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
15202530354045

Find the sample standard deviation s of the data: 23, 39, 15, 31, 49.

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

Exam scores are normally distributed with mean 70 and standard deviation 12. Find the z-score z of a score of 85.

μ = 70σ = 12x = 85
z =
3.ProbabilityMedium
red: 5blue: 8

A bag holds 5 red and 8 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?

5 / 8
P =
4.ProbabilityHard
n = 10, p = 0.25: P(X = 2) = ?k012345678910

A test has 10 multiple-choice questions. A student who guesses gets each one right with probability 0.25, independently. What is the probability P of exactly 2 correct answers?

n = 10p = 0.25k = 2
P =
5.Normal distributionMedium
72 kg75.9 kgT = X₁ + … + X₆?

The masses of boxes are independent and normal with mean 12 kg and standard deviation 0.8 kg. Find the probability P that the total mass of 6 of them is below 75.9 kg. Round z to two decimals.

μ = 12 kgσ = 0.8 kgn = 6< 75.9 kg
P =
6.Normal distributionHard
μ = 178x = ?95%

The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. Find the value x below which 95% of them lie (the 95th percentile).

μ = 178σ = 7P = 95%
x = cm
7.Confidence intervalsMedium
n = 7, s = 10L = ?Ux̄ = 44E90%

A sample of 7 items has mean 44 and standard deviation s = 10. The population is roughly normal. Find the lower limit of the 90% confidence interval for the mean.

n = 7x̄ = 44s = 1090%t(6) = 1.943
lower limit =
8.Confidence intervalsHard
μ₁ − μ₂, t(15) = 2.1310LU = ?x̄₁ − x̄₂E95%

Two fertilisers are tested. Plot group 1: 9 plots, mean yield 33 kg, standard deviation 9.5 kg. Plot group 2: 8 plots, mean 32.5 kg, standard deviation 3 kg. Assume equal variances (pooled t, df = n₁ + n₂ − 2). Find the upper limit of the 95% confidence interval for μ₁ − μ₂.

n₁ = 9x̄₁ = 33s₁ = 9.5n₂ = 8x̄₂ = 32.5s₂ = 395%t(15) = 2.131
upper limit = kg
9.Hypothesis tests & confidence intervalsMedium
t −1.98401.984

A sample of 100 students has a mean score of 73.92 and a sample standard deviation of 12. Find the test statistic t for H₀: μ = 75.

μ₀ = 75x̄ = 73.92s = 12n = 100
t =
10.Hypothesis tests & confidence intervalsHard
z −2.57602.576

A sample of 49 measurements has mean 101.8. The population standard deviation is 5. Find the lower limit L of the 99% confidence interval for the mean.

x̄ = 101.8σ = 5n = 4999%
L =
11.Linear regression & correlationMedium
xy244146168261028xy510102030

The table gives 5 paired observations of x = load in kN and y = extension in mm. Find the coefficient of determination r² of the least-squares line.

n = 5
r² =
12.Linear regression & correlationHard
xy12421932041651469xy24610152025x = 3.5

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. Predict ŷ for x = 3.5.

n = 6x = 3.5
ŷ =

3Solutions

1.Descriptive statistics

Answer: s = 13.3

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 = 157/5 = 31.4

Squared deviations

Σ(x − x̄)² = (−8.4)² + 7.6² + (−16.4)² + (−0.4)² + 17.6² = 707.2

Standard deviation

s = √(707.2/4) = 13.3

Answer

s = 13.3
2.Descriptive statistics

Answer: z = 1.25

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 = (85 − 70)/12 = 1.25

Answer

z = 1.25
3.Probability

Answer: P = 0.1282

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 = 5/13 · 4/12 = 20/156 = 0.1282

Answer

P = 0.1282
4.Probability

Answer: P = 0.2816

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.

Binomial formula

P(X = 2) = 45·0.252·0.758 = 45·0.0625·0.1001 = 0.2816

Answer

P = 0.2816
5.Normal distribution

Answer: P = 0.9767

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 = 72 kg, σT = 0.8·√6 = 1.96 kg

Standardise

z = (75.9 − 72)/1.96 = 1.99

Probability

P = Φ(1.99) = 0.9767

Answer

P = 0.9767
6.Normal distribution

Answer: x = 189.5 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) = 0.95 → z = 1.645

Back to x

x = 178 + 1.645·7 = 189.51 cm

Answer

x = 189.51 cm
7.Confidence intervals

Answer: lower limit = 36.66

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 = 6, t = 1.943

Margin of error

E = 1.943·10/√7 = 7.344

Interval

44 ± 7.344 → [36.66, 51.34]

Answer

lower limit = 36.66
8.Confidence intervals

Answer: upper limit = 7.991 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 = (8·9.5² + 7·3²)/15 = 52.33
sp = 7.234

Margin of error

E = 2.131·7.234·√(1/9 + 1/8) = 7.491

Interval

0.5 ± 7.491 → [−6.991, 7.991]

Answer

upper limit = 7.991 kg
9.Hypothesis tests & confidence intervals

Answer: t = −0.9

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

SE = 12/√100 = 12/10 = 1.2

Test statistic

t = (73.92 − 75)/1.2 = −0.9

Answer

t = −0.9
t0.025; df = 99 = 1.984

|t| < 1.984, so H₀ is not rejected at the 5% level (two-sided).

10.Hypothesis tests & confidence intervals

Answer: L = 99.96

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 = 2.576·5/√49 = 1.84

Interval limit

L = 101.8 − 1.84 = 99.96

Answer

L = 99.96
11.Linear regression & correlation

Answer: r² = 0.9494

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 = 30/5 = 6
ȳ = 88/5 = 17.6

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 120
Syy = Σ(y − ȳ)² = 379.2

Coefficient of determination

r² = 120²/(40·379.2) = 0.9494

Answer

r² = 0.9494

About 94.94% of the variation in y is explained by the line.

12.Linear regression & correlation

Answer: ŷ = 17

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

Regression line

b = −47/17.5 = −2.686
a = 17 − (−2.686)·3.5 = 26.4
ŷ = 26.4 − 2.686·x

Prediction

ŷ = 26.4 + (−2.686)·3.5 = 17

Answer

ŷ = 17

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