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
20253035404550

Find the sample standard deviation s of the data: 49, 23, 37, 31, 17, 51.

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

A quantity is normally distributed with mean 170 and standard deviation 10. What proportion P of values is above 190?

μ = 170σ = 10x = 190
P =
3.ProbabilityMedium
n = 9k = 3N = ?

In how many ways N can a committee of 3 people be chosen from a group of 9? The order does not matter.

n = 9k = 3
N =
4.ProbabilityHard
n = 12, p = 0.25: P(X ≤ 1) = ?k0123456789101112

12 components are tested, and each one fails with probability 0.25, independently. What is the probability P that at most one fails?

n = 12p = 0.25
P =
5.Normal distributionMedium
490574P = ?

The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. What fraction lies between 490 and 574 h? Give P as a decimal.

μ = 500σ = 40x1 = 490x2 = 574
P =
6.Normal distributionHard
0.04 mm0D − d?

Hole diameters D are normal with mean 25.04 mm and standard deviation 0.02 mm. Shaft diameters d are independent and normal with mean 25 mm and standard deviation 0.02 mm. A randomly chosen shaft fits when D − d > 0. Find the probability P that it fits. Round z to two decimals.

hole: μ = 25.04 mm, σ = 0.02 mmshaft: μ = 25 mm, σ = 0.02 mm
P =
7.Confidence intervalsMedium
n = 6, s = 15.5L = ?Ux̄ = 84.5E99%

A sample of 6 items has mean 84.5 and standard deviation s = 15.5. The population is roughly normal. Find the lower limit of the 99% confidence interval for the mean.

n = 6x̄ = 84.5s = 15.599%t(5) = 4.032
lower limit =
8.Confidence intervalsHard
σ = 29, n = ?x̄ − Ex̄ + Ex̄E = 590%

The population standard deviation is 29. How large a sample n do you need so that a 90% confidence interval for the mean has a margin of error of at most 5?

σ = 29E = 590%
n =
9.Hypothesis tests & confidence intervalsMedium
−2.0932.093t = ?df = 19

Two teaching methods are compared. Group 1: 11 students, mean score 65, standard deviation 3. Group 2: 10 students, mean score 62.8, standard deviation 12. Assume equal population variances and use the pooled two-sample t test. Find the test statistic t for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (group 1 minus group 2).

n₁ = 11x̄₁ = 65s₁ = 3n₂ = 10x̄₂ = 62.8s₂ = 12
t =
10.Hypothesis tests & confidence intervalsHard
z −2.57602.576

A sample of 64 measurements has mean 48.7. The population standard deviation is 4. Find the lower limit L of the 99% confidence interval for the mean.

x̄ = 48.7σ = 4n = 6499%
L =
11.Linear regression & correlationMedium
xy363467571676788xy46708090

The table gives 5 paired observations of x = hours studied and y = exam score. Find the coefficient of determination r² of the least-squares line.

n = 5
r² =
12.Linear regression & correlationHard
xy1263295357469511148xy510304050ŷ = 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 = 13.71

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 = 208/6 = 34.67

Squared deviations

Σ(x − x̄)² = 14.33² + (−11.67)² + 2.333² + (−3.667)² + (−17.67)² + 16.33² = 939.3

Standard deviation

s = √(939.3/5) = 13.71

Answer

s = 13.71
2.Descriptive statistics

Answer: P = 0.02275

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 = 2

Standard normal table

Φ(2) = 0.9772
P = 1 − 0.9772 = 0.02275

Answer

P = 0.02275
3.Probability

Answer: N = 84

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.

Combinations

C(9, 3) = (9·8·7)/(3·2·1) = 84

Answer

N = 84
4.Probability

Answer: P = 0.1584

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) = 0.7512 = 0.03168
P(X = 1) = 12·0.25·0.7511 = 0.1267

Add them

P = 0.03168 + 0.1267 = 0.1584

Answer

P = 0.1584
5.Normal distribution

Answer: P = 0.5665

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 = (490 − 500)/40 = −0.25
z₂ = (574 − 500)/40 = 1.85

Area between

P = Φ(1.85) − Φ(−0.25) = 0.9678 − 0.4013 = 0.5665

Answer

P = 0.5665
6.Normal distribution

Answer: P = 0.9207

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 = 25.04 − 25 = 0.04 mm, σ = 0.02828 mm

Standardise

z = (0 − 0.04)/0.02828 = −1.41

Probability

P = 1 − Φ(−1.41) = 0.9207

Answer

P = 0.9207
7.Confidence intervals

Answer: lower limit = 58.99

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 = 5, t = 4.032

Margin of error

E = 4.032·15.5/√6 = 25.51

Interval

84.5 ± 25.51 → [58.99, 110]

Answer

lower limit = 58.99
8.Confidence intervals

Answer: n = 92

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 = (1.645·29/5)² = 91.03

Round up

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

n = 92

Answer

n = 92
9.Hypothesis tests & confidence intervals

Answer: t = 0.5895

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.

Pooled variance

sp² = (10·3² + 9·12²)/19 = 72.95
sp = 8.541

Standard error

SE = 8.541·√(1/11 + 1/10) = 3.732

Test statistic

t = (65 − 62.8)/3.732 = 0.5895

Answer

t = 0.5895

With df = 19, the two-sided 5% critical value is 2.093; |t| is smaller, so H₀ is not rejected.

10.Hypothesis tests & confidence intervals

Answer: L = 47.41

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).

Margin of error

E = 2.576·4/√64 = 1.288

Interval limit

L = 48.7 − 1.288 = 47.41

Answer

L = 47.41
11.Linear regression & correlation

Answer: r² = 0.9307

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 = 365/5 = 73

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = 59
Syy = Σ(y − ȳ)² = 374

Coefficient of determination

r² = 59²/(10·374) = 0.9307

Answer

r² = 0.9307

About 93.07% of the variation in y is explained by the line.

12.Linear regression & correlation

Answer: x = 6.312

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 = 36/6 = 6
ȳ = 235/6 = 39.17

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 187

Regression line

b = 187/70 = 2.671
a = 39.17 − 2.671·6 = 23.14
ŷ = 23.14 + 2.671·x

Solve for x

x = (40 − 23.14)/2.671 = 6.312

Answer

x = 6.312

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