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: 56, 13, 20, 59, 12, 26.

n = 6
s =
2.Descriptive statisticsHard
z −2−1012

A quantity is normally distributed with mean 60 and standard deviation 5. What proportion P of values is below 63?

μ = 60σ = 5x = 63
P =
3.ProbabilityMedium
red: 3blue: 9

A bag holds 3 red and 9 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?

3 / 9
P =
4.ProbabilityHard
xp(x)10.2520.2530.2540.25μ

X has the distribution shown, and Y = 3X + 20. Find the standard deviation σY of Y.

x: 1, 2, 3, 4p(x): 0.25, 0.25, 0.25, 0.25Y = 3X + 20
σY =
5.Normal distributionMedium
−3−2−10123zshaded area between −z and z = 0.6778

Drag z so that the shaded area between −z and z under the standard normal curve is 0.6778. Find z to two decimals.

A = 0.6778
z =
6.Normal distributionHard
μ = 250x = ?99%

The bag weights are normally distributed with mean 250 g and standard deviation 5 g. Find the value x below which 99% of them lie (the 99th percentile).

μ = 250σ = 5P = 99%
x = g
7.Confidence intervalsMedium
x̄E₁ = 8.999%, n = 64x̄E₂ = ?90%, n = 128

A 99% confidence interval for a mean, based on 64 observations and a known σ, has a margin of error of 8.9. What is the margin of error E₂ of a 90% interval based on 128 observations?

99%n₁ = 64E₁ = 8.990%n₂ = 128
E₂ =
8.Confidence intervalsHard
μ₁ − μ₂, t(20) = 1.7250L = ?Ux̄₁ − x̄₂E90%

Two fertilisers are tested. Plot group 1: 11 plots, mean yield 65 kg, standard deviation 3 kg. Plot group 2: 11 plots, mean 60.5 kg, standard deviation 5.5 kg. Assume equal variances (pooled t, df = n₁ + n₂ − 2). Find the lower limit of the 90% confidence interval for μ₁ − μ₂.

n₁ = 11x̄₁ = 65s₁ = 3n₂ = 11x̄₂ = 60.5s₂ = 5.590%t(20) = 1.725
lower limit = kg
9.Hypothesis tests & confidence intervalsMedium
t −1.9901.99

A sample of 81 students has a mean score of 72.6 and a sample standard deviation of 12. Find the test statistic t for H₀: μ = 75.

μ₀ = 75x̄ = 72.6s = 12n = 81
t =
10.Hypothesis tests & confidence intervalsHard
H₀H₁μ₀x̄cμ₀ + 3α = 0.01βσ = 21, power 80%, n = ?

A one-sided z test at α = 0.01 must detect a shift of the mean of 3 units with a power of 80%. The standard deviation is 21. Find the minimum sample size n.

α = 0.01power = 80%δ = 3σ = 21zα = 2.326zβ = 0.842
n =
11.Linear regression & correlationMedium
xy17701982217923902510227107xy202580100

The table gives 6 paired observations of x = temperature in °C and y = ice creams sold. Find the coefficient of determination r² of the least-squares line.

n = 6
r² =
12.Linear regression & correlationHard
xy2741462082410281233xy510102030x = 3

The table gives 6 paired observations of x = load in kN and y = extension in mm. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 3.

n = 6x = 3
ŷ =

3Solutions

1.Descriptive statistics

Answer: s = 21.17

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 = 186/6 = 31

Squared deviations

Σ(x − x̄)² = 25² + (−18)² + (−11)² + 28² + (−19)² + (−5)² = 2240

Standard deviation

s = √(2240/5) = 21.17

Answer

s = 21.17
2.Descriptive statistics

Answer: P = 0.7257

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 = 0.6

Standard normal table

Φ(0.6) = 0.7257
P = 0.7257

Answer

P = 0.7257
3.Probability

Answer: P = 0.04545

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 = 3/12 · 2/11 = 6/132 = 0.04545

Answer

P = 0.04545
4.Probability

Answer: σY = 3.354

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 = 2.5, Var(X) = 1.25, σX = 1.118

Linear change

σY = |3|·1.118 = 3.354

Answer

σY = 3.354

The mean does move: E(Y) = 3·2.5 + 20 = 27.5.

5.Normal distribution

Answer: z = 0.99

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 + 0.6778)/2 = 0.8389

Read the z-table backwards

Find 0.8389 inside the table and read off z.

z = 0.99

Answer

z = 0.99
6.Normal distribution

Answer: x = 261.6 g

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) = 0.99 → z = 2.326

Back to x

x = 250 + 2.326·5 = 261.63 g

Answer

x = 261.63 g
7.Confidence intervals

Answer: E₂ = 4.019

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 = (1.645/2.576)·√(64/128) = 0.4515

New margin

E₂ = 8.9·0.4515 = 4.019

Answer

E₂ = 4.019
8.Confidence intervals

Answer: lower limit = 1.242 kg

Interval for a difference: estimate ± critical value × standard error. For two independent means with known σ: SE = √(σ₁²/n₁ + σ₂²/n₂) and z. With equal unknown variances: pooled s_p and t with n₁ + n₂ − 2 degrees of freedom. For paired data, work with the differences: d̄ ± t·s_d/√n.

Pooled standard deviation

sp² = (10·3² + 10·5.5²)/20 = 19.63
sp = 4.43

Margin of error

E = 1.725·4.43·√(1/11 + 1/11) = 3.258

Interval

4.5 ± 3.258 → [1.242, 7.758]

Answer

lower limit = 1.242 kg
9.Hypothesis tests & confidence intervals

Answer: t = −1.8

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/√81 = 12/9 = 1.333

Test statistic

t = (72.6 − 75)/1.333 = −1.8

Answer

t = −1.8
t0.025; df = 80 = 1.99

|t| < 1.99, so H₀ is not rejected at the 5% level (two-sided).

10.Hypothesis tests & confidence intervals

Answer: n = 492

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 = ((2.326 + 0.842)·21/3)² = 491.8

Round up

n = 492

Answer

n = 492
11.Linear regression & correlation

Answer: r² = 0.935

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 = 132/6 = 22
ȳ = 530/6 = 88.33

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 256
Syy = Σ(y − ȳ)² = 1001

Coefficient of determination

r² = 256²/(70·1001) = 0.935

Answer

r² = 0.935

About 93.5% of the variation in y is explained by the line.

12.Linear regression & correlation

Answer: ŷ = 10.94

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 = 42/6 = 7
ȳ = 126/6 = 21

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 176

Regression line

b = 176/70 = 2.514
a = 21 − 2.514·7 = 3.4
ŷ = 3.4 + 2.514·x

Prediction

ŷ = 3.4 + 2.514·3 = 10.94

Answer

ŷ = 10.94

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