Introductory Statistics: cram pack

Formula sheets, the key method steps for every topic and 12 practice problems, with full solutions at the back.

New problemsBack to the course

In the print window, choose “Save as PDF” to keep a copy on your device.

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: 13, 59, 34, 42, 25, 42.

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

A quantity is normally distributed with mean 60 and standard deviation 15. What proportion P of values is above 48?

μ = 60σ = 15x = 48
P =
3.ProbabilityMedium
red: 8blue: 5

A bag holds 8 red and 5 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?

8 / 5
P =
4.ProbabilityHard
n = 6, p = 0.7: P(X ≤ 1) = ?k0123456

6 components are tested, and each one fails with probability 0.7, independently. What is the probability P that at most one fails?

n = 6p = 0.7
P =
5.Normal distributionMedium
77.5130P = ?

The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. What fraction lies between 77.5 and 130 points? Give P as a decimal.

μ = 100σ = 15x1 = 77.5x2 = 130
P =
6.Normal distributionHard
557085100115130145pointstop 5%

The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. Drag the cut-off so the shaded top 5% is to the right of it. Find the cut-off x.

μ = 100 pointsσ = 15 points5%
x = points
7.Confidence intervalsMedium
x̄E₁ = 4.990%, n = 50x̄E₂ = ?99%, n = 100

A 90% confidence interval for a mean, based on 50 observations and a known σ, has a margin of error of 4.9. What is the margin of error E₂ of a 99% interval based on 100 observations?

90%n₁ = 50E₁ = 4.999%n₂ = 100
E₂ =
8.Confidence intervalsHard
μ₁ − μ₂, t(12) = 2.1790LU = ?x̄₁ − x̄₂E95%

Two fertilisers are tested. Plot group 1: 9 plots, mean yield 26 kg, standard deviation 9.5 kg. Plot group 2: 5 plots, mean 27.5 kg, standard deviation 4 kg. Assume equal variances (pooled t, df = n₁ + n₂ − 2). Find the upper limit of the 95% confidence interval for μ₁ − μ₂.

n₁ = 9x̄₁ = 26s₁ = 9.5n₂ = 5x̄₂ = 27.5s₂ = 495%t(12) = 2.179
upper limit = kg
9.Hypothesis tests & confidence intervalsMedium
−1.541.54p = ?

The mean of a population is claimed to be 250; the population standard deviation is known to be 48. A random sample of 144 observations has a mean of 256.16. Test H₀: μ = 250 against H₁: μ ≠ 250. Find the p-value p.

μ₀ = 250σ = 48n = 144x̄ = 256.16H₁: μ ≠ μ₀
p =
10.Hypothesis tests & confidence intervalsHard
σ = 8, n = ?x̄ − Ex̄ + Ex̄E = 399%

How large must a sample be so that the 99% 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 = 399%
n =
11.Linear regression & correlationMedium
xy350458560666776xy46506070

The table gives 5 paired observations of x = advertising spend in k€ and y = sales in k€. Find the coefficient of determination r² of the least-squares line.

n = 5
r² =
12.Linear regression & correlationHard
xy336438544645747846xy4684045ŷ = 40

The table gives 6 paired observations of x = advertising spend in k€ and y = sales in k€. Use the least-squares line ŷ = a + b·x. For which value x does the line predict ŷ = 40?

n = 6ŷ = 40
x =

3Solutions

1.Descriptive statistics

Answer: s = 15.84

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 = 215/6 = 35.83

Squared deviations

Σ(x − x̄)² = (−22.83)² + 23.17² + (−1.833)² + 6.167² + (−10.83)² + 6.167² = 1255

Standard deviation

s = √(1255/5) = 15.84

Answer

s = 15.84
2.Descriptive statistics

Answer: P = 0.7881

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 = −0.8

Standard normal table

Φ(−0.8) = 0.2119
P = 1 − 0.2119 = 0.7881

Answer

P = 0.7881
3.Probability

Answer: P = 0.359

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 = 8/13 · 7/12 = 56/156 = 0.359

Answer

P = 0.359
4.Probability

Answer: P = 0.01094

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.

Zero and one failure

P(X = 0) = 0.36 = 0.000729
P(X = 1) = 6·0.7·0.35 = 0.01021

Add them

P = 0.000729 + 0.01021 = 0.01094

Answer

P = 0.01094
5.Normal distribution

Answer: P = 0.9104

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-scores

z₁ = (77.5 − 100)/15 = −1.5
z₂ = (130 − 100)/15 = 2

Area between

P = Φ(2) − Φ(−1.5) = 0.9772 − 0.06681 = 0.9104

Answer

P = 0.9104
6.Normal distribution

Answer: x = 124.6 points

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) = 0.95

z from the table

z ≈ 1.64

Back to the original units

x = 100 + 1.64·15 = 124.6 points

Answer

x = 124.6 points
7.Confidence intervals

Answer: E₂ = 5.426

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 = (2.576/1.645)·√(50/100) = 1.107

New margin

E₂ = 4.9·1.107 = 5.426

Answer

E₂ = 5.426
8.Confidence intervals

Answer: upper limit = 8.336 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 = (8·9.5² + 4·4²)/12 = 65.5
sp = 8.093

Margin of error

E = 2.179·8.093·√(1/9 + 1/5) = 9.836

Interval

−1.5 ± 9.836 → [−11.34, 8.336]

Answer

upper limit = 8.336 kg
9.Hypothesis tests & confidence intervals

Answer: p = 0.1236

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 = 48/√144 = 4
z = (256.16 − 250)/4 = 1.54

Tail area

Φ(1.54) = 0.9382
1 − Φ(1.54) = 0.0618

p-value

p = 2·0.0618 = 0.1236

Answer

p = 0.1236

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

10.Hypothesis tests & confidence intervals

Answer: n = 48

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 = (2.576·8/3)² = 47.19

Round up

n = 48

Answer

n = 48
11.Linear regression & correlation

Answer: r² = 0.9574

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 = 25/5 = 5
ȳ = 310/5 = 62

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = 60
Syy = Σ(y − ȳ)² = 376

Coefficient of determination

r² = 60²/(10·376) = 0.9574

Answer

r² = 0.9574

About 95.74% of the variation in y is explained by the line.

12.Linear regression & correlation

Answer: x = 4.303

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 = 33/6 = 5.5
ȳ = 256/6 = 42.67

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = 39

Regression line

b = 39/17.5 = 2.229
a = 42.67 − 2.229·5.5 = 30.41
ŷ = 30.41 + 2.229·x

Solve for x

x = (40 − 30.41)/2.229 = 4.303

Answer

x = 4.303

ExamProblems · examproblems.com/courses/statistics-1