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
253035404550

Find the sample standard deviation s of the data: 30, 21, 45, 52, 46.

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

A quantity is normally distributed with mean 100 and standard deviation 8. What proportion P of values is below 112?

μ = 100σ = 8x = 112
P =
3.ProbabilityMedium
xp(x)00.2510.2520.2530.140.15μ = 1.65

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

x: 0, 1, 2, 3, 4p(x): 0.25, 0.25, 0.25, 0.1, 0.15
Var(X) =
4.ProbabilityHard
n = 11, p = 0.5: P(X = 2) = ?k01234567891011

A test has 11 multiple-choice questions. A student who guesses gets each one right with probability 0.5, independently. What is the probability P of exactly 2 correct answers?

n = 11p = 0.5k = 2
P =
5.Normal distributionMedium
−3−2−10123zshaded area to the right of z = 0.8438

Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.8438. Find z to two decimals.

A = 0.8438
z =
6.Normal distributionHard
0.02 mm0D − d?

Hole diameters D are normal with mean 25.02 mm and standard deviation 0.01 mm. Shaft diameters d are independent and normal with mean 25 mm and standard deviation 0.02 mm. A randomly chosen shaft fits when D − d > 0. Find the probability P that it fits. Round z to two decimals.

hole: μ = 25.02 mm, σ = 0.01 mmshaft: μ = 25 mm, σ = 0.02 mm
P =
7.Confidence intervalsMedium
μ₁ − μ₂0LU = ?x̄₁ − x̄₂E99%

The battery life of two phone models is compared. Model 1: 60 phones, mean 77.5 h, known standard deviation 14 h. Model 2: 50 phones, mean 90.5 h, known standard deviation 7 h. Find the upper limit of the 99% confidence interval for μ₁ − μ₂.

n₁ = 60x̄₁ = 77.5 hσ₁ = 14 hn₂ = 50x̄₂ = 90.5 hσ₂ = 7 h99%
upper limit = h
8.Confidence intervalsHard
σ = 40, n = ?x̄ − Ex̄ + Ex̄E = 390%

The population standard deviation is 40. 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 3?

σ = 40E = 390%
n =
9.Hypothesis tests & confidence intervalsMedium
t −1.9901.99

A sample of 81 students has a mean score of 75.53 and a sample standard deviation of 4. Find the test statistic t for H₀: μ = 75.

μ₀ = 75x̄ = 75.53s = 4n = 81
t =
10.Hypothesis tests & confidence intervalsHard
−1.961.96z = ?

In a trial, 120 of 300 patients on treatment A recover, and 93 of 250 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.

x₁ = 120n₁ = 300x₂ = 93n₂ = 250
z =
11.Linear regression & correlationMedium
xy2514596748831082xy51050607080

The table gives 5 paired observations of x = hours studied and y = exam score. Their sums are given. Find Pearson's correlation coefficient r from these sums.

n = 5Σx = 30Σy = 349Σx² = 220Σy² = 25171Σxy = 2266
r =
12.Linear regression & correlationHard
xy244861581810201225xy5101020x = 3

The table gives 6 paired observations of x = load in kN and y = extension in mm. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 3.

n = 6x = 3
ŷ =

3Solutions

1.Descriptive statistics

Answer: s = 12.83

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 = 194/5 = 38.8

Squared deviations

Σ(x − x̄)² = (−8.8)² + (−17.8)² + 6.2² + 13.2² + 7.2² = 658.8

Standard deviation

s = √(658.8/4) = 12.83

Answer

s = 12.83
2.Descriptive statistics

Answer: P = 0.9332

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 = 1.5

Standard normal table

Φ(1.5) = 0.9332
P = 0.9332

Answer

P = 0.9332
3.Probability

Answer: Var(X) = 1.828

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 + 1·0.25 + 2·0.25 + 3·0.1 + 4·0.15 = 1.65

Mean of the squares

E(X²) = 0·0.25 + 1·0.25 + 4·0.25 + 9·0.1 + 16·0.15 = 4.55

Variance

Var(X) = 4.55 − 1.65² = 1.828

Answer

Var(X) = 1.828
4.Probability

Answer: P = 0.02686

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 = 2) = 55·0.52·0.59 = 55·0.25·0.001953 = 0.02686

Answer

P = 0.02686
5.Normal distribution

Answer: z = −1.01

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 − 0.8438 = 0.1562

Read the z-table backwards

Find 0.1562 inside the table and read off z.

z = −1.01

Answer

z = −1.01
6.Normal distribution

Answer: P = 0.8133

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 difference

μ = 25.02 − 25 = 0.02 mm, σ = 0.02236 mm

Standardise

z = (0 − 0.02)/0.02236 = −0.89

Probability

P = 1 − Φ(−0.89) = 0.8133

Answer

P = 0.8133
7.Confidence intervals

Answer: upper limit = −7.692 h

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.

Point estimate

x̄₁ − x̄₂ = 77.5 − 90.5 = −13

Standard error

SE = √(14²/60 + 7²/50) = 2.061

Margin of error

E = 2.576·2.061 = 5.308

Interval

−13 ± 5.308 → [−18.31, −7.692]

Answer

upper limit = −7.692 h

The interval does not contain 0, so the means differ significantly.

8.Confidence intervals

Answer: n = 482

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 = (1.645·40/3)² = 481.1

Round up

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

n = 482

Answer

n = 482
9.Hypothesis tests & confidence intervals

Answer: t = 1.193

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 = 4/√81 = 4/9 = 0.4444

Test statistic

t = (75.53 − 75)/0.4444 = 1.193

Answer

t = 1.193
t0.025; df = 80 = 1.99

|t| < 1.99, so H₀ is not rejected at the 5% level (two-sided).

10.Hypothesis tests & confidence intervals

Answer: z = 0.6712

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.

Sample proportions

p̂₁ = 120/300 = 0.4
p̂₂ = 93/250 = 0.372
p̂ = 213/550 = 0.3873

Standard error under H₀

SE = √(0.3873·0.6127·(1/300 + 1/250)) = 0.04172

Test statistic

z = (0.4 − 0.372)/0.04172 = 0.6712

Answer

z = 0.6712

|z| < 1.96, so H₀ is not rejected at the 5% level (two-sided).

11.Linear regression & correlation

Answer: r = 0.9551

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 = 2266 − 30·349/5 = 172
Sxx = 220 − 30²/5 = 40
Syy = 25171 − 349²/5 = 810.8

Correlation coefficient

r = 172/√(40·810.8) = 0.9551

Answer

r = 0.9551
12.Linear regression & correlation

Answer: ŷ = 6.771

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 = 90/6 = 15

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 144

Regression line

b = 144/70 = 2.057
a = 15 − 2.057·7 = 0.6
ŷ = 0.6 + 2.057·x

Prediction

ŷ = 0.6 + 2.057·3 = 6.771

Answer

ŷ = 6.771

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