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
20253035404550

Find the sample standard deviation s of the data: 23, 41, 23, 37, 19, 53.

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

A quantity is normally distributed with mean 170 and standard deviation 20. What proportion P of values is below 146?

μ = 170σ = 20x = 146
P =
3.ProbabilityMedium
xp(x)10.220.1530.1540.350.2μ = 3.15

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

x: 1, 2, 3, 4, 5p(x): 0.2, 0.15, 0.15, 0.3, 0.2
σ =
4.ProbabilityHard
n = 7, p = 0.5: P(X ≤ 1) = ?k01234567

7 components are tested, and each one fails with probability 0.5, independently. What is the probability P that at most one fails?

n = 7p = 0.5
P =
5.Normal distributionMedium
45.573P = ?

The exam scores are normally distributed with mean 65 points and standard deviation 10 points. What fraction lies between 45.5 and 73 points? Give P as a decimal.

μ = 65σ = 10x1 = 45.5x2 = 73
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 = 12, s = 18L = ?Ux̄ = 96.5E99%

A sample of 12 items has mean 96.5 and standard deviation s = 18. The population is roughly normal. Find the lower limit of the 99% confidence interval for the mean.

n = 12x̄ = 96.5s = 1899%t(11) = 3.106
lower limit =
8.Confidence intervalsHard
σ = 17, n = ?x̄ − Ex̄ + Ex̄E = 399%

The population standard deviation is 17. How large a sample n do you need so that a 99% confidence interval for the mean has a margin of error of at most 3?

σ = 17E = 399%
n =
9.Hypothesis tests & confidence intervalsMedium
−1.42p = ?

The mean of a population is claimed to be 500; the population standard deviation is known to be 18. A random sample of 144 observations has a mean of 497.87. Test H₀: μ = 500 against H₁: μ < 500. Find the p-value p.

μ₀ = 500σ = 18n = 144x̄ = 497.87H₁: μ < μ₀
p =
10.Hypothesis tests & confidence intervalsHard
μ₀x̄c

A one-sided z test of H₀: μ = 200 against H₁: μ < 200 is done at α = 0.05 with a sample of 25; the standard deviation is known to be 10. In reality the mean is 197.6. Find the power 1 − β of the test in percent.

μ₀ = 200μ = 197.6σ = 10n = 25α = 0.05
1 − β = %
11.Linear regression & correlationMedium
xy264861181310161218xy5101015

The table gives 6 paired observations of x = load in kN and y = extension in mm. Find Pearson's correlation coefficient r.

n = 6
r =
12.Linear regression & correlationHard
xy2374406528571065xy510405060ŷ = 61

The table gives 5 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 = 61?

n = 5ŷ = 61
x =

3Solutions

1.Descriptive statistics

Answer: s = 13.23

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 = 196/6 = 32.67

Squared deviations

Σ(x − x̄)² = (−9.667)² + 8.333² + (−9.667)² + 4.333² + (−13.67)² + 20.33² = 875.3

Standard deviation

s = √(875.3/5) = 13.23

Answer

s = 13.23
2.Descriptive statistics

Answer: P = 0.1151

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 = −1.2

Standard normal table

Φ(−1.2) = 0.1151
P = 0.1151

Answer

P = 0.1151
3.Probability

Answer: σ = 1.424

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 = 1·0.2 + 2·0.15 + 3·0.15 + 4·0.3 + 5·0.2 = 3.15

Mean of the squares

E(X²) = 1·0.2 + 4·0.15 + 9·0.15 + 16·0.3 + 25·0.2 = 11.95

Variance

Var(X) = 11.95 − 3.15² = 2.027
σ = √2.027 = 1.424

Answer

σ = 1.424
4.Probability

Answer: P = 0.0625

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) = 0.57 = 0.007813
P(X = 1) = 7·0.5·0.56 = 0.05469

Add them

P = 0.007813 + 0.05469 = 0.0625

Answer

P = 0.0625
5.Normal distribution

Answer: P = 0.7626

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 = (45.5 − 65)/10 = −1.95
z₂ = (73 − 65)/10 = 0.8

Area between

P = Φ(0.8) − Φ(−1.95) = 0.7881 − 0.02559 = 0.7626

Answer

P = 0.7626
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: lower limit = 80.36

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 = 11, t = 3.106

Margin of error

E = 3.106·18/√12 = 16.14

Interval

96.5 ± 16.14 → [80.36, 112.6]

Answer

lower limit = 80.36
8.Confidence intervals

Answer: n = 214

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 = (2.576·17/3)² = 213.1

Round up

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

n = 214

Answer

n = 214
9.Hypothesis tests & confidence intervals

Answer: p = 0.0778

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 = 18/√144 = 1.5
z = (497.87 − 500)/1.5 = −1.42

Tail area

Φ(1.42) = 0.9222
1 − Φ(1.42) = 0.0778

p-value

p = 0.0778

Answer

p = 0.0778

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

10.Hypothesis tests & confidence intervals

Answer: 1 − β = 32.82 %

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 = 10/√25 = 2
x̄c = 200 − 1.645·2 = 196.7

z under the true mean

z = (196.7 − 197.6)/2 = −0.445

Power

power = P(Z < −0.445) = 32.82%
1 − β = 32.82%

Answer

1 − β = 32.82%
11.Linear regression & correlation

Answer: r = 0.9984

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 = 72/6 = 12

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 86
Syy = Σ(y − ȳ)² = 106

Correlation coefficient

r = 86/√(70·106) = 0.9984

Answer

r = 0.9984
12.Linear regression & correlation

Answer: x = 8.959

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 = 251/5 = 50.2

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 146

Regression line

b = 146/40 = 3.65
a = 50.2 − 3.65·6 = 28.3
ŷ = 28.3 + 3.65·x

Solve for x

x = (61 − 28.3)/3.65 = 8.959

Answer

x = 8.959

ExamProblems · examproblems.com/courses/statistics-1