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: 50, 21, 29, 54, 38, 12, 15.

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

A quantity is normally distributed with mean 170 and standard deviation 20. What proportion P of values is below 130?

μ = 170σ = 20x = 130
P =
3.ProbabilityMedium
n = 9k = 4N = ?

In how many ways N can a committee of 4 people be chosen from a group of 9? The order does not matter.

n = 9k = 4
N =
4.ProbabilityHard
€p(x)−100.85300.1900.05

A game costs €10 to play. You win €40 with probability 0.1 and €100 with probability 0.05; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.

stake = €10€40: p = 0.1€100: p = 0.05
E(G) = €
5.Normal distributionMedium
92.5103.75P = ?

The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. What fraction lies between 92.5 and 103.75 points? Give P as a decimal.

μ = 100σ = 15x1 = 92.5x2 = 103.75
P =
6.Normal distributionHard
157164171178185192199cmtop 10%

The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. Drag the cut-off so the shaded top 10% is to the right of it. Find the cut-off x.

μ = 178 cmσ = 7 cm10%
x = cm
7.Confidence intervalsMedium
n = 13, sd​ = 7.90LUd̄ = −9.2E = ?95%

The blood pressure of 13 patients is measured before and after a treatment. The differences d = before − after (mmHg) have mean −9.2 and standard deviation 7.9. Find the margin of error E of the 95% confidence interval for the mean difference.

n = 13d̄ = −9.2sd = 7.995%t(12) = 2.179
E = mmHg
8.Confidence intervalsHard
σ = 25, n = ?x̄ − Ex̄ + Ex̄E = 590%

The population standard deviation is 25. 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 5?

σ = 25E = 590%
n =
9.Hypothesis tests & confidence intervalsMedium
−2.1792.179t = ?df = 12

Two teaching methods are compared. Group 1: 7 students, mean score 91, standard deviation 3.5. Group 2: 7 students, mean score 81.4, standard deviation 10. Assume equal population variances and use the pooled two-sample t test. Find the test statistic t for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (group 1 minus group 2).

n₁ = 7x̄₁ = 91s₁ = 3.5n₂ = 7x̄₂ = 81.4s₂ = 10
t =
10.Hypothesis tests & confidence intervalsHard
z −1.64501.645

A sample of 16 measurements has mean 91. The population standard deviation is 20. Find the upper limit U of the 90% confidence interval for the mean.

x̄ = 91σ = 20n = 1690%
U =
11.Linear regression & correlationMedium
xy16611872207322882410226103xy20256080100

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
xy181012111824143271563017133196xy202530100150200ŷ = 149

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 = 149?

n = 6ŷ = 149
x =

3Solutions

1.Descriptive statistics

Answer: s = 16.63

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 = 219/7 = 31.29

Squared deviations

Σ(x − x̄)² = 18.71² + (−10.29)² + (−2.286)² + 22.71² + 6.714² + (−19.29)² + (−16.29)² = 1659

Standard deviation

s = √(1659/6) = 16.63

Answer

s = 16.63
2.Descriptive statistics

Answer: P = 0.02275

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 = −2

Standard normal table

Φ(−2) = 0.02275
P = 0.02275

Answer

P = 0.02275
3.Probability

Answer: N = 126

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(9, 4) = (9·8·7·6)/(4·3·2·1) = 126

Answer

N = 126
4.Probability

Answer: E(G) = −1 €

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: 30 (p = 0.1), 90 (p = 0.05), −10 (p = 0.85)

Expected value

E(G) = 30·0.1 + 90·0.05 − 10·0.85 = −1 €

Answer

E(G) = −1 €

A negative value means you lose money on average.

5.Normal distribution

Answer: P = 0.2902

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 = (92.5 − 100)/15 = −0.5
z₂ = (103.75 − 100)/15 = 0.25

Area between

P = Φ(0.25) − Φ(−0.5) = 0.5987 − 0.3085 = 0.2902

Answer

P = 0.2902
6.Normal distribution

Answer: x = 187 cm

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) = 0.9

z from the table

z ≈ 1.28

Back to the original units

x = 178 + 1.28·7 = 187 cm

Answer

x = 187 cm
7.Confidence intervals

Answer: E = 4.774 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 = 12, t = 2.179

Margin of error

E = 2.179·7.9/√13 = 4.774

Answer

E = 4.774 mmHg
8.Confidence intervals

Answer: n = 68

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 = (1.645·25/5)² = 67.65

Round up

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

n = 68

Answer

n = 68
9.Hypothesis tests & confidence intervals

Answer: t = 2.397

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.

Pooled variance

sp² = (6·3.5² + 6·10²)/12 = 56.13
sp = 7.492

Standard error

SE = 7.492·√(1/7 + 1/7) = 4.004

Test statistic

t = (91 − 81.4)/4.004 = 2.397

Answer

t = 2.397

With df = 12, the two-sided 5% critical value is 2.179; |t| is larger, so H₀ is rejected.

10.Hypothesis tests & confidence intervals

Answer: U = 99.22

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 = 1.645·20/√16 = 8.225

Interval limit

U = 91 + 8.225 = 99.22

Answer

U = 99.22
11.Linear regression & correlation

Answer: r = 0.9751

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 = 126/6 = 21
ȳ = 499/6 = 83.17

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 315
Syy = Σ(y − ȳ)² = 1491

Correlation coefficient

r = 315/√(70·1491) = 0.9751

Answer

r = 0.9751
12.Linear regression & correlation

Answer: x = 25.74

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 = 153/6 = 25.5
ȳ = 885/6 = 147.5

Sums of squares and products

Sxx = Σ(x − x̄)² = 157.5
Sxy = Σ(x − x̄)(y − ȳ) = 970.5

Regression line

b = 970.5/157.5 = 6.162
a = 147.5 − 6.162·25.5 = −9.629
ŷ = −9.629 + 6.162·x

Solve for x

x = (149 − (−9.629))/6.162 = 25.74

Answer

x = 25.74

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