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
1015202530354045

Find the sample standard deviation s of the data: 34, 22, 21, 10, 46, 37, 37.

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

Exam scores are normally distributed with mean 60 and standard deviation 5. Find the z-score z of a score of 63.

μ = 60σ = 5x = 63
z =
3.ProbabilityMedium
xp(x)10.320.330.140.3μ = 2.4

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

x: 1, 2, 3, 4p(x): 0.3, 0.3, 0.1, 0.3
σ =
4.ProbabilityHard
€p(x)−50.8150.15450.05

A game costs €5 to play. You win €20 with probability 0.15 and €50 with probability 0.05; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.

stake = €5€20: p = 0.15€50: p = 0.05
E(G) = €
5.Normal distributionMedium
125 kg124.4 kgT = X₁ + … + X₅?

The masses of bags of cement are independent and normal with mean 25 kg and standard deviation 0.5 kg. Find the probability P that the total mass of 5 of them exceeds 124.4 kg. Round z to two decimals.

μ = 25 kgσ = 0.5 kgn = 5> 124.4 kg
P =
6.Normal distributionHard
0.02 mm0D − d?

Hole diameters D are normal with mean 25.02 mm and standard deviation 0.015 mm. Shaft diameters d are independent and normal with mean 25 mm and standard deviation 0.01 mm. A randomly chosen shaft fits when D − d > 0. Find the probability P that it fits. Round z to two decimals.

hole: μ = 25.02 mm, σ = 0.015 mmshaft: μ = 25 mm, σ = 0.01 mm
P =
7.Confidence intervalsMedium
μ₁ − μ₂0LU = ?x̄₁ − x̄₂E90%

The battery life of two phone models is compared. Model 1: 30 phones, mean 114 h, known standard deviation 17 h. Model 2: 50 phones, mean 102 h, known standard deviation 14 h. Find the upper limit of the 90% confidence interval for μ₁ − μ₂.

n₁ = 30x̄₁ = 114 hσ₁ = 17 hn₂ = 50x̄₂ = 102 hσ₂ = 14 h90%
upper limit = h
8.Confidence intervalsHard
σ = 18, n = ?x̄ − Ex̄ + Ex̄E = 490%

The population standard deviation is 18. How large a sample n do you need so that a 90% confidence interval for the mean has a margin of error of at most 4?

σ = 18E = 490%
n =
9.Hypothesis tests & confidence intervalsMedium
−1.321.32p = ?

The mean of a population is claimed to be 250; the population standard deviation is known to be 36. A random sample of 144 observations has a mean of 246.04. Test H₀: μ = 250 against H₁: μ ≠ 250. Find the p-value p.

μ₀ = 250σ = 36n = 144x̄ = 246.04H₁: μ ≠ μ₀
p =
10.Hypothesis tests & confidence intervalsHard
σ = 5, n = ?x̄ − Ex̄ + Ex̄E = 290%

How large must a sample be so that the 90% confidence interval for a mean has a margin of error of at most 2? The standard deviation is 5. Find the smallest sample size n.

σ = 5E = 290%
n =
11.Linear regression & correlationMedium
xy250355457560668xy246506070

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
xy346452553664764xy465060ŷ = 49

The table gives 5 paired observations of x = advertising spend in k€ and y = sales in k€. Use the least-squares line ŷ = a + b·x. For which value x does the line predict ŷ = 49?

n = 5ŷ = 49
x =

3Solutions

1.Descriptive statistics

Answer: s = 12.34

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 = 207/7 = 29.57

Squared deviations

Σ(x − x̄)² = 4.429² + (−7.571)² + (−8.571)² + (−19.57)² + 16.43² + 7.429² + 7.429² = 913.7

Standard deviation

s = √(913.7/6) = 12.34

Answer

s = 12.34
2.Descriptive statistics

Answer: z = 0.6

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 = (63 − 60)/5 = 0.6

Answer

z = 0.6
3.Probability

Answer: σ = 1.2

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 = 1·0.3 + 2·0.3 + 3·0.1 + 4·0.3 = 2.4

Mean of the squares

E(X²) = 1·0.3 + 4·0.3 + 9·0.1 + 16·0.3 = 7.2

Variance

Var(X) = 7.2 − 2.4² = 1.44
σ = √1.44 = 1.2

Answer

σ = 1.2
4.Probability

Answer: E(G) = 0.5 €

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: 15 (p = 0.15), 45 (p = 0.05), −5 (p = 0.8)

Expected value

E(G) = 15·0.15 + 45·0.05 − 5·0.8 = 0.5 €

Answer

E(G) = 0.5 €

A positive value means you win money on average.

5.Normal distribution

Answer: P = 0.7054

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 = 125 kg, σT = 0.5·√5 = 1.118 kg

Standardise

z = (124.4 − 125)/1.118 = −0.54

Probability

P = 1 − Φ(−0.54) = 0.7054

Answer

P = 0.7054
6.Normal distribution

Answer: P = 0.8665

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 = 25.02 − 25 = 0.02 mm, σ = 0.01803 mm

Standardise

z = (0 − 0.02)/0.01803 = −1.11

Probability

P = 1 − Φ(−1.11) = 0.8665

Answer

P = 0.8665
7.Confidence intervals

Answer: upper limit = 18.06 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 = 114 − 102 = 12

Standard error

SE = √(17²/30 + 14²/50) = 3.681

Margin of error

E = 1.645·3.681 = 6.056

Interval

12 ± 6.056 → [5.944, 18.06]

Answer

upper limit = 18.06 h

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

8.Confidence intervals

Answer: n = 55

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.

Solve for n

n = (1.645·18/4)² = 54.8

Round up

A sample size is a whole number, and rounding down would make the margin too large.

n = 55

Answer

n = 55
9.Hypothesis tests & confidence intervals

Answer: p = 0.1868

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 = 36/√144 = 3
z = (246.04 − 250)/3 = −1.32

Tail area

Φ(1.32) = 0.9066
1 − Φ(1.32) = 0.0934

p-value

p = 2·0.0934 = 0.1868

Answer

p = 0.1868

p > 0.05, so H₀ is not rejected at the 5% level.

10.Hypothesis tests & confidence intervals

Answer: n = 17

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 = (1.645·5/2)² = 16.91

Round up

n = 17

Answer

n = 17
11.Linear regression & correlation

Answer: r² = 0.9444

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 = 20/5 = 4
ȳ = 290/5 = 58

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = 41
Syy = Σ(y − ȳ)² = 178

Coefficient of determination

r² = 41²/(10·178) = 0.9444

Answer

r² = 0.9444

About 94.44% of the variation in y is explained by the line.

12.Linear regression & correlation

Answer: x = 3.583

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 = 279/5 = 55.8

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 = 55.8 − 4.8·5 = 31.8
ŷ = 31.8 + 4.8·x

Solve for x

x = (49 − 31.8)/4.8 = 3.583

Answer

x = 3.583

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