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
10152025303540

Find the sample standard deviation s of the data: 28, 11, 40, 10, 32.

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

Exam scores are normally distributed with mean 50 and standard deviation 10. Find the z-score z of a score of 54.

μ = 50σ = 10x = 54
z =
3.ProbabilityMedium
red: 6blue: 3

A bag holds 6 red and 3 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?

6 / 3
P =
4.ProbabilityHard
xp(x)100.3200.2300.25400.25μ

X has the distribution shown, and Y = 0.5X + 5. Find the standard deviation σY of Y.

x: 10, 20, 30, 40p(x): 0.3, 0.2, 0.25, 0.25Y = 0.5X + 5
σY =
5.Normal distributionMedium
340 kg388 kgT = X₁ + … + X₄?

The masses of passengers with luggage are independent and normal with mean 85 kg and standard deviation 12 kg. Find the probability P that the total mass of 4 of them exceeds 388 kg. Round z to two decimals.

μ = 85 kgσ = 12 kgn = 4> 388 kg
P =
6.Normal distributionHard
μ = 32x = ?20%

The delivery times are normally distributed with mean 32 min and standard deviation 6 min. Find the value x below which 20% of them lie (the 20th percentile).

μ = 32σ = 6P = 20%
x = min
7.Confidence intervalsMedium
n = 11, s = 5.5LU = ?x̄ = 105E90%

A sample of 11 items has mean 105 and standard deviation s = 5.5. The population is roughly normal. Find the upper limit of the 90% confidence interval for the mean.

n = 11x̄ = 105s = 5.590%t(10) = 1.812
upper limit =
8.Confidence intervalsHard
σ = 25, n = ?x̄ − Ex̄ + Ex̄E = 499%

The population standard deviation is 25. 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 4?

σ = 25E = 499%
n =
9.Hypothesis tests & confidence intervalsMedium
t −2.13102.131

A sample of 16 students has a mean score of 492.2 and a sample standard deviation of 12. Find the test statistic t for H₀: μ = 500.

μ₀ = 500x̄ = 492.2s = 12n = 16
t =
10.Hypothesis tests & confidence intervalsHard
H₀H₁μ₀x̄cμ₀ + 2.5α = 0.05βσ = 13, power 90%, n = ?

A one-sided z test at α = 0.05 must detect a shift of the mean of 2.5 units with a power of 90%. The standard deviation is 13. Find the minimum sample size n.

α = 0.05power = 90%δ = 2.5σ = 13zα = 1.645zβ = 1.282
n =
11.Linear regression & correlationMedium
xy18552166248227943010333115xy2025306080100120

The table gives 6 paired observations of x = temperature in °C and y = ice creams sold. Their sums are given. Find Pearson's correlation coefficient r from these sums.

n = 6Σx = 153Σy = 515Σx² = 4059Σy² = 46775Σxy = 13767
r =
12.Linear regression & correlationHard
xy2504616648711076xy510506070x = 5

The table gives 5 paired observations of x = hours studied and y = exam score. Use the least-squares line ŷ = a + b·x. A new observation (5, 57) is made. Find its residual e = y − ŷ.

n = 5(5, 57)
e =

3Solutions

1.Descriptive statistics

Answer: s = 13.24

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 = 121/5 = 24.2

Squared deviations

Σ(x − x̄)² = 3.8² + (−13.2)² + 15.8² + (−14.2)² + 7.8² = 700.8

Standard deviation

s = √(700.8/4) = 13.24

Answer

s = 13.24
2.Descriptive statistics

Answer: z = 0.4

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 = (54 − 50)/10 = 0.4

Answer

z = 0.4
3.Probability

Answer: P = 0.4167

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.

First and second draw

P = 6/9 · 5/8 = 30/72 = 0.4167

Answer

P = 0.4167
4.Probability

Answer: σY = 5.804

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.

Spread of X

μ = 24.5, Var(X) = 134.8, σX = 11.61

Linear change

σY = |0.5|·11.61 = 5.804

Answer

σY = 5.804

The mean does move: E(Y) = 0.5·24.5 + 5 = 17.25.

5.Normal distribution

Answer: P = 0.02275

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 = 340 kg, σT = 12·√4 = 24 kg

Standardise

z = (388 − 340)/24 = 2

Probability

P = 1 − Φ(2) = 0.02275

Answer

P = 0.02275
6.Normal distribution

Answer: x = 26.95 min

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) = 0.2 → z = −0.8416

Back to x

x = 32 + −0.8416·6 = 26.95 min

Answer

x = 26.95 min
7.Confidence intervals

Answer: upper limit = 108

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 = 1.812

Margin of error

E = 1.812·5.5/√11 = 3.005

Interval

105 ± 3.005 → [102, 108]

Answer

upper limit = 108
8.Confidence intervals

Answer: n = 260

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 = (2.576·25/4)² = 259.2

Round up

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

n = 260

Answer

n = 260
9.Hypothesis tests & confidence intervals

Answer: t = −2.6

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/√16 = 12/4 = 3

Test statistic

t = (492.2 − 500)/3 = −2.6

Answer

t = −2.6
t0.025; df = 15 = 2.131

|t| > 2.131, so H₀ is rejected at the 5% level (two-sided).

10.Hypothesis tests & confidence intervals

Answer: n = 232

Sample size for a test: to detect a true shift δ with power 1 − β in a one-sided z test at level α, the standard error must satisfy δ = (z_α + z_β)·σ/√n, so n = ((z_α + z_β)·σ/δ)², rounded up.

Solve for n

n = ((1.645 + 1.282)·13/2.5)² = 231.7

Round up

n = 232

Answer

n = 232
11.Linear regression & correlation

Answer: r = 0.9971

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 = 13767 − 153·515/6 = 634.5
Sxx = 4059 − 153²/6 = 157.5
Syy = 46775 − 515²/6 = 2571

Correlation coefficient

r = 634.5/√(157.5·2571) = 0.9971

Answer

r = 0.9971
12.Linear regression & correlation

Answer: e = −4.3

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 = 322/5 = 64.4

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 124

Regression line

b = 124/40 = 3.1
a = 64.4 − 3.1·6 = 45.8
ŷ = 45.8 + 3.1·x

Predicted value

ŷ = 45.8 + 3.1·5 = 61.3

Residual

e = 57 − 61.3 = −4.3

Answer

e = −4.3

The point lies below the line.

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