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
1520253035404550

Find the sample standard deviation s of the data: 15, 18, 54, 11, 46, 42, 48.

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

Exam scores are normally distributed with mean 60 and standard deviation 10. Find the z-score z of a score of 52.

μ = 60σ = 10x = 52
z =
3.ProbabilityMedium
xp(x)10.320.230.1540.35μ = 2.55

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

x: 1, 2, 3, 4p(x): 0.3, 0.2, 0.15, 0.35
σ =
4.ProbabilityHard
n = 12, p = 0.3: P(X ≤ 1) = ?k0123456789101112

12 components are tested, and each one fails with probability 0.3, independently. What is the probability P that at most one fails?

n = 12p = 0.3
P =
5.Normal distributionMedium
−3−2−10123zshaded area to the right of z = 0.9591

Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.9591. Find z to two decimals.

A = 0.9591
z =
6.Normal distributionHard
35455565758595pointsbottom 80%

The exam scores are normally distributed with mean 65 points and standard deviation 10 points. Drag the cut-off so the shaded bottom 80% is to the left of it. Find the cut-off x.

μ = 65 pointsσ = 10 points80%
x = points
7.Confidence intervalsMedium
μ₁ − μ₂0L = ?Ux̄₁ − x̄₂E90%

The battery life of two phone models is compared. Model 1: 80 phones, mean 118.5 h, known standard deviation 12 h. Model 2: 100 phones, mean 117.5 h, known standard deviation 3 h. Find the lower limit of the 90% confidence interval for μ₁ − μ₂.

n₁ = 80x̄₁ = 118.5 hσ₁ = 12 hn₂ = 100x̄₂ = 117.5 hσ₂ = 3 h90%
lower limit = h
8.Confidence intervalsHard
p ≈ 0.2, n = ?p̂ − Ep̂ + Ep̂E = 0.0499%

A pilot study suggests that about 0.2 of products are defective. How many products n must be inspected so that a 99% confidence interval for the proportion has a margin of error of at most 0.04?

p ≈ 0.2E = 0.0499%
n =
9.Hypothesis tests & confidence intervalsMedium
t −1.98401.984

A sample of 100 students has a mean score of 49.1 and a sample standard deviation of 5. Find the test statistic t for H₀: μ = 50.

μ₀ = 50x̄ = 49.1s = 5n = 100
t =
10.Hypothesis tests & confidence intervalsHard
z −1.9601.96

A sample of 64 measurements has mean 102.5. The population standard deviation is 10. Find the upper limit U of the 95% confidence interval for the mean.

x̄ = 102.5σ = 10n = 6495%
U =
11.Linear regression & correlationMedium
xy17572080238126912911532120xy2025306080100120

The table gives 6 paired observations of x = temperature in °C and y = ice creams sold. Find Pearson's correlation coefficient r.

n = 6
r =
12.Linear regression & correlationHard
xy140252358464568xy2440506070ŷ = 67

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 = 67?

n = 5ŷ = 67
x =

3Solutions

1.Descriptive statistics

Answer: s = 18.02

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 = 234/7 = 33.43

Squared deviations

Σ(x − x̄)² = (−18.43)² + (−15.43)² + 20.57² + (−22.43)² + 12.57² + 8.571² + 14.57² = 1948

Standard deviation

s = √(1948/6) = 18.02

Answer

s = 18.02
2.Descriptive statistics

Answer: z = −0.8

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 = (52 − 60)/10 = −0.8

Answer

z = −0.8
3.Probability

Answer: σ = 1.244

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 + 2·0.2 + 3·0.15 + 4·0.35 = 2.55

Mean of the squares

E(X²) = 1·0.3 + 4·0.2 + 9·0.15 + 16·0.35 = 8.05

Variance

Var(X) = 8.05 − 2.55² = 1.548
σ = √1.548 = 1.244

Answer

σ = 1.244
4.Probability

Answer: P = 0.08503

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.

Zero and one failure

P(X = 0) = 0.712 = 0.01384
P(X = 1) = 12·0.3·0.711 = 0.07118

Add them

P = 0.01384 + 0.07118 = 0.08503

Answer

P = 0.08503
5.Normal distribution

Answer: z = −1.74

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 − 0.9591 = 0.0409

Read the z-table backwards

Find 0.0409 inside the table and read off z.

z = −1.74

Answer

z = −1.74
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 = −1.262 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 = 118.5 − 117.5 = 1

Standard error

SE = √(12²/80 + 3²/100) = 1.375

Margin of error

E = 1.645·1.375 = 2.262

Interval

1 ± 2.262 → [−1.262, 3.262]

Answer

lower limit = −1.262 h

The interval contains 0, so the data do not show a significant difference.

8.Confidence intervals

Answer: n = 664

Confidence interval for a proportion: p̂ ± z·√(p̂(1 − p̂)/n), where p̂ = x/n. To reach a margin E you need n ≥ z²·p(1 − p)/E², rounded up; without a prior estimate use p = 0.5, which gives the largest n.

Solve for n

n = 2.576²·0.2·0.8/0.04² = 663.6

Round up

n = 664

Answer

n = 664
9.Hypothesis tests & confidence intervals

Answer: t = −1.8

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 = 5/√100 = 5/10 = 0.5

Test statistic

t = (49.1 − 50)/0.5 = −1.8

Answer

t = −1.8
t0.025; df = 99 = 1.984

|t| < 1.984, so H₀ is not rejected at the 5% level (two-sided).

10.Hypothesis tests & confidence intervals

Answer: U = 105

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 = 1.96·10/√64 = 2.45

Interval limit

U = 102.5 + 2.45 = 105

Answer

U = 105
11.Linear regression & correlation

Answer: r = 0.9724

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 = 147/6 = 24.5
ȳ = 544/6 = 90.67

Sums of squares and products

Sxx = Σ(x − x̄)² = 157.5
Sxy = Σ(x − x̄)(y − ȳ) = 645
Syy = Σ(y − ȳ)² = 2793

Correlation coefficient

r = 645/√(157.5·2793) = 0.9724

Answer

r = 0.9724
12.Linear regression & correlation

Answer: x = 4.559

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 = 282/5 = 56.4

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = 68

Regression line

b = 68/10 = 6.8
a = 56.4 − 6.8·3 = 36
ŷ = 36 + 6.8·x

Solve for x

x = (67 − 36)/6.8 = 4.559

Answer

x = 4.559

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