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
253035404550

Find the sample standard deviation s of the data: 23, 28, 54, 46, 40, 30, 45.

n = 7
s =
2.Descriptive statisticsHard
z −2−1012

Exam scores are normally distributed with mean 500 and standard deviation 25. Find the z-score z of a score of 537.5.

μ = 500σ = 25x = 537.5
z =
3.ProbabilityMedium
xp(x)00.110.3520.130.140.35μ = 2.25

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

x: 0, 1, 2, 3, 4p(x): 0.1, 0.35, 0.1, 0.1, 0.35
σ =
4.ProbabilityHard
€p(x)−30.6930.31470.01

A game costs €3 to play. You win €6 with probability 0.3 and €150 with probability 0.01; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.

stake = €3€6: p = 0.3€150: p = 0.01
E(G) = €
5.Normal distributionMedium
640 g649.6 gT = X₁ + … + X₁₆?

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 16 of them exceeds 649.6 g. Round z to two decimals.

μ = 40 gσ = 1.2 gn = 16> 649.6 g
P =
6.Normal distributionHard
10 g0X − Y?

The mass X of a product from line 1 and Y from line 2 are independent and normal: X has mean 500 g and standard deviation 8 g, Y has mean 490 g and standard deviation 6 g. Find the probability P that X is greater than Y. Round z to two decimals.

μX = 500 g, σX = 8 gμY = 490 g, σY = 6 g
P =
7.Confidence intervalsMedium
n = 25, s = 10LU = ?x̄ = 45.5E95%

A sample of 25 items has mean 45.5 and standard deviation s = 10. The population is roughly normal. Find the upper limit of the 95% confidence interval for the mean.

n = 25x̄ = 45.5s = 1095%t(24) = 2.064
upper limit =
8.Confidence intervalsHard
n = 11σ = ?Us = 11.195%

The diameters of 11 machined shafts have a sample standard deviation of 11.1 µm. Assuming the diameters are normally distributed, find the lower limit σ of the 95% confidence interval for the population standard deviation. With df = 10, 2.5% of the χ² distribution lies below 3.247 and 2.5% lies above 20.483.

n = 11s = 11.1 µmχ² = 3.247, 20.483
σ = µm
9.Hypothesis tests & confidence intervalsMedium
t −2.06402.064

A sample of 25 students has a mean score of 197.8 and a sample standard deviation of 8. Find the test statistic t for H₀: μ = 200.

μ₀ = 200x̄ = 197.8s = 8n = 25
t =
10.Hypothesis tests & confidence intervalsHard
H₀H₁μ₀x̄cμ₀ + 2α = 0.05βσ = 17, power 80%, n = ?

A one-sided z test at α = 0.05 must detect a shift of the mean of 2 units with a power of 80%. The standard deviation is 17. Find the minimum sample size n.

α = 0.05power = 80%δ = 2σ = 17zα = 1.645zβ = 0.842
n =
11.Linear regression & correlationMedium
xy139237340449548xy24404550

The table gives 5 paired observations of x = advertising spend in k€ and y = sales in k€. Find the coefficient of determination r² of the least-squares line.

n = 5
r² =
12.Linear regression & correlationHard
xy18732087229424942610628123xy202580100120x = 27

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 = 27.

n = 6x = 27
ŷ =

3Solutions

1.Descriptive statistics

Answer: s = 11.27

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 = 266/7 = 38

Squared deviations

Σ(x − x̄)² = (−15)² + (−10)² + 16² + 8² + 2² + (−8)² + 7² = 762

Standard deviation

s = √(762/6) = 11.27

Answer

s = 11.27
2.Descriptive statistics

Answer: z = 1.5

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 = (537.5 − 500)/25 = 1.5

Answer

z = 1.5
3.Probability

Answer: σ = 1.479

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 = 0·0.1 + 1·0.35 + 2·0.1 + 3·0.1 + 4·0.35 = 2.25

Mean of the squares

E(X²) = 0·0.1 + 1·0.35 + 4·0.1 + 9·0.1 + 16·0.35 = 7.25

Variance

Var(X) = 7.25 − 2.25² = 2.188
σ = √2.188 = 1.479

Answer

σ = 1.479
4.Probability

Answer: E(G) = 0.3 €

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: 3 (p = 0.3), 147 (p = 0.01), −3 (p = 0.69)

Expected value

E(G) = 3·0.3 + 147·0.01 − 3·0.69 = 0.3 €

Answer

E(G) = 0.3 €

A positive value means you win money on average.

5.Normal distribution

Answer: P = 0.02275

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 = 640 g, σT = 1.2·√16 = 4.8 g

Standardise

z = (649.6 − 640)/4.8 = 2

Probability

P = 1 − Φ(2) = 0.02275

Answer

P = 0.02275
6.Normal distribution

Answer: P = 0.8413

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 = 500 − 490 = 10 g, σ = 10 g

Standardise

z = (0 − 10)/10 = −1

Probability

P = 1 − Φ(−1) = 0.8413

Answer

P = 0.8413
7.Confidence intervals

Answer: upper limit = 49.63

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 = 24, t = 2.064

Margin of error

E = 2.064·10/√25 = 4.128

Interval

45.5 ± 4.128 → [41.37, 49.63]

Answer

upper limit = 49.63
8.Confidence intervals

Answer: σ = 7.756 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 = 10·11.1² = 1232

Limits for σ²

1232/20.483 = 60.15
1232/3.247 = 379.5

Limit for σ

σ = √60.15 = 7.756 µm

Answer

σ = 7.756 µm

95% interval for σ: [7.756, 19.48] µm.

9.Hypothesis tests & confidence intervals

Answer: t = −1.375

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/√25 = 8/5 = 1.6

Test statistic

t = (197.8 − 200)/1.6 = −1.375

Answer

t = −1.375
t0.025; df = 24 = 2.064

|t| < 2.064, so H₀ is not rejected at the 5% level (two-sided).

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 = ((1.645 + 0.842)·17/2)² = 446.9

Round up

n = 447

Answer

n = 447
11.Linear regression & correlation

Answer: r² = 0.7426

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 = 213/5 = 42.6

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = 30
Syy = Σ(y − ȳ)² = 121.2

Coefficient of determination

r² = 30²/(10·121.2) = 0.7426

Answer

r² = 0.7426

About 74.26% of the variation in y is explained by the line.

12.Linear regression & correlation

Answer: ŷ = 113.7

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 = 138/6 = 23
ȳ = 577/6 = 96.17

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 307

Regression line

b = 307/70 = 4.386
a = 96.17 − 4.386·23 = −4.705
ŷ = −4.705 + 4.386·x

Prediction

ŷ = −4.705 + 4.386·27 = 113.7

Answer

ŷ = 113.7

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