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
222426283032343638

Find the sample standard deviation s of the data: 22, 39, 35, 21, 35.

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

Exam scores are normally distributed with mean 60 and standard deviation 8. Find the z-score z of a score of 70.

μ = 60σ = 8x = 70
z =
3.ProbabilityMedium
xp(x)00.3510.120.230.35μ = 1.55

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

x: 0, 1, 2, 3p(x): 0.35, 0.1, 0.2, 0.35
Var(X) =
4.ProbabilityHard
n = 6, p = 0.2: P(X = 2) = ?k0123456

A test has 6 multiple-choice questions. A student who guesses gets each one right with probability 0.2, independently. What is the probability P of exactly 2 correct answers?

n = 6p = 0.2k = 2
P =
5.Normal distributionMedium
−3−2−10123zshaded area between −z and z = 0.9051

Drag z so that the shaded area between −z and z under the standard normal curve is 0.9051. Find z to two decimals.

A = 0.9051
z =
6.Normal distributionHard
μ = 32x = ?5%

The delivery times are normally distributed with mean 32 min and standard deviation 6 min. Find the value x below which 5% of them lie (the 5th percentile).

μ = 32σ = 6P = 5%
x = min
7.Confidence intervalsMedium
x̄E₁ = 11.199%, n = 36x̄E₂ = ?90%, n = 36

A 99% confidence interval for a mean, based on 36 observations and a known σ, has a margin of error of 11.1. What is the margin of error E₂ of a 90% interval from the same data?

99%n₁ = 36E₁ = 11.190%
E₂ =
8.Confidence intervalsHard
μ₁ − μ₂, t(10) = 2.2280LU = ?x̄₁ − x̄₂E95%

Two fertilisers are tested. Plot group 1: 6 plots, mean yield 86.5 kg, standard deviation 4.5 kg. Plot group 2: 6 plots, mean 78 kg, standard deviation 7.5 kg. Assume equal variances (pooled t, df = n₁ + n₂ − 2). Find the upper limit of the 95% confidence interval for μ₁ − μ₂.

n₁ = 6x̄₁ = 86.5s₁ = 4.5n₂ = 6x̄₂ = 78s₂ = 7.595%t(10) = 2.228
upper limit = kg
9.Hypothesis tests & confidence intervalsMedium
−2.3652.365t = ?df = 7

8 employees take a typing test before and after a training course. For each person the difference d = after − before (words per minute) is recorded. The mean difference is 6.22 and the standard deviation of the differences is 5.5. Find the paired test statistic t for H₀: μd = 0.

n = 8d̄ = 6.22sd = 5.5
t =
10.Hypothesis tests & confidence intervalsHard
σ = 15, n = ?x̄ − Ex̄ + Ex̄E = 199%

How large must a sample be so that the 99% confidence interval for a mean has a margin of error of at most 1? The standard deviation is 15. Find the smallest sample size n.

σ = 15E = 199%
n =
11.Linear regression & correlationMedium
xy126234341448552653xy246304050

The table gives 6 paired observations of x = advertising spend in k€ and y = sales in k€. Find Pearson's correlation coefficient r.

n = 6
r =
12.Linear regression & correlationHard
xy245355457566670xy246506070x = 4.5

The table gives 5 paired observations of x = hours studied and y = exam score. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 4.5.

n = 5x = 4.5
ŷ =

3Solutions

1.Descriptive statistics

Answer: s = 8.295

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 = 152/5 = 30.4

Squared deviations

Σ(x − x̄)² = (−8.4)² + 8.6² + 4.6² + (−9.4)² + 4.6² = 275.2

Standard deviation

s = √(275.2/4) = 8.295

Answer

s = 8.295
2.Descriptive statistics

Answer: z = 1.25

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 = (70 − 60)/8 = 1.25

Answer

z = 1.25
3.Probability

Answer: Var(X) = 1.648

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 + 1·0.1 + 2·0.2 + 3·0.35 = 1.55

Mean of the squares

E(X²) = 0·0.35 + 1·0.1 + 4·0.2 + 9·0.35 = 4.05

Variance

Var(X) = 4.05 − 1.55² = 1.648

Answer

Var(X) = 1.648
4.Probability

Answer: P = 0.2458

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.

Binomial formula

P(X = 2) = 15·0.22·0.84 = 15·0.04·0.4096 = 0.2458

Answer

P = 0.2458
5.Normal distribution

Answer: z = 1.67

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 + 0.9051)/2 = 0.9525

Read the z-table backwards

Find 0.9525 inside the table and read off z.

z = 1.67

Answer

z = 1.67
6.Normal distribution

Answer: x = 22.13 min

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) = 0.05 → z = −1.645

Back to x

x = 32 + −1.645·6 = 22.13 min

Answer

x = 22.13 min
7.Confidence intervals

Answer: E₂ = 7.088

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)·√(36/36) = 0.6386

New margin

E₂ = 11.1·0.6386 = 7.088

Answer

E₂ = 7.088
8.Confidence intervals

Answer: upper limit = 16.46 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 = (5·4.5² + 5·7.5²)/10 = 38.25
sp = 6.185

Margin of error

E = 2.228·6.185·√(1/6 + 1/6) = 7.956

Interval

8.5 ± 7.956 → [0.5444, 16.46]

Answer

upper limit = 16.46 kg
9.Hypothesis tests & confidence intervals

Answer: t = 3.199

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 of the mean difference

SE = 5.5/√8 = 1.945

Test statistic

t = 6.22/1.945 = 3.199

Answer

t = 3.199

With df = 7, the two-sided 5% critical value is 2.365; |t| is larger, so H₀ is rejected.

10.Hypothesis tests & confidence intervals

Answer: n = 1494

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 = (2.576·15/1)² = 1493

Round up

n = 1494

Answer

n = 1494
11.Linear regression & correlation

Answer: r = 0.975

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 = 21/6 = 3.5
ȳ = 254/6 = 42.33

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = 98
Syy = Σ(y − ȳ)² = 577.3

Correlation coefficient

r = 98/√(17.5·577.3) = 0.975

Answer

r = 0.975
12.Linear regression & correlation

Answer: ŷ = 61.65

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 = 20/5 = 4
ȳ = 293/5 = 58.6

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = 61

Regression line

b = 61/10 = 6.1
a = 58.6 − 6.1·4 = 34.2
ŷ = 34.2 + 6.1·x

Prediction

ŷ = 34.2 + 6.1·4.5 = 61.65

Answer

ŷ = 61.65

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