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
1520253035

Find the sample standard deviation s of the data: 29, 36, 30, 12, 37.

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

A quantity is normally distributed with mean 50 and standard deviation 10. What proportion P of values is above 35?

μ = 50σ = 10x = 35
P =
3.ProbabilityMedium
xp(x)10.320.130.2540.35μ = 2.65

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

x: 1, 2, 3, 4p(x): 0.3, 0.1, 0.25, 0.35
σ =
4.ProbabilityHard
n = 8, p = 0.7: P(X ≤ 1) = ?k012345678

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

n = 8p = 0.7
P =
5.Normal distributionMedium
245.25256P = ?

The bag weights are normally distributed with mean 250 g and standard deviation 5 g. What fraction lies between 245.25 and 256 g? Give P as a decimal.

μ = 250σ = 5x1 = 245.25x2 = 256
P =
6.Normal distributionHard
μ = 178x = ?5%

The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. Find the value x below which 5% of them lie (the 5th percentile).

μ = 178σ = 7P = 5%
x = cm
7.Confidence intervalsMedium
x̄E₁ = 4.395%, n = 36x̄E₂ = ?90%, n = 9

A 95% confidence interval for a mean, based on 36 observations and a known σ, has a margin of error of 4.3. What is the margin of error E₂ of a 90% interval based on 9 observations?

95%n₁ = 36E₁ = 4.390%n₂ = 9
E₂ =
8.Confidence intervalsHard
n = 30σ = ?Us = 5.295%

The diameters of 30 machined shafts have a sample standard deviation of 5.2 µm. Assuming the diameters are normally distributed, find the lower limit σ of the 95% confidence interval for the population standard deviation. With df = 29, 2.5% of the χ² distribution lies below 16.047 and 2.5% lies above 45.722.

n = 30s = 5.2 µmχ² = 16.047, 45.722
σ = µm
9.Hypothesis tests & confidence intervalsMedium
−1.961.96z = ?

Two production lines make the same part. Line 1: a sample of 60 parts has mean length 115.5 mm, and the line's known standard deviation is 14 mm. Line 2: a sample of 50 parts has mean 118.8 mm, known standard deviation 11 mm. Find the test statistic z for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (line 1 minus line 2).

n₁ = 60x̄₁ = 115.5 mmσ₁ = 14 mmn₂ = 50x̄₂ = 118.8 mmσ₂ = 11 mm
z =
10.Hypothesis tests & confidence intervalsHard
σ = 15, n = ?x̄ − Ex̄ + Ex̄E = 395%

How large must a sample be so that the 95% confidence interval for a mean has a margin of error of at most 3? The standard deviation is 15. Find the smallest sample size n.

σ = 15E = 395%
n =
11.Linear regression & correlationMedium
xy15481754196621662372xy1520506070

The table gives 5 paired observations of x = temperature in °C and y = ice creams sold. Find the coefficient of determination r² of the least-squares line.

n = 5
r² =
12.Linear regression & correlationHard
xy123224319418516xy2420ŷ = 23

The table gives 5 paired observations of x = car age in years and y = price in k€. Use the least-squares line ŷ = a + b·x. For which value x does the line predict ŷ = 23?

n = 5ŷ = 23
x =

3Solutions

1.Descriptive statistics

Answer: s = 10.03

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 = 144/5 = 28.8

Squared deviations

Σ(x − x̄)² = 0.2² + 7.2² + 1.2² + (−16.8)² + 8.2² = 402.8

Standard deviation

s = √(402.8/4) = 10.03

Answer

s = 10.03
2.Descriptive statistics

Answer: P = 0.9332

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 = −1.5

Standard normal table

Φ(−1.5) = 0.06681
P = 1 − 0.06681 = 0.9332

Answer

P = 0.9332
3.Probability

Answer: σ = 1.236

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 = 1·0.3 + 2·0.1 + 3·0.25 + 4·0.35 = 2.65

Mean of the squares

E(X²) = 1·0.3 + 4·0.1 + 9·0.25 + 16·0.35 = 8.55

Variance

Var(X) = 8.55 − 2.65² = 1.528
σ = √1.528 = 1.236

Answer

σ = 1.236
4.Probability

Answer: P = 0.00129

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) = 0.38 = 6.561 × 10⁻⁵
P(X = 1) = 8·0.7·0.37 = 0.001225

Add them

P = 6.561 × 10⁻⁵ + 0.001225 = 0.00129

Answer

P = 0.00129
5.Normal distribution

Answer: P = 0.7139

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 = (245.25 − 250)/5 = −0.95
z₂ = (256 − 250)/5 = 1.2

Area between

P = Φ(1.2) − Φ(−0.95) = 0.8849 − 0.1711 = 0.7139

Answer

P = 0.7139
6.Normal distribution

Answer: x = 166.5 cm

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 = 178 + −1.645·7 = 166.49 cm

Answer

x = 166.49 cm
7.Confidence intervals

Answer: E₂ = 7.218

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 = (1.645/1.96)·√(36/9) = 1.679

New margin

E₂ = 4.3·1.679 = 7.218

Answer

E₂ = 7.218
8.Confidence intervals

Answer: σ = 4.141 µm

Interval for a variance: (n − 1)s²/σ² has a χ² distribution with n − 1 degrees of freedom. The 95% interval for σ² runs from (n − 1)s² divided by the upper χ² value (2.5% above it) to (n − 1)s² divided by the lower χ² value (2.5% below it); the interval for σ is the square root of both limits. It is not symmetric around s.

Sum of squares

(n − 1)s² = 29·5.2² = 784.2

Limits for σ²

784.2/45.722 = 17.15
784.2/16.047 = 48.87

Limit for σ

σ = √17.15 = 4.141 µm

Answer

σ = 4.141 µm

95% interval for σ: [4.141, 6.99] µm.

9.Hypothesis tests & confidence intervals

Answer: z = −1.384

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.

Standard error of the difference

SE = √(14²/60 + 11²/50) = √5.687 = 2.385

Test statistic

z = (115.5 − 118.8)/2.385 = −1.384

Answer

z = −1.384

|z| < 1.96, so H₀ is not rejected at the 5% level (two-sided).

10.Hypothesis tests & confidence intervals

Answer: n = 97

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 = (1.96·15/3)² = 96.04

Round up

n = 97

Answer

n = 97
11.Linear regression & correlation

Answer: r² = 0.9259

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 = 95/5 = 19
ȳ = 306/5 = 61.2

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 120
Syy = Σ(y − ȳ)² = 388.8

Coefficient of determination

r² = 120²/(40·388.8) = 0.9259

Answer

r² = 0.9259

About 92.59% of the variation in y is explained by the line.

12.Linear regression & correlation

Answer: x = 1.5

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 = 15/5 = 3
ȳ = 100/5 = 20

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = −20

Regression line

b = −20/10 = −2
a = 20 − (−2)·3 = 26
ŷ = 26 − 2·x

Solve for x

x = (23 − 26)/(−2) = 1.5

Answer

x = 1.5

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