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: 22, 18, 37, 50, 25, 50.

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

Exam scores are normally distributed with mean 70 and standard deviation 8. Find the z-score z of a score of 54.

μ = 70σ = 8x = 54
z =
3.ProbabilityMedium
xp(x)10.220.1530.340.350.05μ = 2.85

For the distribution shown, find the variance Var(X) of X.

x: 1, 2, 3, 4, 5p(x): 0.2, 0.15, 0.3, 0.3, 0.05
Var(X) =
4.ProbabilityHard
n = 9, p = 0.5: P(X ≤ 1) = ?k0123456789

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

n = 9p = 0.5
P =
5.Normal distributionMedium
32 min35.8 minX + Y?

X is normal with mean 12 min and standard deviation 1.5 min; Y is independent and normal with mean 20 min and standard deviation 2 min. Find the probability P that X + Y is more than 35.8 min. Round z to two decimals.

μX = 12, σX = 1.5μY = 20, σY = 2X + Y > 35.8 min
P =
6.Normal distributionHard
0.02 mm0D − d?

Hole diameters D are normal with mean 25.02 mm and standard deviation 0.015 mm. Shaft diameters d are independent and normal with mean 25 mm and standard deviation 0.01 mm. A randomly chosen shaft fits when D − d > 0. Find the probability P that it fits. Round z to two decimals.

hole: μ = 25.02 mm, σ = 0.015 mmshaft: μ = 25 mm, σ = 0.01 mm
P =
7.Confidence intervalsMedium
x̄E₁ = 2.499%, n = 64x̄E₂ = ?99%, n = 576

A 99% confidence interval for a mean, based on 64 observations and a known σ, has a margin of error of 2.4. What is the margin of error E₂ if the same study uses 576 observations instead?

99%n₁ = 64E₁ = 2.4n₂ = 576
E₂ =
8.Confidence intervalsHard
σ = 27, n = ?x̄ − Ex̄ + Ex̄E = 499%

The population standard deviation is 27. 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 4?

σ = 27E = 499%
n =
9.Hypothesis tests & confidence intervalsMedium
2.38p = ?

The mean of a population is claimed to be 250; the population standard deviation is known to be 15. A random sample of 36 observations has a mean of 255.95. Test H₀: μ = 250 against H₁: μ > 250. Find the p-value p.

μ₀ = 250σ = 15n = 36x̄ = 255.95H₁: μ > μ₀
p =
10.Hypothesis tests & confidence intervalsHard
σ = 12, n = ?x̄ − Ex̄ + Ex̄E = 299%

How large must a sample be so that the 99% confidence interval for a mean has a margin of error of at most 2? The standard deviation is 12. Find the smallest sample size n.

σ = 12E = 299%
n =
11.Linear regression & correlationMedium
xy145242352449553xy244550

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
xy259363469580683xy246607080x = 3.5

The table gives 5 paired observations of x = hours studied and y = exam score. Use the least-squares line ŷ = a + b·x. A new observation (3.5, 63) is made. Find its residual e = y − ŷ.

n = 5(3.5, 63)
e =

3Solutions

1.Descriptive statistics

Answer: s = 14.15

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 = 202/6 = 33.67

Squared deviations

Σ(x − x̄)² = (−11.67)² + (−15.67)² + 3.333² + 16.33² + (−8.667)² + 16.33² = 1001

Standard deviation

s = √(1001/5) = 14.15

Answer

s = 14.15
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 = (54 − 70)/8 = −2

Answer

z = −2
3.Probability

Answer: Var(X) = 1.428

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 + 3·0.3 + 4·0.3 + 5·0.05 = 2.85

Mean of the squares

E(X²) = 1·0.2 + 4·0.15 + 9·0.3 + 16·0.3 + 25·0.05 = 9.55

Variance

Var(X) = 9.55 − 2.85² = 1.428

Answer

Var(X) = 1.428
4.Probability

Answer: P = 0.01953

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) = 0.59 = 0.001953
P(X = 1) = 9·0.5·0.58 = 0.01758

Add them

P = 0.001953 + 0.01758 = 0.01953

Answer

P = 0.01953
5.Normal distribution

Answer: P = 0.06426

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 sum

μ = 32 min, σ = √(1.5² + 2²) = 2.5 min

Standardise

z = (35.8 − 32)/2.5 = 1.52

Upper tail

P = 1 − Φ(1.52) = 1 − 0.9357 = 0.06426

Answer

P = 0.06426
6.Normal distribution

Answer: P = 0.8665

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 = 25.02 − 25 = 0.02 mm, σ = 0.01803 mm

Standardise

z = (0 − 0.02)/0.01803 = −1.11

Probability

P = 1 − Φ(−1.11) = 0.8665

Answer

P = 0.8665
7.Confidence intervals

Answer: E₂ = 0.8

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 = (2.576/2.576)·√(64/576) = 0.3333

New margin

E₂ = 2.4·0.3333 = 0.8

Answer

E₂ = 0.8
8.Confidence intervals

Answer: n = 303

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 = (2.576·27/4)² = 302.3

Round up

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

n = 303

Answer

n = 303
9.Hypothesis tests & confidence intervals

Answer: p = 0.008656

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 = 15/√36 = 2.5
z = (255.95 − 250)/2.5 = 2.38

Tail area

Φ(2.38) = 0.9913
1 − Φ(2.38) = 0.0087

p-value

p = 0.008656

Answer

p = 0.008656

p ≤ 0.05, so H₀ is rejected at the 5% level.

10.Hypothesis tests & confidence intervals

Answer: n = 239

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 = (2.576·12/2)² = 238.9

Round up

n = 239

Answer

n = 239
11.Linear regression & correlation

Answer: r = 0.7807

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 = 241/5 = 48.2

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = 23
Syy = Σ(y − ȳ)² = 86.8

Correlation coefficient

r = 23/√(10·86.8) = 0.7807

Answer

r = 0.7807
12.Linear regression & correlation

Answer: e = −4.55

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 = 20/5 = 4
ȳ = 354/5 = 70.8

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = 65

Regression line

b = 65/10 = 6.5
a = 70.8 − 6.5·4 = 44.8
ŷ = 44.8 + 6.5·x

Predicted value

ŷ = 44.8 + 6.5·3.5 = 67.55

Residual

e = 63 − 67.55 = −4.55

Answer

e = −4.55

The point lies below the line.

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