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
121416182022

Find the sample standard deviation s of the data: 22, 19, 15, 21, 11.

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

A quantity is normally distributed with mean 50 and standard deviation 10. What proportion P of values is above 35?

μ = 50σ = 10x = 35
P =
3.ProbabilityMedium
xp(x)10.3520.130.240.2550.1μ = 2.65

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

x: 1, 2, 3, 4, 5p(x): 0.35, 0.1, 0.2, 0.25, 0.1
σ =
4.ProbabilityHard
n = 12, p = 0.6: P(X = 2) = ?k0123456789101112

A test has 12 multiple-choice questions. A student who guesses gets each one right with probability 0.6, independently. What is the probability P of exactly 2 correct answers?

n = 12p = 0.6k = 2
P =
5.Normal distributionMedium
−3−2−10123zshaded area to the right of z = 0.1271

Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.1271. Find z to two decimals.

A = 0.1271
z =
6.Normal distributionHard
557085100115130145pointsbottom 85%

The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. Drag the cut-off so the shaded bottom 85% is to the left of it. Find the cut-off x.

μ = 100 pointsσ = 15 points85%
x = points
7.Confidence intervalsMedium
μ₁ − μ₂0L = ?Ux̄₁ − x̄₂E95%

The battery life of two phone models is compared. Model 1: 40 phones, mean 86 h, known standard deviation 4 h. Model 2: 30 phones, mean 79.5 h, known standard deviation 16 h. Find the lower limit of the 95% confidence interval for μ₁ − μ₂.

n₁ = 40x̄₁ = 86 hσ₁ = 4 hn₂ = 30x̄₂ = 79.5 hσ₂ = 16 h95%
lower limit = h
8.Confidence intervalsHard
n = 20σ = ?Us = 4.895%

The diameters of 20 machined shafts have a sample standard deviation of 4.8 µm. Assuming the diameters are normally distributed, find the lower limit σ of the 95% confidence interval for the population standard deviation. With df = 19, 2.5% of the χ² distribution lies below 8.907 and 2.5% lies above 32.852.

n = 20s = 4.8 µmχ² = 8.907, 32.852
σ = µm
9.Hypothesis tests & confidence intervalsMedium
t −2.13102.131

A sample of 16 students has a mean score of 494.8 and a sample standard deviation of 8. Find the test statistic t for H₀: μ = 500.

μ₀ = 500x̄ = 494.8s = 8n = 16
t =
10.Hypothesis tests & confidence intervalsHard
H₀H₁μ₀x̄cμ₀ + 5α = 0.05βσ = 13, power 95%, n = ?

A one-sided z test at α = 0.05 must detect a shift of the mean of 5 units with a power of 95%. The standard deviation is 13. Find the minimum sample size n.

α = 0.05power = 95%δ = 5σ = 13zα = 1.645zβ = 1.645
n =
11.Linear regression & correlationMedium
xy264176248311035xy510102030

The table gives 5 paired observations of x = load in kN and y = extension in mm. Their sums are given. Find Pearson's correlation coefficient r from these sums.

n = 5Σx = 30Σy = 113Σx² = 220Σy² = 3087Σxy = 822
r =
12.Linear regression & correlationHard
xy343445558655762xy465060ŷ = 55

The table gives 5 paired observations of x = hours studied and y = exam score. Use the least-squares line ŷ = a + b·x. For which value x does the line predict ŷ = 55?

n = 5ŷ = 55
x =

3Solutions

1.Descriptive statistics

Answer: s = 4.561

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 = 88/5 = 17.6

Squared deviations

Σ(x − x̄)² = 4.4² + 1.4² + (−2.6)² + 3.4² + (−6.6)² = 83.2

Standard deviation

s = √(83.2/4) = 4.561

Answer

s = 4.561
2.Descriptive statistics

Answer: P = 0.9332

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 = −1.5

Standard normal table

Φ(−1.5) = 0.06681
P = 1 − 0.06681 = 0.9332

Answer

P = 0.9332
3.Probability

Answer: σ = 1.424

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 = 1·0.35 + 2·0.1 + 3·0.2 + 4·0.25 + 5·0.1 = 2.65

Mean of the squares

E(X²) = 1·0.35 + 4·0.1 + 9·0.2 + 16·0.25 + 25·0.1 = 9.05

Variance

Var(X) = 9.05 − 2.65² = 2.027
σ = √2.027 = 1.424

Answer

σ = 1.424
4.Probability

Answer: P = 0.002491

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) = 66·0.62·0.410 = 66·0.36·0.0001049 = 0.002491

Answer

P = 0.002491
5.Normal distribution

Answer: z = 1.14

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 − 0.1271 = 0.8729

Read the z-table backwards

Find 0.8729 inside the table and read off z.

z = 1.14

Answer

z = 1.14
6.Normal distribution

Answer: x = 115.6 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) = 0.85

z from the table

z ≈ 1.04

Back to the original units

x = 100 + 1.04·15 = 115.6 points

Answer

x = 115.6 points
7.Confidence intervals

Answer: lower limit = 0.6418 h

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.

Point estimate

x̄₁ − x̄₂ = 86 − 79.5 = 6.5

Standard error

SE = √(4²/40 + 16²/30) = 2.989

Margin of error

E = 1.96·2.989 = 5.858

Interval

6.5 ± 5.858 → [0.6418, 12.36]

Answer

lower limit = 0.6418 h

The interval does not contain 0, so the means differ significantly.

8.Confidence intervals

Answer: σ = 3.65 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 = 19·4.8² = 437.8

Limits for σ²

437.8/32.852 = 13.33
437.8/8.907 = 49.15

Limit for σ

σ = √13.33 = 3.65 µm

Answer

σ = 3.65 µm

95% interval for σ: [3.65, 7.011] µm.

9.Hypothesis tests & confidence intervals

Answer: t = −2.6

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 = 8/√16 = 8/4 = 2

Test statistic

t = (494.8 − 500)/2 = −2.6

Answer

t = −2.6
t0.025; df = 15 = 2.131

|t| > 2.131, so H₀ is rejected at the 5% level (two-sided).

10.Hypothesis tests & confidence intervals

Answer: n = 74

Sample size for a test: to detect a true shift δ with power 1 − β in a one-sided z test at level α, the standard error must satisfy δ = (z_α + z_β)·σ/√n, so n = ((z_α + z_β)·σ/δ)², rounded up.

Solve for n

n = ((1.645 + 1.645)·13/5)² = 73.17

Round up

n = 74

Answer

n = 74
11.Linear regression & correlation

Answer: r = 0.986

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.

Centred sums

Sxy = 822 − 30·113/5 = 144
Sxx = 220 − 30²/5 = 40
Syy = 3087 − 113²/5 = 533.2

Correlation coefficient

r = 144/√(40·533.2) = 0.986

Answer

r = 0.986
12.Linear regression & correlation

Answer: x = 5.5

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 = 263/5 = 52.6

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = 48

Regression line

b = 48/10 = 4.8
a = 52.6 − 4.8·5 = 28.6
ŷ = 28.6 + 4.8·x

Solve for x

x = (55 − 28.6)/4.8 = 5.5

Answer

x = 5.5

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