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
152025303540

Find the sample standard deviation s of the data: 24, 31, 11, 27, 41, 44, 41.

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

A quantity is normally distributed with mean 70 and standard deviation 15. What proportion P of values is below 88.75?

μ = 70σ = 15x = 88.75
P =
3.ProbabilityMedium
xp(x)00.1510.3520.1530.1540.2μ = 1.9

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

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

A test has 11 multiple-choice questions. A student who guesses gets each one right with probability 0.25, independently. What is the probability P of exactly 1 correct answers?

n = 11p = 0.25k = 1
P =
5.Normal distributionMedium
160 g163.6 gT = X₁ + … + X₄?

The masses of bolts are independent and normal with mean 40 g and standard deviation 1.2 g. Find the probability P that the total mass of 4 of them exceeds 163.6 g. Round z to two decimals.

μ = 40 gσ = 1.2 gn = 4> 163.6 g
P =
6.Normal distributionHard
μ = 178x = ?75%

The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. Find the value x below which 75% of them lie (the 75th percentile).

μ = 178σ = 7P = 75%
x = cm
7.Confidence intervalsMedium
n = 13, sd​ = 4.20LU = ?d̄ = −8.4E99%

The blood pressure of 13 patients is measured before and after a treatment. The differences d = before − after (mmHg) have mean −8.4 and standard deviation 4.2. Find the upper limit of the 99% confidence interval for the mean difference.

n = 13d̄ = −8.4sd = 4.299%t(12) = 3.055
upper limit = mmHg
8.Confidence intervalsHard
σ = 32, n = ?x̄ − Ex̄ + Ex̄E = 299%

The population standard deviation is 32. How large a sample n do you need so that a 99% confidence interval for the mean has a margin of error of at most 2?

σ = 32E = 299%
n =
9.Hypothesis tests & confidence intervalsMedium
t −2.01102.011

A sample of 49 students has a mean score of 116.5 and a sample standard deviation of 8. Find the test statistic t for H₀: μ = 120.

μ₀ = 120x̄ = 116.5s = 8n = 49
t =
10.Hypothesis tests & confidence intervalsHard
z −1.9601.96

A sample of 100 measurements has mean 498.4. The population standard deviation is 15. Find the upper limit U of the 95% confidence interval for the mean.

x̄ = 498.4σ = 15n = 10095%
U =
11.Linear regression & correlationMedium
xy128229326423523xy24242628

The table gives 5 paired observations of x = car age in years and y = price in k€. Their sums are given. Find Pearson's correlation coefficient r from these sums.

n = 5Σx = 15Σy = 129Σx² = 55Σy² = 3359Σxy = 371
r =
12.Linear regression & correlationHard
xy3555597699801191xy5106080x = 4

The table gives 5 paired observations of x = hours studied and y = exam score. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 4.

n = 5x = 4
ŷ =

3Solutions

1.Descriptive statistics

Answer: s = 11.79

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 = 219/7 = 31.29

Squared deviations

Σ(x − x̄)² = (−7.286)² + (−0.2857)² + (−20.29)² + (−4.286)² + 9.714² + 12.71² + 9.714² = 833.4

Standard deviation

s = √(833.4/6) = 11.79

Answer

s = 11.79
2.Descriptive statistics

Answer: P = 0.8944

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 = 1.25

Standard normal table

Φ(1.25) = 0.8944
P = 0.8944

Answer

P = 0.8944
3.Probability

Answer: Var(X) = 1.89

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 = 0·0.15 + 1·0.35 + 2·0.15 + 3·0.15 + 4·0.2 = 1.9

Mean of the squares

E(X²) = 0·0.15 + 1·0.35 + 4·0.15 + 9·0.15 + 16·0.2 = 5.5

Variance

Var(X) = 5.5 − 1.9² = 1.89

Answer

Var(X) = 1.89
4.Probability

Answer: P = 0.1549

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 = 1) = 11·0.251·0.7510 = 11·0.25·0.05631 = 0.1549

Answer

P = 0.1549
5.Normal distribution

Answer: P = 0.06681

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 = 160 g, σT = 1.2·√4 = 2.4 g

Standardise

z = (163.6 − 160)/2.4 = 1.5

Probability

P = 1 − Φ(1.5) = 0.06681

Answer

P = 0.06681
6.Normal distribution

Answer: x = 182.7 cm

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) = 0.75 → z = 0.6745

Back to x

x = 178 + 0.6745·7 = 182.72 cm

Answer

x = 182.72 cm
7.Confidence intervals

Answer: upper limit = −4.841 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 = 12, t = 3.055

Margin of error

E = 3.055·4.2/√13 = 3.559

Interval

−8.4 ± 3.559 → [−11.96, −4.841]

Answer

upper limit = −4.841 mmHg
8.Confidence intervals

Answer: n = 1699

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 = (2.576·32/2)² = 1699

Round up

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

n = 1699

Answer

n = 1699
9.Hypothesis tests & confidence intervals

Answer: t = −3.063

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 = 8/√49 = 8/7 = 1.143

Test statistic

t = (116.5 − 120)/1.143 = −3.063

Answer

t = −3.063
t0.025; df = 48 = 2.011

|t| > 2.011, so H₀ is rejected at the 5% level (two-sided).

10.Hypothesis tests & confidence intervals

Answer: U = 501.3

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 = 1.96·15/√100 = 2.94

Interval limit

U = 498.4 + 2.94 = 501.3

Answer

U = 501.3
11.Linear regression & correlation

Answer: r = −0.9117

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 = 371 − 15·129/5 = −16
Sxx = 55 − 15²/5 = 10
Syy = 3359 − 129²/5 = 30.8

Correlation coefficient

r = −16/√(10·30.8) = −0.9117

Answer

r = −0.9117
12.Linear regression & correlation

Answer: ŷ = 56.85

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 = 35/5 = 7
ȳ = 354/5 = 70.8

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 186

Regression line

b = 186/40 = 4.65
a = 70.8 − 4.65·7 = 38.25
ŷ = 38.25 + 4.65·x

Prediction

ŷ = 38.25 + 4.65·4 = 56.85

Answer

ŷ = 56.85

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