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: 50, 30, 54, 42, 21, 25, 31.

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 50.

μ = 60σ = 5x = 50
z =
3.ProbabilityMedium
n = 12k = 4N = ?

In how many ways N can a committee of 4 people be chosen from a group of 12? The order does not matter.

n = 12k = 4
N =
4.ProbabilityHard
xp(x)00.320.3540.1560.2μ

X has the distribution shown, and Y = −2X − 3. Find the standard deviation σY of Y.

x: 0, 2, 4, 6p(x): 0.3, 0.35, 0.15, 0.2Y = −2X − 3
σY =
5.Normal distributionMedium
200 mm199.5 mmX + Y?

X is normal with mean 120 mm and standard deviation 0.3 mm; Y is independent and normal with mean 80 mm and standard deviation 0.4 mm. Find the probability P that X + Y is more than 199.5 mm. Round z to two decimals.

μX = 120, σX = 0.3μY = 80, σY = 0.4X + Y > 199.5 mm
P =
6.Normal distributionHard
0.04 mm0D − d?

Hole diameters D are normal with mean 25.04 mm and standard deviation 0.02 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 = 25.04 mm, σ = 0.02 mmshaft: μ = 25 mm, σ = 0.01 mm
P =
7.Confidence intervalsMedium
n = 8, s = 18.5LU = ?x̄ = 55.5E90%

A sample of 8 items has mean 55.5 and standard deviation s = 18.5. The population is roughly normal. Find the upper limit of the 90% confidence interval for the mean.

n = 8x̄ = 55.5s = 18.590%t(7) = 1.895
upper limit =
8.Confidence intervalsHard
μ₁ − μ₂, t(15) = 2.1310LU = ?x̄₁ − x̄₂E95%

Two fertilisers are tested. Plot group 1: 6 plots, mean yield 85.5 kg, standard deviation 9 kg. Plot group 2: 11 plots, mean 91.5 kg, standard deviation 7.5 kg. Assume equal variances (pooled t, df = n₁ + n₂ − 2). Find the upper limit of the 95% confidence interval for μ₁ − μ₂.

n₁ = 6x̄₁ = 85.5s₁ = 9n₂ = 11x̄₂ = 91.5s₂ = 7.595%t(15) = 2.131
upper limit = kg
9.Hypothesis tests & confidence intervalsMedium
t −2.06402.064

A sample of 25 students has a mean score of 69.4 and a sample standard deviation of 20. Find the test statistic t for H₀: μ = 75.

μ₀ = 75x̄ = 69.4s = 20n = 25
t =
10.Hypothesis tests & confidence intervalsHard
z −1.9601.96

A sample of 25 measurements has mean 105.3. The population standard deviation is 12. Find the lower limit L of the 95% confidence interval for the mean.

x̄ = 105.3σ = 12n = 2595%
L =
11.Linear regression & correlationMedium
xy241350451559667xy24640506070

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
xy156517881996211092311025121xy15202580100120ŷ = 119

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. For which value x does the line predict ŷ = 119?

n = 6ŷ = 119
x =

3Solutions

1.Descriptive statistics

Answer: s = 12.67

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 = 253/7 = 36.14

Squared deviations

Σ(x − x̄)² = 13.86² + (−6.143)² + 17.86² + 5.857² + (−15.14)² + (−11.14)² + (−5.143)² = 962.9

Standard deviation

s = √(962.9/6) = 12.67

Answer

s = 12.67
2.Descriptive statistics

Answer: z = −2

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 = (50 − 60)/5 = −2

Answer

z = −2
3.Probability

Answer: N = 495

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.

Combinations

C(12, 4) = (12·11·10·9)/(4·3·2·1) = 495

Answer

N = 495
4.Probability

Answer: σY = 4.359

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 = 2.5, Var(X) = 4.75, σX = 2.179

Linear change

σY = |−2|·2.179 = 4.359

Answer

σY = 4.359

The mean does move: E(Y) = −2·2.5 − 3 = −8.

5.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 sum

μ = 200 mm, σ = √(0.3² + 0.4²) = 0.5 mm

Standardise

z = (199.5 − 200)/0.5 = −1

Upper tail

P = 1 − Φ(−1) = 1 − 0.1587 = 0.8413

Answer

P = 0.8413
6.Normal distribution

Answer: P = 0.9633

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 = 25.04 − 25 = 0.04 mm, σ = 0.02236 mm

Standardise

z = (0 − 0.04)/0.02236 = −1.79

Probability

P = 1 − Φ(−1.79) = 0.9633

Answer

P = 0.9633
7.Confidence intervals

Answer: upper limit = 67.89

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 = 7, t = 1.895

Margin of error

E = 1.895·18.5/√8 = 12.39

Interval

55.5 ± 12.39 → [43.11, 67.89]

Answer

upper limit = 67.89
8.Confidence intervals

Answer: upper limit = 2.686 kg

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.

Pooled standard deviation

sp² = (5·9² + 10·7.5²)/15 = 64.5
sp = 8.031

Margin of error

E = 2.131·8.031·√(1/6 + 1/11) = 8.686

Interval

−6 ± 8.686 → [−14.69, 2.686]

Answer

upper limit = 2.686 kg
9.Hypothesis tests & confidence intervals

Answer: t = −1.4

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 = 20/√25 = 20/5 = 4

Test statistic

t = (69.4 − 75)/4 = −1.4

Answer

t = −1.4
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: L = 100.6

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).

Margin of error

E = 1.96·12/√25 = 4.704

Interval limit

L = 105.3 − 4.704 = 100.6

Answer

L = 100.6
11.Linear regression & correlation

Answer: r² = 0.961

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 = 268/5 = 53.6

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = 61
Syy = Σ(y − ȳ)² = 387.2

Coefficient of determination

r² = 61²/(10·387.2) = 0.961

Answer

r² = 0.961

About 96.1% of the variation in y is explained by the line.

12.Linear regression & correlation

Answer: x = 24.06

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 = 120/6 = 20
ȳ = 589/6 = 98.17

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 359

Regression line

b = 359/70 = 5.129
a = 98.17 − 5.129·20 = −4.405
ŷ = −4.405 + 5.129·x

Solve for x

x = (119 − (−4.405))/5.129 = 24.06

Answer

x = 24.06

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