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
15202530354045

Find the sample standard deviation s of the data: 22, 14, 27, 42, 45, 42.

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

A quantity is normally distributed with mean 50 and standard deviation 12. What proportion P of values is above 62?

μ = 50σ = 12x = 62
P =
3.ProbabilityMedium
xp(x)00.0510.3520.330.3μ = 1.85

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

x: 0, 1, 2, 3p(x): 0.05, 0.35, 0.3, 0.3
σ =
4.ProbabilityHard
n = 11, p = 0.6: P(X ≤ 1) = ?k01234567891011

11 components are tested, and each one fails with probability 0.6, independently. What is the probability P that at most one fails?

n = 11p = 0.6
P =
5.Normal distributionMedium
93.25127.75P = ?

The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. What fraction lies between 93.25 and 127.75 points? Give P as a decimal.

μ = 100σ = 15x1 = 93.25x2 = 127.75
P =
6.Normal distributionHard
35455565758595pointstop 70%

The exam scores are normally distributed with mean 65 points and standard deviation 10 points. Drag the cut-off so the shaded top 70% is to the right of it. Find the cut-off x.

μ = 65 pointsσ = 10 points70%
x = points
7.Confidence intervalsMedium
n = 10, sd​ = 20LUd̄ = −6.7E = ?95%

The blood pressure of 10 patients is measured before and after a treatment. The differences d = before − after (mmHg) have mean −6.7 and standard deviation 2. Find the margin of error E of the 95% confidence interval for the mean difference.

n = 10d̄ = −6.7sd = 295%t(9) = 2.262
E = mmHg
8.Confidence intervalsHard
p ≈ 0.2, n = ?p̂ − Ep̂ + Ep̂E = 0.0190%

A pilot study suggests that about 0.2 of products are defective. How many products n must be inspected so that a 90% confidence interval for the proportion has a margin of error of at most 0.01?

p ≈ 0.2E = 0.0190%
n =
9.Hypothesis tests & confidence intervalsMedium
t −1.98401.984

A sample of 100 students has a mean score of 201.1 and a sample standard deviation of 5. Find the test statistic t for H₀: μ = 200.

μ₀ = 200x̄ = 201.1s = 5n = 100
t =
10.Hypothesis tests & confidence intervalsHard
μ₀x̄c

A one-sided z test of H₀: μ = 50 against H₁: μ < 50 is done at α = 0.05 with a sample of 25; the standard deviation is known to be 7.5. In reality the mean is 45.8. Find the probability of a Type II error β in percent.

μ₀ = 50μ = 45.8σ = 7.5n = 25α = 0.05
β = %
11.Linear regression & correlationMedium
xy1433435487539551165xy5105060

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
xy15401751194821592363xy1520405060x = 16

The table gives 5 paired observations of x = temperature in °C and y = ice creams sold. Use the least-squares line ŷ = a + b·x. A new observation (16, 47) is made. Find its residual e = y − ŷ.

n = 5(16, 47)
e =

3Solutions

1.Descriptive statistics

Answer: s = 12.79

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 = 192/6 = 32

Squared deviations

Σ(x − x̄)² = (−10)² + (−18)² + (−5)² + 10² + 13² + 10² = 818

Standard deviation

s = √(818/5) = 12.79

Answer

s = 12.79
2.Descriptive statistics

Answer: P = 0.1587

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 = 1

Standard normal table

Φ(1) = 0.8413
P = 1 − 0.8413 = 0.1587

Answer

P = 0.1587
3.Probability

Answer: σ = 0.9097

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 = 0·0.05 + 1·0.35 + 2·0.3 + 3·0.3 = 1.85

Mean of the squares

E(X²) = 0·0.05 + 1·0.35 + 4·0.3 + 9·0.3 = 4.25

Variance

Var(X) = 4.25 − 1.85² = 0.8275
σ = √0.8275 = 0.9097

Answer

σ = 0.9097
4.Probability

Answer: P = 0.000734

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) = 0.411 = 4.194 × 10⁻⁵
P(X = 1) = 11·0.6·0.410 = 0.0006921

Add them

P = 4.194 × 10⁻⁵ + 0.0006921 = 0.000734

Answer

P = 0.000734
5.Normal distribution

Answer: P = 0.6415

Normal distribution: convert a value x to a z-score, z = (x − μ)/σ, which counts standard deviations from the mean. A z-table (or Φ(z)) gives the area to the left of z, which is P(X < x). The area to the right is 1 − Φ(z), and the area between two values is the difference of their two areas.

z-scores

z₁ = (93.25 − 100)/15 = −0.45
z₂ = (127.75 − 100)/15 = 1.85

Area between

P = Φ(1.85) − Φ(−0.45) = 0.9678 − 0.3264 = 0.6415

Answer

P = 0.6415
6.Normal distribution

Answer: x = 59.8 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) = 0.3

z from the table

z ≈ −0.52

Back to the original units

x = 65 − 0.52·10 = 59.8 points

Answer

x = 59.8 points
7.Confidence intervals

Answer: E = 1.431 mmHg

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.

Critical value

df = 9, t = 2.262

Margin of error

E = 2.262·2/√10 = 1.431

Answer

E = 1.431 mmHg
8.Confidence intervals

Answer: n = 4330

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 = 1.645²·0.2·0.8/0.01² = 4330

Round up

n = 4330

Answer

n = 4330
9.Hypothesis tests & confidence intervals

Answer: t = 2.2

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 = (201.1 − 200)/0.5 = 2.2

Answer

t = 2.2
t0.025; df = 99 = 1.984

|t| > 1.984, so H₀ is rejected at the 5% level (two-sided).

10.Hypothesis tests & confidence intervals

Answer: β = 12.4 %

Power of a test: power = 1 − β is the probability of rejecting H₀ when it is false. Find the cutoff x̄c from μ₀ and z_α, then compute the probability of landing beyond x̄c using the true mean: z = (x̄c − μ)/(σ/√n).

Critical sample mean

SE = 7.5/√25 = 1.5
x̄c = 50 − 1.645·1.5 = 47.53

z under the true mean

z = (47.53 − 45.8)/1.5 = 1.155

Type II error

power = P(Z < 1.155) = 87.6%
β = 100% − 87.6% = 12.4%

Answer

β = 12.4%
11.Linear regression & correlation

Answer: r = 0.9608

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 = 36/6 = 6
ȳ = 307/6 = 51.17

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 151
Syy = Σ(y − ȳ)² = 352.8

Correlation coefficient

r = 151/√(70·352.8) = 0.9608

Answer

r = 0.9608
12.Linear regression & correlation

Answer: e = 2.9

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 = 95/5 = 19
ȳ = 261/5 = 52.2

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 108

Regression line

b = 108/40 = 2.7
a = 52.2 − 2.7·19 = 0.9
ŷ = 0.9 + 2.7·x

Predicted value

ŷ = 0.9 + 2.7·16 = 44.1

Residual

e = 47 − 44.1 = 2.9

Answer

e = 2.9

The point lies above the line.

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