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, 24, 49, 26, 52, 40.

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

Exam scores are normally distributed with mean 70 and standard deviation 8. Find the z-score z of a score of 80.

μ = 70σ = 8x = 80
z =
3.ProbabilityMedium
xp(x)00.3520.3540.1560.15μ = 2.2

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

x: 0, 2, 4, 6p(x): 0.35, 0.35, 0.15, 0.15
Var(X) =
4.ProbabilityHard
n = 7, p = 0.7: P(X = 5) = ?k01234567

A test has 7 multiple-choice questions. A student who guesses gets each one right with probability 0.7, independently. What is the probability P of exactly 5 correct answers?

n = 7p = 0.7k = 5
P =
5.Normal distributionMedium
200 kg202.1 kgT = X₁ + … + X₈?

The masses of bags of cement are independent and normal with mean 25 kg and standard deviation 0.5 kg. Find the probability P that the total mass of 8 of them exceeds 202.1 kg. Round z to two decimals.

μ = 25 kgσ = 0.5 kgn = 8> 202.1 kg
P =
6.Normal distributionHard
4 s0X − Y?

Runner A's lap time X and runner B's lap time Y are independent and normal: X has mean 52 s and standard deviation 4 s, Y has mean 48 s and standard deviation 3 s. Find the probability P that X is greater than Y. Round z to two decimals.

μX = 52 s, σX = 4 sμY = 48 s, σY = 3 s
P =
7.Confidence intervalsMedium
n = 15, sd​ = 6.10L = ?Ud̄ = 2.6E99%

The blood pressure of 15 patients is measured before and after a treatment. The differences d = before − after (mmHg) have mean 2.6 and standard deviation 6.1. Find the lower limit of the 99% confidence interval for the mean difference.

n = 15d̄ = 2.6sd = 6.199%t(14) = 2.977
lower limit = mmHg
8.Confidence intervalsHard
σ = 17, n = ?x̄ − Ex̄ + Ex̄E = 590%

The population standard deviation is 17. 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 5?

σ = 17E = 590%
n =
9.Hypothesis tests & confidence intervalsMedium
−1.961.96z = ?

Two production lines make the same part. Line 1: a sample of 100 parts has mean length 79 mm, and the line's known standard deviation is 12 mm. Line 2: a sample of 100 parts has mean 84 mm, known standard deviation 13 mm. Find the test statistic z for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (line 1 minus line 2).

n₁ = 100x̄₁ = 79 mmσ₁ = 12 mmn₂ = 100x̄₂ = 84 mmσ₂ = 13 mm
z =
10.Hypothesis tests & confidence intervalsHard
z −2.57602.576

A sample of 16 measurements has mean 52.1. The population standard deviation is 12. Find the lower limit L of the 99% confidence interval for the mean.

x̄ = 52.1σ = 12n = 1699%
L =
11.Linear regression & correlationMedium
xy124220322418519xy2418202224

The table gives 5 paired observations of x = car age in years and y = price in k€. Find the coefficient of determination r² of the least-squares line.

n = 5
r² =
12.Linear regression & correlationHard
xy356457571675777xy466070ŷ = 58

The table gives 5 paired observations of x = hours studied and y = exam score. Use the least-squares line ŷ = a + b·x. For which value x does the line predict ŷ = 58?

n = 5ŷ = 58
x =

3Solutions

1.Descriptive statistics

Answer: s = 14.05

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 = 249/6 = 41.5

Squared deviations

Σ(x − x̄)² = 16.5² + (−17.5)² + 7.5² + (−15.5)² + 10.5² + (−1.5)² = 987.5

Standard deviation

s = √(987.5/5) = 14.05

Answer

s = 14.05
2.Descriptive statistics

Answer: z = 1.25

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 = (80 − 70)/8 = 1.25

Answer

z = 1.25
3.Probability

Answer: Var(X) = 4.36

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 + 2·0.35 + 4·0.15 + 6·0.15 = 2.2

Mean of the squares

E(X²) = 0·0.35 + 4·0.35 + 16·0.15 + 36·0.15 = 9.2

Variance

Var(X) = 9.2 − 2.2² = 4.36

Answer

Var(X) = 4.36
4.Probability

Answer: P = 0.3177

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 = 5) = 21·0.75·0.32 = 21·0.1681·0.09 = 0.3177

Answer

P = 0.3177
5.Normal distribution

Answer: P = 0.06944

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 total

μT = 200 kg, σT = 0.5·√8 = 1.414 kg

Standardise

z = (202.1 − 200)/1.414 = 1.48

Probability

P = 1 − Φ(1.48) = 0.06944

Answer

P = 0.06944
6.Normal distribution

Answer: P = 0.7881

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 = 52 − 48 = 4 s, σ = 5 s

Standardise

z = (0 − 4)/5 = −0.8

Probability

P = 1 − Φ(−0.8) = 0.7881

Answer

P = 0.7881
7.Confidence intervals

Answer: lower limit = −2.089 mmHg

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.

Critical value

df = 14, t = 2.977

Margin of error

E = 2.977·6.1/√15 = 4.689

Interval

2.6 ± 4.689 → [−2.089, 7.289]

Answer

lower limit = −2.089 mmHg
8.Confidence intervals

Answer: n = 32

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 = (1.645·17/5)² = 31.28

Round up

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

n = 32

Answer

n = 32
9.Hypothesis tests & confidence intervals

Answer: z = −2.826

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.

Standard error of the difference

SE = √(12²/100 + 13²/100) = √3.13 = 1.769

Test statistic

z = (79 − 84)/1.769 = −2.826

Answer

z = −2.826

|z| > 1.96, so H₀ is rejected at the 5% level (two-sided).

10.Hypothesis tests & confidence intervals

Answer: L = 44.37

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 = 2.576·12/√16 = 7.728

Interval limit

L = 52.1 − 7.728 = 44.37

Answer

L = 44.37
11.Linear regression & correlation

Answer: r² = 0.6207

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 = 15/5 = 3
ȳ = 103/5 = 20.6

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = −12
Syy = Σ(y − ȳ)² = 23.2

Coefficient of determination

r² = −12²/(10·23.2) = 0.6207

Answer

r² = 0.6207

About 62.07% of the variation in y is explained by the line.

12.Linear regression & correlation

Answer: x = 3.467

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 = 25/5 = 5
ȳ = 336/5 = 67.2

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = 60

Regression line

b = 60/10 = 6
a = 67.2 − 6·5 = 37.2
ŷ = 37.2 + 6·x

Solve for x

x = (58 − 37.2)/6 = 3.467

Answer

x = 3.467

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