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
20253035404550

Find the sample standard deviation s of the data: 21, 17, 51, 28, 37, 42.

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

A quantity is normally distributed with mean 70 and standard deviation 12. What proportion P of values is above 77.2?

μ = 70σ = 12x = 77.2
P =
3.ProbabilityMedium
xp(x)00.1520.2540.3560.25μ = 3.4

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

x: 0, 2, 4, 6p(x): 0.15, 0.25, 0.35, 0.25
σ =
4.ProbabilityHard
€p(x)−10.7940.290.01

A game costs €1 to play. You win €5 with probability 0.2 and €10 with probability 0.01; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.

stake = €1€5: p = 0.2€10: p = 0.01
E(G) = €
5.Normal distributionMedium
320 g323.4 gT = X₁ + … + X₈?

The masses of bolts are independent and normal with mean 40 g and standard deviation 1.2 g. Find the probability P that the total mass of 8 of them exceeds 323.4 g. Round z to two decimals.

μ = 40 gσ = 1.2 gn = 8> 323.4 g
P =
6.Normal distributionHard
35455565758595pointstop 20%

The exam scores are normally distributed with mean 65 points and standard deviation 10 points. Drag the cut-off so the shaded top 20% is to the right of it. Find the cut-off x.

μ = 65 pointsσ = 10 points20%
x = points
7.Confidence intervalsMedium
n = 12, s = 5.5L = ?Ux̄ = 80.5E99%

A sample of 12 items has mean 80.5 and standard deviation s = 5.5. The population is roughly normal. Find the lower limit of the 99% confidence interval for the mean.

n = 12x̄ = 80.5s = 5.599%t(11) = 3.106
lower limit =
8.Confidence intervalsHard
n = 25σ = ?Us = 2.595%

The diameters of 25 machined shafts have a sample standard deviation of 2.5 µm. Assuming the diameters are normally distributed, find the lower limit σ of the 95% confidence interval for the population standard deviation. With df = 24, 2.5% of the χ² distribution lies below 12.401 and 2.5% lies above 39.364.

n = 25s = 2.5 µmχ² = 12.401, 39.364
σ = µm
9.Hypothesis tests & confidence intervalsMedium
−2.1452.145t = ?df = 14

15 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 −4.23 and the standard deviation of the differences is 8.2. Find the paired test statistic t for H₀: μd = 0.

n = 15d̄ = −4.23sd = 8.2
t =
10.Hypothesis tests & confidence intervalsHard
σ = 8, n = ?x̄ − Ex̄ + Ex̄E = 0.595%

How large must a sample be so that the 95% confidence interval for a mean has a margin of error of at most 0.5? The standard deviation is 8. Find the smallest sample size n.

σ = 8E = 0.595%
n =
11.Linear regression & correlationMedium
xy235344451556660764xy246405060

The table gives 6 paired observations of x = advertising spend in k€ and y = sales in k€. Find Pearson's correlation coefficient r.

n = 6
r =
12.Linear regression & correlationHard
xy124221322417515xy24152025ŷ = 17

The table gives 5 paired observations of x = car age in years and y = price in k€. Use the least-squares line ŷ = a + b·x. For which value x does the line predict ŷ = 17?

n = 5ŷ = 17
x =

3Solutions

1.Descriptive statistics

Answer: s = 13

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 = 196/6 = 32.67

Squared deviations

Σ(x − x̄)² = (−11.67)² + (−15.67)² + 18.33² + (−4.667)² + 4.333² + 9.333² = 845.3

Standard deviation

s = √(845.3/5) = 13

Answer

s = 13
2.Descriptive statistics

Answer: P = 0.2743

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 = 0.6

Standard normal table

Φ(0.6) = 0.7257
P = 1 − 0.7257 = 0.2743

Answer

P = 0.2743
3.Probability

Answer: σ = 2.01

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 = 0·0.15 + 2·0.25 + 4·0.35 + 6·0.25 = 3.4

Mean of the squares

E(X²) = 0·0.15 + 4·0.25 + 16·0.35 + 36·0.25 = 15.6

Variance

Var(X) = 15.6 − 3.4² = 4.04
σ = √4.04 = 2.01

Answer

σ = 2.01
4.Probability

Answer: E(G) = 0.1 €

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.

Net outcomes

G: 4 (p = 0.2), 9 (p = 0.01), −1 (p = 0.79)

Expected value

E(G) = 4·0.2 + 9·0.01 − 1·0.79 = 0.1 €

Answer

E(G) = 0.1 €

A positive value means you win money on average.

5.Normal distribution

Answer: P = 0.1587

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 = 320 g, σT = 1.2·√8 = 3.394 g

Standardise

z = (323.4 − 320)/3.394 = 1

Probability

P = 1 − Φ(1) = 0.1587

Answer

P = 0.1587
6.Normal distribution

Answer: x = 73.4 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) = 0.8

z from the table

z ≈ 0.84

Back to the original units

x = 65 + 0.84·10 = 73.4 points

Answer

x = 73.4 points
7.Confidence intervals

Answer: lower limit = 75.57

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 = 11, t = 3.106

Margin of error

E = 3.106·5.5/√12 = 4.931

Interval

80.5 ± 4.931 → [75.57, 85.43]

Answer

lower limit = 75.57
8.Confidence intervals

Answer: σ = 1.952 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 = 24·2.5² = 150

Limits for σ²

150/39.364 = 3.811
150/12.401 = 12.1

Limit for σ

σ = √3.811 = 1.952 µm

Answer

σ = 1.952 µm

95% interval for σ: [1.952, 3.478] µm.

9.Hypothesis tests & confidence intervals

Answer: t = −1.998

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 = 8.2/√15 = 2.117

Test statistic

t = −4.23/2.117 = −1.998

Answer

t = −1.998

With df = 14, the two-sided 5% critical value is 2.145; |t| is smaller, so H₀ is not rejected.

10.Hypothesis tests & confidence intervals

Answer: n = 984

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 = (1.96·8/0.5)² = 983.4

Round up

n = 984

Answer

n = 984
11.Linear regression & correlation

Answer: r = 0.9849

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 = 27/6 = 4.5
ȳ = 310/6 = 51.67

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = 99
Syy = Σ(y − ȳ)² = 577.3

Correlation coefficient

r = 99/√(17.5·577.3) = 0.9849

Answer

r = 0.9849
12.Linear regression & correlation

Answer: x = 4.273

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 = 15/5 = 3
ȳ = 99/5 = 19.8

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = −22

Regression line

b = −22/10 = −2.2
a = 19.8 − (−2.2)·3 = 26.4
ŷ = 26.4 − 2.2·x

Solve for x

x = (17 − 26.4)/(−2.2) = 4.273

Answer

x = 4.273

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