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: 48, 42, 36, 35, 12.

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

A quantity is normally distributed with mean 500 and standard deviation 50. What proportion P of values is above 600?

μ = 500σ = 50x = 600
P =
3.ProbabilityMedium
red: 3blue: 8

A bag holds 3 red and 8 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?

3 / 8
P =
4.ProbabilityHard
xp(x)10.3520.1530.1540.350.05μ

X has the distribution shown, and Y = 5X − 3. Find the standard deviation σY of Y.

x: 1, 2, 3, 4, 5p(x): 0.35, 0.15, 0.15, 0.3, 0.05Y = 5X − 3
σY =
5.Normal distributionMedium
65.573.5P = ?

The exam scores are normally distributed with mean 65 points and standard deviation 10 points. What fraction lies between 65.5 and 73.5 points? Give P as a decimal.

μ = 65σ = 10x1 = 65.5x2 = 73.5
P =
6.Normal distributionHard
4 s0X − Y?

Runner A's lap time X and runner B's lap time Y are independent and normal: X has mean 52 s and standard deviation 4 s, Y has mean 48 s and standard deviation 3 s. Find the probability P that X is greater than Y. Round z to two decimals.

μX = 52 s, σX = 4 sμY = 48 s, σY = 3 s
P =
7.Confidence intervalsMedium
n = 7, s = 18LU = ?x̄ = 15E95%

A sample of 7 items has mean 15 and standard deviation s = 18. The population is roughly normal. Find the upper limit of the 95% confidence interval for the mean.

n = 7x̄ = 15s = 1895%t(6) = 2.447
upper limit =
8.Confidence intervalsHard
n = 30Lσ = ?s = 6.895%

The diameters of 30 machined shafts have a sample standard deviation of 6.8 µm. Assuming the diameters are normally distributed, find the upper limit σ of the 95% confidence interval for the population standard deviation. With df = 29, 2.5% of the χ² distribution lies below 16.047 and 2.5% lies above 45.722.

n = 30s = 6.8 µmχ² = 16.047, 45.722
σ = µm
9.Hypothesis tests & confidence intervalsMedium
−1.92p = ?

The mean of a population is claimed to be 250; the population standard deviation is known to be 18. A random sample of 36 observations has a mean of 244.24. Test H₀: μ = 250 against H₁: μ < 250. Find the p-value p.

μ₀ = 250σ = 18n = 36x̄ = 244.24H₁: μ < μ₀
p =
10.Hypothesis tests & confidence intervalsHard
H₀H₁μ₀x̄cμ₀ + 3α = 0.01βσ = 20, power 80%, n = ?

A one-sided z test at α = 0.01 must detect a shift of the mean of 3 units with a power of 80%. The standard deviation is 20. Find the minimum sample size n.

α = 0.01power = 80%δ = 3σ = 20zα = 2.326zβ = 0.842
n =
11.Linear regression & correlationMedium
xy284126178231028xy510102030

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

n = 5
r =
12.Linear regression & correlationHard
xy1575189321110241232714630161xy15202530100150x = 22.5

The table gives 6 paired observations of x = temperature in °C and y = ice creams sold. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 22.5.

n = 6x = 22.5
ŷ =

3Solutions

1.Descriptive statistics

Answer: s = 13.67

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 = 173/5 = 34.6

Squared deviations

Σ(x − x̄)² = 13.4² + 7.4² + 1.4² + 0.4² + (−22.6)² = 747.2

Standard deviation

s = √(747.2/4) = 13.67

Answer

s = 13.67
2.Descriptive statistics

Answer: P = 0.02275

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 = 2

Standard normal table

Φ(2) = 0.9772
P = 1 − 0.9772 = 0.02275

Answer

P = 0.02275
3.Probability

Answer: P = 0.05455

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 = 3/11 · 2/10 = 6/110 = 0.05455

Answer

P = 0.05455
4.Probability

Answer: σY = 6.796

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 = 2.55, Var(X) = 1.848, σX = 1.359

Linear change

σY = |5|·1.359 = 6.796

Answer

σY = 6.796

The mean does move: E(Y) = 5·2.55 − 3 = 9.75.

5.Normal distribution

Answer: P = 0.2824

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-scores

z₁ = (65.5 − 65)/10 = 0.05
z₂ = (73.5 − 65)/10 = 0.85

Area between

P = Φ(0.85) − Φ(0.05) = 0.8023 − 0.5199 = 0.2824

Answer

P = 0.2824
6.Normal distribution

Answer: P = 0.7881

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 = 52 − 48 = 4 s, σ = 5 s

Standardise

z = (0 − 4)/5 = −0.8

Probability

P = 1 − Φ(−0.8) = 0.7881

Answer

P = 0.7881
7.Confidence intervals

Answer: upper limit = 31.65

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 = 2.447

Margin of error

E = 2.447·18/√7 = 16.65

Interval

15 ± 16.65 → [−1.648, 31.65]

Answer

upper limit = 31.65
8.Confidence intervals

Answer: σ = 9.141 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 = 29·6.8² = 1341

Limits for σ²

1341/45.722 = 29.33
1341/16.047 = 83.56

Limit for σ

σ = √83.56 = 9.141 µm

Answer

σ = 9.141 µm

95% interval for σ: [5.416, 9.141] µm.

9.Hypothesis tests & confidence intervals

Answer: p = 0.02743

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 = 18/√36 = 3
z = (244.24 − 250)/3 = −1.92

Tail area

Φ(1.92) = 0.9726
1 − Φ(1.92) = 0.0274

p-value

p = 0.02743

Answer

p = 0.02743

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

10.Hypothesis tests & confidence intervals

Answer: n = 447

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 = ((2.326 + 0.842)·20/3)² = 446.1

Round up

n = 447

Answer

n = 447
11.Linear regression & correlation

Answer: r = 0.9979

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) = 102
Syy = Σ(y − ȳ)² = 261.2

Correlation coefficient

r = 102/√(40·261.2) = 0.9979

Answer

r = 0.9979
12.Linear regression & correlation

Answer: ŷ = 118

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 = 135/6 = 22.5
ȳ = 708/6 = 118

Sums of squares and products

Sxx = Σ(x − x̄)² = 157.5
Sxy = Σ(x − x̄)(y − ȳ) = 903

Regression line

b = 903/157.5 = 5.733
a = 118 − 5.733·22.5 = −11
ŷ = −11 + 5.733·x

Prediction

ŷ = −11 + 5.733·22.5 = 118

Answer

ŷ = 118

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