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
152025303540455055

Find the sample standard deviation s of the data: 56, 52, 41, 32, 15, 37.

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

Exam scores are normally distributed with mean 60 and standard deviation 12. Find the z-score z of a score of 45.6.

μ = 60σ = 12x = 45.6
z =
3.ProbabilityMedium
xp(x)00.310.3520.35μ = 1.05

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

x: 0, 1, 2p(x): 0.3, 0.35, 0.35
σ =
4.ProbabilityHard
n = 6, p = 0.4: P(X ≤ 1) = ?k0123456

6 components are tested, and each one fails with probability 0.4, independently. What is the probability P that at most one fails?

n = 6p = 0.4
P =
5.Normal distributionMedium
462520P = ?

The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. What fraction lies between 462 and 520 h? Give P as a decimal.

μ = 500σ = 40x1 = 462x2 = 520
P =
6.Normal distributionHard
μ = 100x = ?97.5%

The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. Find the value x below which 97.5% of them lie (the 97.5th percentile).

μ = 100σ = 15P = 97.5%
x = points
7.Confidence intervalsMedium
n = 16, s = 11.5LUx̄ = 61E = ?90%

A sample of 16 items has mean 61 and standard deviation s = 11.5. The population is roughly normal. Find the margin of error E of a 90% confidence interval for the mean.

n = 16x̄ = 61s = 11.590%t(15) = 1.753
E =
8.Confidence intervalsHard
n = 25σ = ?Us = 6.395%

The diameters of 25 machined shafts have a sample standard deviation of 6.3 µm. Assuming the diameters are normally distributed, find the lower limit σ of the 95% confidence interval for the population standard deviation. With df = 24, 2.5% of the χ² distribution lies below 12.401 and 2.5% lies above 39.364.

n = 25s = 6.3 µmχ² = 12.401, 39.364
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 74.5 mm, and the line's known standard deviation is 12 mm. Line 2: a sample of 30 parts has mean 71 mm, known standard deviation 17 mm. Find the test statistic z for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (line 1 minus line 2).

n₁ = 60x̄₁ = 74.5 mmσ₁ = 12 mmn₂ = 30x̄₂ = 71 mmσ₂ = 17 mm
z =
10.Hypothesis tests & confidence intervalsHard
z −2.57602.576

A sample of 25 measurements has mean 496.2. The population standard deviation is 10. Find the upper limit U of the 99% confidence interval for the mean.

x̄ = 496.2σ = 10n = 2599%
U =
11.Linear regression & correlationMedium
xy124221317417513610xy24610152025

The table gives 6 paired observations of x = car age in years and y = price in k€. Find the coefficient of determination r² of the least-squares line.

n = 6
r² =
12.Linear regression & correlationHard
xy15381739194821422350xy1520404550x = 20

The table gives 5 paired observations of x = temperature in °C and y = ice creams sold. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 20.

n = 5x = 20
ŷ =

3Solutions

1.Descriptive statistics

Answer: s = 14.77

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 = 233/6 = 38.83

Squared deviations

Σ(x − x̄)² = 17.17² + 13.17² + 2.167² + (−6.833)² + (−23.83)² + (−1.833)² = 1091

Standard deviation

s = √(1091/5) = 14.77

Answer

s = 14.77
2.Descriptive statistics

Answer: z = −1.2

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 = (45.6 − 60)/12 = −1.2

Answer

z = −1.2
3.Probability

Answer: σ = 0.8047

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 = 0·0.3 + 1·0.35 + 2·0.35 = 1.05

Mean of the squares

E(X²) = 0·0.3 + 1·0.35 + 4·0.35 = 1.75

Variance

Var(X) = 1.75 − 1.05² = 0.6475
σ = √0.6475 = 0.8047

Answer

σ = 0.8047
4.Probability

Answer: P = 0.2333

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) = 0.66 = 0.04666
P(X = 1) = 6·0.4·0.65 = 0.1866

Add them

P = 0.04666 + 0.1866 = 0.2333

Answer

P = 0.2333
5.Normal distribution

Answer: P = 0.5204

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 = (462 − 500)/40 = −0.95
z₂ = (520 − 500)/40 = 0.5

Area between

P = Φ(0.5) − Φ(−0.95) = 0.6915 − 0.1711 = 0.5204

Answer

P = 0.5204
6.Normal distribution

Answer: x = 129.4 points

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) = 0.975 → z = 1.96

Back to x

x = 100 + 1.96·15 = 129.4 points

Answer

x = 129.4 points
7.Confidence intervals

Answer: E = 5.04

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 = 15, t = 1.753

Margin of error

E = 1.753·11.5/√16 = 5.04

Answer

E = 5.04
8.Confidence intervals

Answer: σ = 4.919 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 = 24·6.3² = 952.6

Limits for σ²

952.6/39.364 = 24.2
952.6/12.401 = 76.81

Limit for σ

σ = √24.2 = 4.919 µm

Answer

σ = 4.919 µm

95% interval for σ: [4.919, 8.764] µm.

9.Hypothesis tests & confidence intervals

Answer: z = 1.009

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 = √(12²/60 + 17²/30) = √12.03 = 3.469

Test statistic

z = (74.5 − 71)/3.469 = 1.009

Answer

z = 1.009

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

10.Hypothesis tests & confidence intervals

Answer: U = 501.4

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 = 2.576·10/√25 = 5.152

Interval limit

U = 496.2 + 5.152 = 501.4

Answer

U = 501.4
11.Linear regression & correlation

Answer: r² = 0.971

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 = 102/6 = 17

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −47
Syy = Σ(y − ȳ)² = 130

Coefficient of determination

r² = −47²/(17.5·130) = 0.971

Answer

r² = 0.971

About 97.1% of the variation in y is explained by the line.

12.Linear regression & correlation

Answer: ŷ = 44.75

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 = 217/5 = 43.4

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 54

Regression line

b = 54/40 = 1.35
a = 43.4 − 1.35·19 = 17.75
ŷ = 17.75 + 1.35·x

Prediction

ŷ = 17.75 + 1.35·20 = 44.75

Answer

ŷ = 44.75

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