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
2025303540455055

Find the sample standard deviation s of the data: 49, 20, 44, 56, 36, 17.

n = 6
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 35.

μ = 50σ = 10x = 35
z =
3.ProbabilityMedium
xp(x)10.2520.330.340.15μ = 2.35

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

x: 1, 2, 3, 4p(x): 0.25, 0.3, 0.3, 0.15
Var(X) =
4.ProbabilityHard
€p(x)−50.78100.22450.02

A game costs €5 to play. You win €15 with probability 0.2 and €250 with probability 0.02; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.

stake = €5€15: p = 0.2€250: p = 0.02
E(G) = €
5.Normal distributionMedium
−3−2−10123zshaded area to the right of z = 0.0655

Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.0655. Find z to two decimals.

A = 0.0655
z =
6.Normal distributionHard
10 g0X − Y?

The mass X of a product from line 1 and Y from line 2 are independent and normal: X has mean 500 g and standard deviation 8 g, Y has mean 490 g and standard deviation 6 g. Find the probability P that X is greater than Y. Round z to two decimals.

μX = 500 g, σX = 8 gμY = 490 g, σY = 6 g
P =
7.Confidence intervalsMedium
n = 6, s = 9LUx̄ = 92E = ?99%

A sample of 6 items has mean 92 and standard deviation s = 9. The population is roughly normal. Find the margin of error E of a 99% confidence interval for the mean.

n = 6x̄ = 92s = 999%t(5) = 4.032
E =
8.Confidence intervalsHard
σ = 23, n = ?x̄ − Ex̄ + Ex̄E = 290%

The population standard deviation is 23. 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 2?

σ = 23E = 290%
n =
9.Hypothesis tests & confidence intervalsMedium
−0.86p = ?

The mean of a population is claimed to be 100; the population standard deviation is known to be 48. A random sample of 144 observations has a mean of 96.56. Test H₀: μ = 100 against H₁: μ < 100. Find the p-value p.

μ₀ = 100σ = 48n = 144x̄ = 96.56H₁: μ < μ₀
p =
10.Hypothesis tests & confidence intervalsHard
z −2.57602.576

A sample of 81 measurements has mean 77.4. The population standard deviation is 12. Find the lower limit L of the 99% confidence interval for the mean.

x̄ = 77.4σ = 12n = 8199%
L =
11.Linear regression & correlationMedium
xy238346451554668xy24640506070

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

n = 5Σx = 20Σy = 257Σx² = 90Σy² = 13701Σxy = 1096
r =
12.Linear regression & correlationHard
xy187921103241182713530149xy202530100150x = 28.5

The table gives 5 paired observations of x = temperature in °C and y = ice creams sold. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 28.5.

n = 5x = 28.5
ŷ =

3Solutions

1.Descriptive statistics

Answer: s = 15.77

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 = 222/6 = 37

Squared deviations

Σ(x − x̄)² = 12² + (−17)² + 7² + 19² + (−1)² + (−20)² = 1244

Standard deviation

s = √(1244/5) = 15.77

Answer

s = 15.77
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 = (35 − 50)/10 = −1.5

Answer

z = −1.5
3.Probability

Answer: Var(X) = 1.027

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 = 1·0.25 + 2·0.3 + 3·0.3 + 4·0.15 = 2.35

Mean of the squares

E(X²) = 1·0.25 + 4·0.3 + 9·0.3 + 16·0.15 = 6.55

Variance

Var(X) = 6.55 − 2.35² = 1.027

Answer

Var(X) = 1.027
4.Probability

Answer: E(G) = 3 €

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.

Net outcomes

G: 10 (p = 0.2), 245 (p = 0.02), −5 (p = 0.78)

Expected value

E(G) = 10·0.2 + 245·0.02 − 5·0.78 = 3 €

Answer

E(G) = 3 €

A positive value means you win money on average.

5.Normal distribution

Answer: z = 1.51

From an area back to z: Φ(z) is the area to the left of z. For a left-tail area A, find z with Φ(z) = A in the table. A right-tail area A means Φ(z) = 1 − A. A central area A between −z and z leaves (1 − A)/2 in each tail, so Φ(z) = (1 + A)/2. For a value x, use x = μ + zσ.

Area to the left of z

Φ(z) = 1 − 0.0655 = 0.9345

Read the z-table backwards

Find 0.9345 inside the table and read off z.

z = 1.51

Answer

z = 1.51
6.Normal distribution

Answer: P = 0.8413

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 = 500 − 490 = 10 g, σ = 10 g

Standardise

z = (0 − 10)/10 = −1

Probability

P = 1 − Φ(−1) = 0.8413

Answer

P = 0.8413
7.Confidence intervals

Answer: E = 14.81

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 = 5, t = 4.032

Margin of error

E = 4.032·9/√6 = 14.81

Answer

E = 14.81
8.Confidence intervals

Answer: n = 358

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 = (1.645·23/2)² = 357.9

Round up

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

n = 358

Answer

n = 358
9.Hypothesis tests & confidence intervals

Answer: p = 0.1949

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.

Test statistic

SE = σ/√n = 48/√144 = 4
z = (96.56 − 100)/4 = −0.86

Tail area

Φ(0.86) = 0.8051
1 − Φ(0.86) = 0.1949

p-value

p = 0.1949

Answer

p = 0.1949

p > 0.05, so H₀ is not rejected at the 5% level.

10.Hypothesis tests & confidence intervals

Answer: L = 73.97

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/√81 = 3.435

Interval limit

L = 77.4 − 3.435 = 73.97

Answer

L = 73.97
11.Linear regression & correlation

Answer: r = 0.9702

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 = 1096 − 20·257/5 = 68
Sxx = 90 − 20²/5 = 10
Syy = 13701 − 257²/5 = 491.2

Correlation coefficient

r = 68/√(10·491.2) = 0.9702

Answer

r = 0.9702
12.Linear regression & correlation

Answer: ŷ = 142.6

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 = 120/5 = 24
ȳ = 584/5 = 116.8

Sums of squares and products

Sxx = Σ(x − x̄)² = 90
Sxy = Σ(x − x̄)(y − ȳ) = 516

Regression line

b = 516/90 = 5.733
a = 116.8 − 5.733·24 = −20.8
ŷ = −20.8 + 5.733·x

Prediction

ŷ = −20.8 + 5.733·28.5 = 142.6

Answer

ŷ = 142.6

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