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
1520253035404550

Find the sample standard deviation s of the data: 51, 12, 52, 39, 31, 46.

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

Exam scores are normally distributed with mean 50 and standard deviation 20. Find the z-score z of a score of 62.

μ = 50σ = 20x = 62
z =
3.ProbabilityMedium
red: 8blue: 6

A bag holds 8 red and 6 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?

8 / 6
P =
4.ProbabilityHard
n = 11, p = 0.5: P(X ≤ 1) = ?k01234567891011

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

n = 11p = 0.5
P =
5.Normal distributionMedium
−3−2−10123zshaded area to the right of z = 0.0853

Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.0853. Find z to two decimals.

A = 0.0853
z =
6.Normal distributionHard
μ = 178x = ?90%

The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. Find the value x below which 90% of them lie (the 90th percentile).

μ = 178σ = 7P = 90%
x = cm
7.Confidence intervalsMedium
x̄E₁ = 1.790%, n = 25x̄E₂ = ?99%, n = 25

A 90% confidence interval for a mean, based on 25 observations and a known σ, has a margin of error of 1.7. What is the margin of error E₂ of a 99% interval from the same data?

90%n₁ = 25E₁ = 1.799%
E₂ =
8.Confidence intervalsHard
μ₁ − μ₂, t(24) = 1.7110LU = ?x̄₁ − x̄₂E90%

Two fertilisers are tested. Plot group 1: 5 plots, mean yield 61 kg, standard deviation 9.5 kg. Plot group 2: 21 plots, mean 51.5 kg, standard deviation 10 kg. Assume equal variances (pooled t, df = n₁ + n₂ − 2). Find the upper limit of the 90% confidence interval for μ₁ − μ₂.

n₁ = 5x̄₁ = 61s₁ = 9.5n₂ = 21x̄₂ = 51.5s₂ = 1090%t(24) = 1.711
upper limit = kg
9.Hypothesis tests & confidence intervalsMedium
−2.4472.447t = ?df = 6

7 employees take a typing test before and after a training course. For each person the difference d = after − before (words per minute) is recorded. The mean difference is 11.64 and the standard deviation of the differences is 11. Find the paired test statistic t for H₀: μd = 0.

n = 7d̄ = 11.64sd = 11
t =
10.Hypothesis tests & confidence intervalsHard
σ = 12, n = ?x̄ − Ex̄ + Ex̄E = 2.599%

How large must a sample be so that the 99% confidence interval for a mean has a margin of error of at most 2.5? The standard deviation is 12. Find the smallest sample size n.

σ = 12E = 2.599%
n =
11.Linear regression & correlationMedium
xy345450552662765xy465060

The table gives 5 paired observations of x = advertising spend in k€ and y = sales in k€. Find the coefficient of determination r² of the least-squares line.

n = 5
r² =
12.Linear regression & correlationHard
xy2841762582910381246xy5102040x = 11

The table gives 6 paired observations of x = load in kN and y = extension in mm. Use the least-squares line ŷ = a + b·x. A new observation (11, 46) is made. Find its residual e = y − ŷ.

n = 6(11, 46)
e =

3Solutions

1.Descriptive statistics

Answer: s = 15.19

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 = 231/6 = 38.5

Squared deviations

Σ(x − x̄)² = 12.5² + (−26.5)² + 13.5² + 0.5² + (−7.5)² + 7.5² = 1154

Standard deviation

s = √(1154/5) = 15.19

Answer

s = 15.19
2.Descriptive statistics

Answer: z = 0.6

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 = (62 − 50)/20 = 0.6

Answer

z = 0.6
3.Probability

Answer: P = 0.3077

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.

First and second draw

P = 8/14 · 7/13 = 56/182 = 0.3077

Answer

P = 0.3077
4.Probability

Answer: P = 0.005859

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) = 0.511 = 0.0004883
P(X = 1) = 11·0.5·0.510 = 0.005371

Add them

P = 0.0004883 + 0.005371 = 0.005859

Answer

P = 0.005859
5.Normal distribution

Answer: z = 1.37

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 of z

Φ(z) = 1 − 0.0853 = 0.9147

Read the z-table backwards

Find 0.9147 inside the table and read off z.

z = 1.37

Answer

z = 1.37
6.Normal distribution

Answer: x = 187 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) = 0.9 → z = 1.282

Back to x

x = 178 + 1.282·7 = 186.97 cm

Answer

x = 186.97 cm
7.Confidence intervals

Answer: E₂ = 2.662

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 = (2.576/1.645)·√(25/25) = 1.566

New margin

E₂ = 1.7·1.566 = 2.662

Answer

E₂ = 2.662
8.Confidence intervals

Answer: upper limit = 17.94 kg

Interval for a difference: estimate ± critical value × standard error. For two independent means with known σ: SE = √(σ₁²/n₁ + σ₂²/n₂) and z. With equal unknown variances: pooled s_p and t with n₁ + n₂ − 2 degrees of freedom. For paired data, work with the differences: d̄ ± t·s_d/√n.

Pooled standard deviation

sp² = (4·9.5² + 20·10²)/24 = 98.38
sp = 9.918

Margin of error

E = 1.711·9.918·√(1/5 + 1/21) = 8.445

Interval

9.5 ± 8.445 → [1.055, 17.94]

Answer

upper limit = 17.94 kg
9.Hypothesis tests & confidence intervals

Answer: t = 2.8

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).

Standard error of the mean difference

SE = 11/√7 = 4.158

Test statistic

t = 11.64/4.158 = 2.8

Answer

t = 2.8

With df = 6, the two-sided 5% critical value is 2.447; |t| is larger, so H₀ is rejected.

10.Hypothesis tests & confidence intervals

Answer: n = 153

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 = (2.576·12/2.5)² = 152.9

Round up

n = 153

Answer

n = 153
11.Linear regression & correlation

Answer: r² = 0.9562

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 = 274/5 = 54.8

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = 52
Syy = Σ(y − ȳ)² = 282.8

Coefficient of determination

r² = 52²/(10·282.8) = 0.9562

Answer

r² = 0.9562

About 95.62% of the variation in y is explained by the line.

12.Linear regression & correlation

Answer: e = 4.148

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 = 42/6 = 7
ȳ = 163/6 = 27.17

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 257

Regression line

b = 257/70 = 3.671
a = 27.17 − 3.671·7 = 1.467
ŷ = 1.467 + 3.671·x

Predicted value

ŷ = 1.467 + 3.671·11 = 41.85

Residual

e = 46 − 41.85 = 4.148

Answer

e = 4.148

The point lies above the line.

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