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
3035404550

Find the sample standard deviation s of the data: 54, 40, 29, 27, 54, 31, 27.

n = 7
s =
2.Descriptive statisticsHard
z −2−1012

Exam scores are normally distributed with mean 60 and standard deviation 10. Find the z-score z of a score of 45.

μ = 60σ = 10x = 45
z =
3.ProbabilityMedium
xp(x)10.0520.1530.2540.350.25μ = 3.55

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

x: 1, 2, 3, 4, 5p(x): 0.05, 0.15, 0.25, 0.3, 0.25
σ =
4.ProbabilityHard
n = 9, p = 0.6: P(X ≤ 1) = ?k0123456789

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

n = 9p = 0.6
P =
5.Normal distributionMedium
750 g735.2 gX + Y?

X is normal with mean 250 g and standard deviation 4 g; Y is independent and normal with mean 500 g and standard deviation 9 g. Find the probability P that X + Y is more than 735.2 g. Round z to two decimals.

μX = 250, σX = 4μY = 500, σY = 9X + Y > 735.2 g
P =
6.Normal distributionHard
0.03 mm0D − d?

Hole diameters D are normal with mean 25.03 mm and standard deviation 0.015 mm. Shaft diameters d are independent and normal with mean 25 mm and standard deviation 0.01 mm. A randomly chosen shaft fits when D − d > 0. Find the probability P that it fits. Round z to two decimals.

hole: μ = 25.03 mm, σ = 0.015 mmshaft: μ = 25 mm, σ = 0.01 mm
P =
7.Confidence intervalsMedium
n = 11, s = 1.5L = ?Ux̄ = 26E99%

A sample of 11 items has mean 26 and standard deviation s = 1.5. The population is roughly normal. Find the lower limit of the 99% confidence interval for the mean.

n = 11x̄ = 26s = 1.599%t(10) = 3.169
lower limit =
8.Confidence intervalsHard
n = 16σ = ?Us = 7.595%

The diameters of 16 machined shafts have a sample standard deviation of 7.5 µm. Assuming the diameters are normally distributed, find the lower limit σ of the 95% confidence interval for the population standard deviation. With df = 15, 2.5% of the χ² distribution lies below 6.262 and 2.5% lies above 27.488.

n = 16s = 7.5 µmχ² = 6.262, 27.488
σ = µm
9.Hypothesis tests & confidence intervalsMedium
t −2.06402.064

A sample of 25 students has a mean score of 55.76 and a sample standard deviation of 12. Find the test statistic t for H₀: μ = 50.

μ₀ = 50x̄ = 55.76s = 12n = 25
t =
10.Hypothesis tests & confidence intervalsHard
−1.961.96z = ?

In a trial, 120 of 300 patients on treatment A recover, and 64 of 200 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.

x₁ = 120n₁ = 300x₂ = 64n₂ = 200
z =
11.Linear regression & correlationMedium
xy254351463559667xy24650556065

The table gives 5 paired observations of x = hours studied and y = exam score. Find Pearson's correlation coefficient r.

n = 5
r =
12.Linear regression & correlationHard
xy294146158241028xy510102030x = 7

The table gives 5 paired observations of x = load in kN and y = extension in mm. Use the least-squares line ŷ = a + b·x. A new observation (7, 15) is made. Find its residual e = y − ŷ.

n = 5(7, 15)
e =

3Solutions

1.Descriptive statistics

Answer: s = 12.15

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 = 262/7 = 37.43

Squared deviations

Σ(x − x̄)² = 16.57² + 2.571² + (−8.429)² + (−10.43)² + 16.57² + (−6.429)² + (−10.43)² = 885.7

Standard deviation

s = √(885.7/6) = 12.15

Answer

s = 12.15
2.Descriptive statistics

Answer: z = −1.5

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 = (45 − 60)/10 = −1.5

Answer

z = −1.5
3.Probability

Answer: σ = 1.161

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 = 1·0.05 + 2·0.15 + 3·0.25 + 4·0.3 + 5·0.25 = 3.55

Mean of the squares

E(X²) = 1·0.05 + 4·0.15 + 9·0.25 + 16·0.3 + 25·0.25 = 13.95

Variance

Var(X) = 13.95 − 3.55² = 1.348
σ = √1.348 = 1.161

Answer

σ = 1.161
4.Probability

Answer: P = 0.003801

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) = 0.49 = 0.0002621
P(X = 1) = 9·0.6·0.48 = 0.003539

Add them

P = 0.0002621 + 0.003539 = 0.003801

Answer

P = 0.003801
5.Normal distribution

Answer: P = 0.9332

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 sum

μ = 750 g, σ = √(4² + 9²) = 9.849 g

Standardise

z = (735.2 − 750)/9.849 = −1.5

Upper tail

P = 1 − Φ(−1.5) = 1 − 0.06681 = 0.9332

Answer

P = 0.9332
6.Normal distribution

Answer: P = 0.9515

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 = 25.03 − 25 = 0.03 mm, σ = 0.01803 mm

Standardise

z = (0 − 0.03)/0.01803 = −1.66

Probability

P = 1 − Φ(−1.66) = 0.9515

Answer

P = 0.9515
7.Confidence intervals

Answer: lower limit = 24.57

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 = 10, t = 3.169

Margin of error

E = 3.169·1.5/√11 = 1.433

Interval

26 ± 1.433 → [24.57, 27.43]

Answer

lower limit = 24.57
8.Confidence intervals

Answer: σ = 5.54 µm

Interval for a variance: (n − 1)s²/σ² has a χ² distribution with n − 1 degrees of freedom. The 95% interval for σ² runs from (n − 1)s² divided by the upper χ² value (2.5% above it) to (n − 1)s² divided by the lower χ² value (2.5% below it); the interval for σ is the square root of both limits. It is not symmetric around s.

Sum of squares

(n − 1)s² = 15·7.5² = 843.8

Limits for σ²

843.8/27.488 = 30.7
843.8/6.262 = 134.7

Limit for σ

σ = √30.7 = 5.54 µm

Answer

σ = 5.54 µm

95% interval for σ: [5.54, 11.61] µm.

9.Hypothesis tests & confidence intervals

Answer: t = 2.4

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).

Standard error

SE = 12/√25 = 12/5 = 2.4

Test statistic

t = (55.76 − 50)/2.4 = 2.4

Answer

t = 2.4
t0.025; df = 24 = 2.064

|t| > 2.064, so H₀ is rejected at the 5% level (two-sided).

10.Hypothesis tests & confidence intervals

Answer: z = 1.817

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 = 64/200 = 0.32
p̂ = 184/500 = 0.368

Standard error under H₀

SE = √(0.368·0.632·(1/300 + 1/200)) = 0.04402

Test statistic

z = (0.4 − 0.32)/0.04402 = 1.817

Answer

z = 1.817

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

11.Linear regression & correlation

Answer: r = 0.8275

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 = 20/5 = 4
ȳ = 294/5 = 58.8

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = 34
Syy = Σ(y − ȳ)² = 168.8

Correlation coefficient

r = 34/√(10·168.8) = 0.8275

Answer

r = 0.8275
12.Linear regression & correlation

Answer: e = −5.4

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 = 30/5 = 6
ȳ = 90/5 = 18

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 96

Regression line

b = 96/40 = 2.4
a = 18 − 2.4·6 = 3.6
ŷ = 3.6 + 2.4·x

Predicted value

ŷ = 3.6 + 2.4·7 = 20.4

Residual

e = 15 − 20.4 = −5.4

Answer

e = −5.4

The point lies below the line.

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