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
152025303540455055

Find the sample standard deviation s of the data: 15, 40, 18, 58, 58, 30, 36.

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

A quantity is normally distributed with mean 50 and standard deviation 10. What proportion P of values is above 30?

μ = 50σ = 10x = 30
P =
3.ProbabilityMedium
xp(x)10.1520.3530.1540.35μ = 2.7

For the distribution shown, find the standard deviation σ of X.

x: 1, 2, 3, 4p(x): 0.15, 0.35, 0.15, 0.35
σ =
4.ProbabilityHard
€p(x)−50.83200.152450.02

A game costs €5 to play. You win €25 with probability 0.15 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€25: p = 0.15€250: p = 0.02
E(G) = €
5.Normal distributionMedium
150 kg149.4 kgT = X₁ + … + X₆?

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 6 of them is below 149.4 kg. Round z to two decimals.

μ = 25 kgσ = 0.5 kgn = 6< 149.4 kg
P =
6.Normal distributionHard
235240245250255260265gtop 40%

The bag weights are normally distributed with mean 250 g and standard deviation 5 g. Drag the cut-off so the shaded top 40% is to the right of it. Find the cut-off x.

μ = 250 gσ = 5 g40%
x = g
7.Confidence intervalsMedium
x̄E₁ = 6.790%, n = 25x̄E₂ = ?95%, n = 225

A 90% confidence interval for a mean, based on 25 observations and a known σ, has a margin of error of 6.7. What is the margin of error E₂ of a 95% interval based on 225 observations?

90%n₁ = 25E₁ = 6.795%n₂ = 225
E₂ =
8.Confidence intervalsHard
σ = 17, n = ?x̄ − Ex̄ + Ex̄E = 190%

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 1?

σ = 17E = 190%
n =
9.Hypothesis tests & confidence intervalsMedium
−1.961.96z = ?

Two production lines make the same part. Line 1: a sample of 30 parts has mean length 71.5 mm, and the line's known standard deviation is 12 mm. Line 2: a sample of 30 parts has mean 69.4 mm, known standard deviation 4 mm. Find the test statistic z for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (line 1 minus line 2).

n₁ = 30x̄₁ = 71.5 mmσ₁ = 12 mmn₂ = 30x̄₂ = 69.4 mmσ₂ = 4 mm
z =
10.Hypothesis tests & confidence intervalsHard
σ = 8, n = ?x̄ − Ex̄ + Ex̄E = 395%

How large must a sample be so that the 95% confidence interval for a mean has a margin of error of at most 3? The standard deviation is 8. Find the smallest sample size n.

σ = 8E = 395%
n =
11.Linear regression & correlationMedium
xy2344678121014xy51051015

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

n = 5
r =
12.Linear regression & correlationHard
xy1587189721113241292715630167xy15202530100150x = 28.5

The table gives 6 paired observations of x = temperature in °C and y = ice creams sold. Use the least-squares line ŷ = a + b·x. A new observation (28.5, 164) is made. Find its residual e = y − ŷ.

n = 6(28.5, 164)
e =

3Solutions

1.Descriptive statistics

Answer: s = 17.24

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 = 255/7 = 36.43

Squared deviations

Σ(x − x̄)² = (−21.43)² + 3.571² + (−18.43)² + 21.57² + 21.57² + (−6.429)² + (−0.4286)² = 1784

Standard deviation

s = √(1784/6) = 17.24

Answer

s = 17.24
2.Descriptive statistics

Answer: P = 0.9772

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 = −2

Standard normal table

Φ(−2) = 0.02275
P = 1 − 0.02275 = 0.9772

Answer

P = 0.9772
3.Probability

Answer: σ = 1.1

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 = 1·0.15 + 2·0.35 + 3·0.15 + 4·0.35 = 2.7

Mean of the squares

E(X²) = 1·0.15 + 4·0.35 + 9·0.15 + 16·0.35 = 8.5

Variance

Var(X) = 8.5 − 2.7² = 1.21
σ = √1.21 = 1.1

Answer

σ = 1.1
4.Probability

Answer: E(G) = 3.75 €

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: 20 (p = 0.15), 245 (p = 0.02), −5 (p = 0.83)

Expected value

E(G) = 20·0.15 + 245·0.02 − 5·0.83 = 3.75 €

Answer

E(G) = 3.75 €

A positive value means you win money on average.

5.Normal distribution

Answer: P = 0.3121

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 = 150 kg, σT = 0.5·√6 = 1.225 kg

Standardise

z = (149.4 − 150)/1.225 = −0.49

Probability

P = Φ(−0.49) = 0.3121

Answer

P = 0.3121
6.Normal distribution

Answer: x = 251.3 g

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

Φ(z) = 0.6

z from the table

z ≈ 0.25

Back to the original units

x = 250 + 0.25·5 = 251.3 g

Answer

x = 251.3 g
7.Confidence intervals

Answer: E₂ = 2.661

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.

Ratio of margins

E₂/E₁ = (1.96/1.645)·√(25/225) = 0.3972

New margin

E₂ = 6.7·0.3972 = 2.661

Answer

E₂ = 2.661
8.Confidence intervals

Answer: n = 783

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 = (1.645·17/1)² = 782

Round up

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

n = 783

Answer

n = 783
9.Hypothesis tests & confidence intervals

Answer: z = 0.9093

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 = √(12²/30 + 4²/30) = √5.333 = 2.309

Test statistic

z = (71.5 − 69.4)/2.309 = 0.9093

Answer

z = 0.9093

|z| < 1.96, so H₀ is not rejected at the 5% level (two-sided).

10.Hypothesis tests & confidence intervals

Answer: n = 28

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 = (1.96·8/3)² = 27.32

Round up

n = 28

Answer

n = 28
11.Linear regression & correlation

Answer: r = 0.9785

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 = 40/5 = 8

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 60
Syy = Σ(y − ȳ)² = 94

Correlation coefficient

r = 60/√(40·94) = 0.9785

Answer

r = 0.9785
12.Linear regression & correlation

Answer: e = 5.281

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 = 135/6 = 22.5
ȳ = 749/6 = 124.8

Sums of squares and products

Sxx = Σ(x − x̄)² = 157.5
Sxy = Σ(x − x̄)(y − ȳ) = 889.5

Regression line

b = 889.5/157.5 = 5.648
a = 124.8 − 5.648·22.5 = −2.238
ŷ = −2.238 + 5.648·x

Predicted value

ŷ = −2.238 + 5.648·28.5 = 158.7

Residual

e = 164 − 158.7 = 5.281

Answer

e = 5.281

The point lies above the line.

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