.16
σ = √4.16 = 2.04

Answer

σ = 2.04
4.Probability

Answer: P = 0.2362

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 = 3) = 220·0.23·0.89 = 220·0.008·0.1342 = 0.2362

Answer

P = 0.2362
5.Normal distribution

Answer: z = 1.98

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 + 0.9523)/2 = 0.9761

Read the z-table backwards

Find 0.9761 inside the table and read off z.

z = 1.98

Answer

z = 1.98
6.Normal distribution

Answer: x = 449 h

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) = 0.1

z from the table

z ≈ −1.28

Back to the original units

x = 500 − 1.28·40 = 449 h

Answer

x = 449 h
7.Confidence intervals

Answer: E₂ = 2.234

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 = (2.576/1.96)·√(36/144) = 0.6571

New margin

E₂ = 3.4·0.6571 = 2.234

Answer

E₂ = 2.234
8.Confidence intervals

Answer: lower limit = −0.8395 kg

Interval for a difference: estimate ± critical value × standard error. For two independent means with known σ: SE = √(σ₁²/n₁ + σ₂²/n₂) and z. With equal unknown variances: pooled s_p and t with n₁ + n₂ − 2 degrees of freedom. For paired data, work with the differences: d̄ ± t·s_d/√n.

Pooled standard deviation

sp² = (16·5² + 13·1.5²)/29 = 14.8
sp = 3.847

Margin of error

E = 2.045·3.847·√(1/17 + 1/14) = 2.839

Interval

2 ± 2.839 → [−0.8395, 4.839]

Answer

lower limit = −0.8395 kg
9.Hypothesis tests & confidence intervals

Answer: p = 0.03754

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 = 32/√64 = 4
z = (72.88 − 80)/4 = −1.78

Tail area

Φ(1.78) = 0.9625
1 − Φ(1.78) = 0.0375

p-value

p = 0.03754

Answer

p = 0.03754

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

10.Hypothesis tests & confidence intervals

Answer: n = 1083

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 = (1.645·20/1)² = 1082

Round up

n = 1083

Answer

n = 1083
11.Linear regression & correlation

Answer: r² = 0.9797

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 = 42/6 = 7
ȳ = 375/6 = 62.5

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 267
Syy = Σ(y − ȳ)² = 1040

Coefficient of determination

r² = 267²/(70·1040) = 0.9797

Answer

r² = 0.9797

About 97.97% of the variation in y is explained by the line.

12.Linear regression & correlation

Answer: x = 25.68

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 = 673/6 = 112.2

Sums of squares and products

Sxx = Σ(x − x̄)² = 157.5
Sxy = Σ(x − x̄)(y − ȳ) = 742.5

Regression line

b = 742.5/157.5 = 4.714
a = 112.2 − 4.714·25.5 = −8.048
ŷ = −8.048 + 4.714·x

Solve for x

x = (113 − (−8.048))/4.714 = 25.68

Answer

x = 25.68

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