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: 33, 39, 13, 37, 38.

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

Exam scores are normally distributed with mean 60 and standard deviation 5. Find the z-score z of a score of 70.

μ = 60σ = 5x = 70
z =
3.ProbabilityMedium
xp(x)00.1510.2520.130.340.2μ = 2.15

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

x: 0, 1, 2, 3, 4p(x): 0.15, 0.25, 0.1, 0.3, 0.2
σ =
4.ProbabilityHard
xp(x)10.220.0530.340.150.35μ

X has the distribution shown, and Y = 2X + 20. Find the standard deviation σY of Y.

x: 1, 2, 3, 4, 5p(x): 0.2, 0.05, 0.3, 0.1, 0.35Y = 2X + 20
σY =
5.Normal distributionMedium
480 g475.8 gT = X₁ + … + X₁₂?

The masses of bolts are independent and normal with mean 40 g and standard deviation 1.2 g. Find the probability P that the total mass of 12 of them exceeds 475.8 g. Round z to two decimals.

μ = 40 gσ = 1.2 gn = 12> 475.8 g
P =
6.Normal distributionHard
35455565758595pointsbottom 10%

The exam scores are normally distributed with mean 65 points and standard deviation 10 points. Drag the cut-off so the shaded bottom 10% is to the left of it. Find the cut-off x.

μ = 65 pointsσ = 10 points10%
x = points
7.Confidence intervalsMedium
x̄E₁ = 690%, n = 25x̄E₂ = ?90%, n = 100

A 90% confidence interval for a mean, based on 25 observations and a known σ, has a margin of error of 6. What is the margin of error E₂ if the same study uses 100 observations instead?

90%n₁ = 25E₁ = 6n₂ = 100
E₂ =
8.Confidence intervalsHard
n = 11Lσ = ?s = 1.795%

The diameters of 11 machined shafts have a sample standard deviation of 1.7 µm. Assuming the diameters are normally distributed, find the upper limit σ of the 95% confidence interval for the population standard deviation. With df = 10, 2.5% of the χ² distribution lies below 3.247 and 2.5% lies above 20.483.

n = 11s = 1.7 µmχ² = 3.247, 20.483
σ = µm
9.Hypothesis tests & confidence intervalsMedium
−1.961.96z = ?

Two production lines make the same part. Line 1: a sample of 40 parts has mean length 53 mm, and the line's known standard deviation is 10 mm. Line 2: a sample of 50 parts has mean 49.9 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 = 40x̄₁ = 53 mmσ₁ = 10 mmn₂ = 50x̄₂ = 49.9 mmσ₂ = 11 mm
z =
10.Hypothesis tests & confidence intervalsHard
z −1.9601.96

A sample of 16 measurements has mean 499.8. The population standard deviation is 4. Find the lower limit L of the 95% confidence interval for the mean.

x̄ = 499.8σ = 4n = 1695%
L =
11.Linear regression & correlationMedium
xy350462568666777xy4650607080

The table gives 5 paired observations of x = advertising spend in k€ and y = sales in k€. Find Pearson's correlation coefficient r.

n = 5
r =
12.Linear regression & correlationHard
xy126226322421520619xy2462025ŷ = 21

The table gives 6 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 = 21?

n = 6ŷ = 21
x =

3Solutions

1.Descriptive statistics

Answer: s = 10.86

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 = 160/5 = 32

Squared deviations

Σ(x − x̄)² = 1² + 7² + (−19)² + 5² + 6² = 472

Standard deviation

s = √(472/4) = 10.86

Answer

s = 10.86
2.Descriptive statistics

Answer: z = 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 = (70 − 60)/5 = 2

Answer

z = 2
3.Probability

Answer: σ = 1.388

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 = 0·0.15 + 1·0.25 + 2·0.1 + 3·0.3 + 4·0.2 = 2.15

Mean of the squares

E(X²) = 0·0.15 + 1·0.25 + 4·0.1 + 9·0.3 + 16·0.2 = 6.55

Variance

Var(X) = 6.55 − 2.15² = 1.928
σ = √1.928 = 1.388

Answer

σ = 1.388
4.Probability

Answer: σY = 2.985

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.

Spread of X

μ = 3.35, Var(X) = 2.227, σX = 1.492

Linear change

σY = |2|·1.492 = 2.985

Answer

σY = 2.985

The mean does move: E(Y) = 2·3.35 + 20 = 26.7.

5.Normal distribution

Answer: P = 0.8438

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 total

μT = 480 g, σT = 1.2·√12 = 4.157 g

Standardise

z = (475.8 − 480)/4.157 = −1.01

Probability

P = 1 − Φ(−1.01) = 0.8438

Answer

P = 0.8438
6.Normal distribution

Answer: x = 52.2 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) = 0.1

z from the table

z ≈ −1.28

Back to the original units

x = 65 − 1.28·10 = 52.2 points

Answer

x = 52.2 points
7.Confidence intervals

Answer: E₂ = 3

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 = (1.645/1.645)·√(25/100) = 0.5

New margin

E₂ = 6·0.5 = 3

Answer

E₂ = 3
8.Confidence intervals

Answer: σ = 2.983 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 = 10·1.7² = 28.9

Limits for σ²

28.9/20.483 = 1.411
28.9/3.247 = 8.901

Limit for σ

σ = √8.901 = 2.983 µm

Answer

σ = 2.983 µm

95% interval for σ: [1.188, 2.983] µm.

9.Hypothesis tests & confidence intervals

Answer: z = 1.398

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 = √(10²/40 + 11²/50) = √4.92 = 2.218

Test statistic

z = (53 − 49.9)/2.218 = 1.398

Answer

z = 1.398

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

10.Hypothesis tests & confidence intervals

Answer: L = 497.8

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 = 1.96·4/√16 = 1.96

Interval limit

L = 499.8 − 1.96 = 497.8

Answer

L = 497.8
11.Linear regression & correlation

Answer: r = 0.9321

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 = 323/5 = 64.6

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = 58
Syy = Σ(y − ȳ)² = 387.2

Correlation coefficient

r = 58/√(10·387.2) = 0.9321

Answer

r = 0.9321
12.Linear regression & correlation

Answer: x = 4.364

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 = 134/6 = 22.33

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −27

Regression line

b = −27/17.5 = −1.543
a = 22.33 − (−1.543)·3.5 = 27.73
ŷ = 27.73 − 1.543·x

Solve for x

x = (21 − 27.73)/(−1.543) = 4.364

Answer

x = 4.364

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