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
15202530354045

Find the sample standard deviation s of the data: 49, 28, 38, 36, 12.

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

Exam scores are normally distributed with mean 60 and standard deviation 15. Find the z-score z of a score of 90.

μ = 60σ = 15x = 90
z =
3.ProbabilityMedium
xp(x)00.320.340.0560.35μ = 2.9

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

x: 0, 2, 4, 6p(x): 0.3, 0.3, 0.05, 0.35
σ =
4.ProbabilityHard
€p(x)−50.88100.1450.02

A game costs €5 to play. You win €15 with probability 0.1 and €50 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 = 0.1€50: p = 0.02
E(G) = €
5.Normal distributionMedium
1020 kg957.6 kgT = X₁ + … + X₁₂?

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 12 of them exceeds 957.6 kg. Round z to two decimals.

μ = 85 kgσ = 12 kgn = 12> 957.6 kg
P =
6.Normal distributionHard
4 s0X − Y?

Runner A's lap time X and runner B's lap time Y are independent and normal: X has mean 52 s and standard deviation 4 s, Y has mean 48 s and standard deviation 3 s. Find the probability P that X is greater than Y. Round z to two decimals.

μX = 52 s, σX = 4 sμY = 48 s, σY = 3 s
P =
7.Confidence intervalsMedium
n = 21, s = 7.5LU = ?x̄ = 74E99%

A sample of 21 items has mean 74 and standard deviation s = 7.5. The population is roughly normal. Find the upper limit of the 99% confidence interval for the mean.

n = 21x̄ = 74s = 7.599%t(20) = 2.845
upper limit =
8.Confidence intervalsHard
σ = 40, n = ?x̄ − Ex̄ + Ex̄E = 295%

The population standard deviation is 40. 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 2?

σ = 40E = 295%
n =
9.Hypothesis tests & confidence intervalsMedium
−1.841.84p = ?

The mean of a population is claimed to be 250; the population standard deviation is known to be 36. A random sample of 144 observations has a mean of 244.48. Test H₀: μ = 250 against H₁: μ ≠ 250. Find the p-value p.

μ₀ = 250σ = 36n = 144x̄ = 244.48H₁: μ ≠ μ₀
p =
10.Hypothesis tests & confidence intervalsHard
μ₀x̄c

A one-sided z test of H₀: μ = 100 against H₁: μ < 100 is done at α = 0.05 with a sample of 100; the standard deviation is known to be 5. In reality the mean is 98.35. Find the power 1 − β of the test in percent.

μ₀ = 100μ = 98.35σ = 5n = 100α = 0.05
1 − β = %
11.Linear regression & correlationMedium
xy24496128121015xy51051015

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
xy125220321419516xy242025x = 3.5

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 = 3.5.

n = 5x = 3.5
ŷ =

3Solutions

1.Descriptive statistics

Answer: s = 13.74

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 = 163/5 = 32.6

Squared deviations

Σ(x − x̄)² = 16.4² + (−4.6)² + 5.4² + 3.4² + (−20.6)² = 755.2

Standard deviation

s = √(755.2/4) = 13.74

Answer

s = 13.74
2.Descriptive statistics

Answer: z = 2

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 = (90 − 60)/15 = 2

Answer

z = 2
3.Probability

Answer: σ = 2.488

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 = 0·0.3 + 2·0.3 + 4·0.05 + 6·0.35 = 2.9

Mean of the squares

E(X²) = 0·0.3 + 4·0.3 + 16·0.05 + 36·0.35 = 14.6

Variance

Var(X) = 14.6 − 2.9² = 6.19
σ = √6.19 = 2.488

Answer

σ = 2.488
4.Probability

Answer: E(G) = −2.5 €

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 = 0.1), 45 (p = 0.02), −5 (p = 0.88)

Expected value

E(G) = 10·0.1 + 45·0.02 − 5·0.88 = −2.5 €

Answer

E(G) = −2.5 €

A negative value means you lose money on average.

5.Normal distribution

Answer: P = 0.9332

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 = 1020 kg, σT = 12·√12 = 41.57 kg

Standardise

z = (957.6 − 1020)/41.57 = −1.5

Probability

P = 1 − Φ(−1.5) = 0.9332

Answer

P = 0.9332
6.Normal distribution

Answer: P = 0.7881

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 = 52 − 48 = 4 s, σ = 5 s

Standardise

z = (0 − 4)/5 = −0.8

Probability

P = 1 − Φ(−0.8) = 0.7881

Answer

P = 0.7881
7.Confidence intervals

Answer: upper limit = 78.66

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 = 20, t = 2.845

Margin of error

E = 2.845·7.5/√21 = 4.656

Interval

74 ± 4.656 → [69.34, 78.66]

Answer

upper limit = 78.66
8.Confidence intervals

Answer: n = 1537

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 = (1.96·40/2)² = 1537

Round up

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

n = 1537

Answer

n = 1537
9.Hypothesis tests & confidence intervals

Answer: p = 0.06577

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 = 36/√144 = 3
z = (244.48 − 250)/3 = −1.84

Tail area

Φ(1.84) = 0.9671
1 − Φ(1.84) = 0.0329

p-value

p = 2·0.0329 = 0.06577

Answer

p = 0.06577

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

10.Hypothesis tests & confidence intervals

Answer: 1 − β = 95.1 %

Power of a test: power = 1 − β is the probability of rejecting H₀ when it is false. Find the cutoff x̄c from μ₀ and z_α, then compute the probability of landing beyond x̄c using the true mean: z = (x̄c − μ)/(σ/√n).

Critical sample mean

SE = 5/√100 = 0.5
x̄c = 100 − 1.645·0.5 = 99.18

z under the true mean

z = (99.18 − 98.35)/0.5 = 1.655

Power

power = P(Z < 1.655) = 95.1%
1 − β = 95.1%

Answer

1 − β = 95.1%
11.Linear regression & correlation

Answer: r = 0.9504

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 = 52/5 = 10.4

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 50
Syy = Σ(y − ȳ)² = 69.2

Correlation coefficient

r = 50/√(40·69.2) = 0.9504

Answer

r = 0.9504
12.Linear regression & correlation

Answer: ŷ = 19.25

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 = 101/5 = 20.2

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = −19

Regression line

b = −19/10 = −1.9
a = 20.2 − (−1.9)·3 = 25.9
ŷ = 25.9 − 1.9·x

Prediction

ŷ = 25.9 + (−1.9)·3.5 = 19.25

Answer

ŷ = 19.25

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