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
1015202530354045

Find the sample standard deviation s of the data: 32, 32, 48, 30, 46, 10, 42.

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

A quantity is normally distributed with mean 70 and standard deviation 20. What proportion P of values is above 90?

μ = 70σ = 20x = 90
P =
3.ProbabilityMedium
xp(x)00.3510.3520.1530.15μ = 1.1

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

x: 0, 1, 2, 3p(x): 0.35, 0.35, 0.15, 0.15
Var(X) =
4.ProbabilityHard
€p(x)−100.79400.21900.01

A game costs €10 to play. You win €50 with probability 0.2 and €200 with probability 0.01; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.

stake = €10€50: p = 0.2€200: p = 0.01
E(G) = €
5.Normal distributionMedium
640 g642.4 gT = X₁ + … + X₁₆?

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 16 of them is below 642.4 g. Round z to two decimals.

μ = 40 gσ = 1.2 gn = 16< 642.4 g
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 = 30, s = 18LU = ?x̄ = 110E90%

A sample of 30 items has mean 110 and standard deviation s = 18. The population is roughly normal. Find the upper limit of the 90% confidence interval for the mean.

n = 30x̄ = 110s = 1890%t(29) = 1.699
upper limit =
8.Confidence intervalsHard
p ≈ 0.15, n = ?p̂ − Ep̂ + Ep̂E = 0.0299%

A pilot study suggests that about 0.15 of products are defective. How many products n must be inspected so that a 99% confidence interval for the proportion has a margin of error of at most 0.02?

p ≈ 0.15E = 0.0299%
n =
9.Hypothesis tests & confidence intervalsMedium
t −1.9901.99

A sample of 81 students has a mean score of 501.9 and a sample standard deviation of 8. Find the test statistic t for H₀: μ = 500.

μ₀ = 500x̄ = 501.9s = 8n = 81
t =
10.Hypothesis tests & confidence intervalsHard
μ₀x̄c

A one-sided z test of H₀: μ = 100 against H₁: μ > 100 is done at α = 0.10 with a sample of 64; the standard deviation is known to be 4. In reality the mean is 101.45. Find the power 1 − β of the test in percent.

μ₀ = 100μ = 101.45σ = 4n = 64α = 0.10
1 − β = %
11.Linear regression & correlationMedium
xy264106108131019xy5105101520

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

n = 5Σx = 30Σy = 58Σx² = 220Σy² = 766Σxy = 406
r =
12.Linear regression & correlationHard
xy18612159247027943094xy2025306080x = 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. A new observation (28.5, 94) is made. Find its residual e = y − ŷ.

n = 5(28.5, 94)
e =

3Solutions

1.Descriptive statistics

Answer: s = 12.93

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 = 240/7 = 34.29

Squared deviations

Σ(x − x̄)² = (−2.286)² + (−2.286)² + 13.71² + (−4.286)² + 11.71² + (−24.29)² + 7.714² = 1003

Standard deviation

s = √(1003/6) = 12.93

Answer

s = 12.93
2.Descriptive statistics

Answer: P = 0.1587

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

Standard normal table

Φ(1) = 0.8413
P = 1 − 0.8413 = 0.1587

Answer

P = 0.1587
3.Probability

Answer: Var(X) = 1.09

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 = 0·0.35 + 1·0.35 + 2·0.15 + 3·0.15 = 1.1

Mean of the squares

E(X²) = 0·0.35 + 1·0.35 + 4·0.15 + 9·0.15 = 2.3

Variance

Var(X) = 2.3 − 1.1² = 1.09

Answer

Var(X) = 1.09
4.Probability

Answer: E(G) = 2 €

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: 40 (p = 0.2), 190 (p = 0.01), −10 (p = 0.79)

Expected value

E(G) = 40·0.2 + 190·0.01 − 10·0.79 = 2 €

Answer

E(G) = 2 €

A positive value means you win money on average.

5.Normal distribution

Answer: P = 0.6915

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 = 640 g, σT = 1.2·√16 = 4.8 g

Standardise

z = (642.4 − 640)/4.8 = 0.5

Probability

P = Φ(0.5) = 0.6915

Answer

P = 0.6915
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 = 115.6

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 = 29, t = 1.699

Margin of error

E = 1.699·18/√30 = 5.583

Interval

110 ± 5.583 → [104.4, 115.6]

Answer

upper limit = 115.6
8.Confidence intervals

Answer: n = 2116

Confidence interval for a proportion: p̂ ± z·√(p̂(1 − p̂)/n), where p̂ = x/n. To reach a margin E you need n ≥ z²·p(1 − p)/E², rounded up; without a prior estimate use p = 0.5, which gives the largest n.

Solve for n

n = 2.576²·0.15·0.85/0.02² = 2115

Round up

n = 2116

Answer

n = 2116
9.Hypothesis tests & confidence intervals

Answer: t = 2.137

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/√81 = 8/9 = 0.8889

Test statistic

t = (501.9 − 500)/0.8889 = 2.137

Answer

t = 2.137
t0.025; df = 80 = 1.99

|t| > 1.99, so H₀ is rejected at the 5% level (two-sided).

10.Hypothesis tests & confidence intervals

Answer: 1 − β = 94.72 %

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 = 4/√64 = 0.5
x̄c = 100 + 1.282·0.5 = 100.6

z under the true mean

z = (100.6 − 101.45)/0.5 = −1.618

Power

power = P(Z > −1.618) = 94.72%
1 − β = 94.72%

Answer

1 − β = 94.72%
11.Linear regression & correlation

Answer: r = 0.9499

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 = 406 − 30·58/5 = 58
Sxx = 220 − 30²/5 = 40
Syy = 766 − 58²/5 = 93.2

Correlation coefficient

r = 58/√(40·93.2) = 0.9499

Answer

r = 0.9499
12.Linear regression & correlation

Answer: e = 3.25

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 = 120/5 = 24
ȳ = 378/5 = 75.6

Sums of squares and products

Sxx = Σ(x − x̄)² = 90
Sxy = Σ(x − x̄)(y − ȳ) = 303

Regression line

b = 303/90 = 3.367
a = 75.6 − 3.367·24 = −5.2
ŷ = −5.2 + 3.367·x

Predicted value

ŷ = −5.2 + 3.367·28.5 = 90.75

Residual

e = 94 − 90.75 = 3.25

Answer

e = 3.25

The point lies above the line.

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