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
30354045505560

Find the sample standard deviation s of the data: 37, 53, 53, 60, 29, 28.

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

A quantity is normally distributed with mean 70 and standard deviation 12. What proportion P of values is below 55.6?

μ = 70σ = 12x = 55.6
P =
3.ProbabilityMedium
xp(x)00.210.3520.0530.2540.15μ = 1.8

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

x: 0, 1, 2, 3, 4p(x): 0.2, 0.35, 0.05, 0.25, 0.15
Var(X) =
4.ProbabilityHard
n = 11, p = 0.1: P(X = 5) = ?k01234567891011

A test has 11 multiple-choice questions. A student who guesses gets each one right with probability 0.1, independently. What is the probability P of exactly 5 correct answers?

n = 11p = 0.1k = 5
P =
5.Normal distributionMedium
240.25254.25P = ?

The bag weights are normally distributed with mean 250 g and standard deviation 5 g. What fraction lies between 240.25 and 254.25 g? Give P as a decimal.

μ = 250σ = 5x1 = 240.25x2 = 254.25
P =
6.Normal distributionHard
μ = 65x = ?95%

The exam scores are normally distributed with mean 65 points and standard deviation 10 points. Find the value x below which 95% of them lie (the 95th percentile).

μ = 65σ = 10P = 95%
x = points
7.Confidence intervalsMedium
n = 16, sd​ = 6.70LU = ?d̄ = 2.9E99%

The blood pressure of 16 patients is measured before and after a treatment. The differences d = before − after (mmHg) have mean 2.9 and standard deviation 6.7. Find the upper limit of the 99% confidence interval for the mean difference.

n = 16d̄ = 2.9sd = 6.799%t(15) = 2.947
upper limit = mmHg
8.Confidence intervalsHard
σ = 30, n = ?x̄ − Ex̄ + Ex̄E = 595%

The population standard deviation is 30. How large a sample n do you need so that a 95% confidence interval for the mean has a margin of error of at most 5?

σ = 30E = 595%
n =
9.Hypothesis tests & confidence intervalsMedium
−2.1452.145t = ?df = 14

Two teaching methods are compared. Group 1: 5 students, mean score 88.5, standard deviation 5. Group 2: 11 students, mean score 97.6, standard deviation 7. Assume equal population variances and use the pooled two-sample t test. Find the test statistic t for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (group 1 minus group 2).

n₁ = 5x̄₁ = 88.5s₁ = 5n₂ = 11x̄₂ = 97.6s₂ = 7
t =
10.Hypothesis tests & confidence intervalsHard
σ = 15, n = ?x̄ − Ex̄ + Ex̄E = 390%

How large must a sample be so that the 90% confidence interval for a mean has a margin of error of at most 3? The standard deviation is 15. Find the smallest sample size n.

σ = 15E = 390%
n =
11.Linear regression & correlationMedium
xy27496108141016xy5101015

The table gives 5 paired observations of x = load in kN and y = extension in mm. Their sums are given. Find Pearson's correlation coefficient r from these sums.

n = 5Σx = 30Σy = 56Σx² = 220Σy² = 682Σxy = 382
r =
12.Linear regression & correlationHard
xy128224327423524xy24242628x = 2.5

The table gives 5 paired observations of x = car age in years and y = price in k€. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 2.5.

n = 5x = 2.5
ŷ =

3Solutions

1.Descriptive statistics

Answer: s = 13.75

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 = 260/6 = 43.33

Squared deviations

Σ(x − x̄)² = (−6.333)² + 9.667² + 9.667² + 16.67² + (−14.33)² + (−15.33)² = 945.3

Standard deviation

s = √(945.3/5) = 13.75

Answer

s = 13.75
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: Var(X) = 1.96

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 = 0·0.2 + 1·0.35 + 2·0.05 + 3·0.25 + 4·0.15 = 1.8

Mean of the squares

E(X²) = 0·0.2 + 1·0.35 + 4·0.05 + 9·0.25 + 16·0.15 = 5.2

Variance

Var(X) = 5.2 − 1.8² = 1.96

Answer

Var(X) = 1.96
4.Probability

Answer: P = 0.002455

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) = 462·0.15·0.96 = 462·1 × 10⁻⁵·0.5314 = 0.002455

Answer

P = 0.002455
5.Normal distribution

Answer: P = 0.7767

Normal distribution: convert a value x to a z-score, z = (x − μ)/σ, which counts standard deviations from the mean. A z-table (or Φ(z)) gives the area to the left of z, which is P(X < x). The area to the right is 1 − Φ(z), and the area between two values is the difference of their two areas.

z-scores

z₁ = (240.25 − 250)/5 = −1.95
z₂ = (254.25 − 250)/5 = 0.85

Area between

P = Φ(0.85) − Φ(−1.95) = 0.8023 − 0.02559 = 0.7767

Answer

P = 0.7767
6.Normal distribution

Answer: x = 81.45 points

Normal distribution: convert a value x to a z-score, z = (x − μ)/σ, which counts standard deviations from the mean. A z-table (or Φ(z)) gives the area to the left of z, which is P(X < x). The area to the right is 1 − Φ(z), and the area between two values is the difference of their two areas.

z for the percentile

Φ(z) = 0.95 → z = 1.645

Back to x

x = 65 + 1.645·10 = 81.45 points

Answer

x = 81.45 points
7.Confidence intervals

Answer: upper limit = 7.836 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 = 15, t = 2.947

Margin of error

E = 2.947·6.7/√16 = 4.936

Interval

2.9 ± 4.936 → [−2.036, 7.836]

Answer

upper limit = 7.836 mmHg
8.Confidence intervals

Answer: n = 139

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 = (1.96·30/5)² = 138.3

Round up

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

n = 139

Answer

n = 139
9.Hypothesis tests & confidence intervals

Answer: t = −2.599

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.

Pooled variance

sp² = (4·5² + 10·7²)/14 = 42.14
sp = 6.492

Standard error

SE = 6.492·√(1/5 + 1/11) = 3.501

Test statistic

t = (88.5 − 97.6)/3.501 = −2.599

Answer

t = −2.599

With df = 14, the two-sided 5% critical value is 2.145; |t| is larger, so H₀ is rejected.

10.Hypothesis tests & confidence intervals

Answer: n = 68

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 = (1.645·15/3)² = 67.65

Round up

n = 68

Answer

n = 68
11.Linear regression & correlation

Answer: r = 0.9825

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 = 382 − 30·56/5 = 46
Sxx = 220 − 30²/5 = 40
Syy = 682 − 56²/5 = 54.8

Correlation coefficient

r = 46/√(40·54.8) = 0.9825

Answer

r = 0.9825
12.Linear regression & correlation

Answer: ŷ = 25.65

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 = 126/5 = 25.2

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = −9

Regression line

b = −9/10 = −0.9
a = 25.2 − (−0.9)·3 = 27.9
ŷ = 27.9 − 0.9·x

Prediction

ŷ = 27.9 + (−0.9)·2.5 = 25.65

Answer

ŷ = 25.65

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