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
2025303540

Find the sample standard deviation s of the data: 42, 25, 19, 36, 23.

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

A quantity is normally distributed with mean 100 and standard deviation 20. What proportion P of values is above 76?

μ = 100σ = 20x = 76
P =
3.ProbabilityMedium
xp(x)100.2200.35300.1400.35μ = 26

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

x: 10, 20, 30, 40p(x): 0.2, 0.35, 0.1, 0.35
Var(X) =
4.ProbabilityHard
n = 12, p = 0.1: P(X = 4) = ?k0123456789101112

A test has 12 multiple-choice questions. A student who guesses gets each one right with probability 0.1, independently. What is the probability P of exactly 4 correct answers?

n = 12p = 0.1k = 4
P =
5.Normal distributionMedium
−3−2−10123zshaded area to the right of z = 0.0202

Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.0202. Find z to two decimals.

A = 0.0202
z =
6.Normal distributionHard
μ = 250x = ?20%

The bag weights are normally distributed with mean 250 g and standard deviation 5 g. Find the value x below which 20% of them lie (the 20th percentile).

μ = 250σ = 5P = 20%
x = g
7.Confidence intervalsMedium
n = 30, s = 2LU = ?x̄ = 57.5E90%

A sample of 30 items has mean 57.5 and standard deviation s = 2. The population is roughly normal. Find the upper limit of the 90% confidence interval for the mean.

n = 30x̄ = 57.5s = 290%t(29) = 1.699
upper limit =
8.Confidence intervalsHard
n = 15σ = ?Us = 8.995%

The diameters of 15 machined shafts have a sample standard deviation of 8.9 µm. Assuming the diameters are normally distributed, find the lower limit σ of the 95% confidence interval for the population standard deviation. With df = 14, 2.5% of the χ² distribution lies below 5.629 and 2.5% lies above 26.119.

n = 15s = 8.9 µmχ² = 5.629, 26.119
σ = µm
9.Hypothesis tests & confidence intervalsMedium
t −2.13102.131

A sample of 16 students has a mean score of 103 and a sample standard deviation of 10. Find the test statistic t for H₀: μ = 100.

μ₀ = 100x̄ = 103s = 10n = 16
t =
10.Hypothesis tests & confidence intervalsHard
σ = 10, n = ?x̄ − Ex̄ + Ex̄E = 199%

How large must a sample be so that the 99% confidence interval for a mean has a margin of error of at most 1? The standard deviation is 10. Find the smallest sample size n.

σ = 10E = 199%
n =
11.Linear regression & correlationMedium
xy243351457559668xy246506070

The table gives 5 paired observations of x = advertising spend in k€ and y = sales in k€. Their sums are given. Find Pearson's correlation coefficient r from these sums.

n = 5Σx = 20Σy = 278Σx² = 90Σy² = 15804Σxy = 1170
r =
12.Linear regression & correlationHard
xy123219318414513611xy2461520x = 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. A new observation (3.5, 12) is made. Find its residual e = y − ŷ.

n = 6(3.5, 12)
e =

3Solutions

1.Descriptive statistics

Answer: s = 9.618

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 = 145/5 = 29

Squared deviations

Σ(x − x̄)² = 13² + (−4)² + (−10)² + 7² + (−6)² = 370

Standard deviation

s = √(370/4) = 9.618

Answer

s = 9.618
2.Descriptive statistics

Answer: P = 0.8849

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 = −1.2

Standard normal table

Φ(−1.2) = 0.1151
P = 1 − 0.1151 = 0.8849

Answer

P = 0.8849
3.Probability

Answer: Var(X) = 134

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 = 10·0.2 + 20·0.35 + 30·0.1 + 40·0.35 = 26

Mean of the squares

E(X²) = 100·0.2 + 400·0.35 + 900·0.1 + 1600·0.35 = 810

Variance

Var(X) = 810 − 26² = 134

Answer

Var(X) = 134
4.Probability

Answer: P = 0.02131

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 = 4) = 495·0.14·0.98 = 495·0.0001·0.4305 = 0.02131

Answer

P = 0.02131
5.Normal distribution

Answer: z = 2.05

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 − 0.0202 = 0.9798

Read the z-table backwards

Find 0.9798 inside the table and read off z.

z = 2.05

Answer

z = 2.05
6.Normal distribution

Answer: x = 245.8 g

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) = 0.2 → z = −0.8416

Back to x

x = 250 + −0.8416·5 = 245.79 g

Answer

x = 245.79 g
7.Confidence intervals

Answer: upper limit = 58.12

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 = 29, t = 1.699

Margin of error

E = 1.699·2/√30 = 0.6204

Interval

57.5 ± 0.6204 → [56.88, 58.12]

Answer

upper limit = 58.12
8.Confidence intervals

Answer: σ = 6.516 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 = 14·8.9² = 1109

Limits for σ²

1109/26.119 = 42.46
1109/5.629 = 197

Limit for σ

σ = √42.46 = 6.516 µm

Answer

σ = 6.516 µm

95% interval for σ: [6.516, 14.04] µm.

9.Hypothesis tests & confidence intervals

Answer: t = 1.2

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 = 10/√16 = 10/4 = 2.5

Test statistic

t = (103 − 100)/2.5 = 1.2

Answer

t = 1.2
t0.025; df = 15 = 2.131

|t| < 2.131, so H₀ is not rejected at the 5% level (two-sided).

10.Hypothesis tests & confidence intervals

Answer: n = 664

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 = (2.576·10/1)² = 663.6

Round up

n = 664

Answer

n = 664
11.Linear regression & correlation

Answer: r = 0.9843

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 = 1170 − 20·278/5 = 58
Sxx = 90 − 20²/5 = 10
Syy = 15804 − 278²/5 = 347.2

Correlation coefficient

r = 58/√(10·347.2) = 0.9843

Answer

r = 0.9843
12.Linear regression & correlation

Answer: e = −4.333

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 = 98/6 = 16.33

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −41

Regression line

b = −41/17.5 = −2.343
a = 16.33 − (−2.343)·3.5 = 24.53
ŷ = 24.53 − 2.343·x

Predicted value

ŷ = 24.53 + (−2.343)·3.5 = 16.33

Residual

e = 12 − 16.33 = −4.333

Answer

e = −4.333

The point lies below the line.

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