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
25303540455055

Find the sample standard deviation s of the data: 59, 44, 31, 47, 25, 50.

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

A quantity is normally distributed with mean 500 and standard deviation 40. What proportion P of values is above 550?

μ = 500σ = 40x = 550
P =
3.ProbabilityMedium
xp(x)00.210.3520.330.140.05μ = 1.45

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

x: 0, 1, 2, 3, 4p(x): 0.2, 0.35, 0.3, 0.1, 0.05
σ =
4.ProbabilityHard
€p(x)−50.750.25950.05

A game costs €5 to play. You win €10 with probability 0.25 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 = €5€10: p = 0.25€100: p = 0.05
E(G) = €
5.Normal distributionMedium
−3−2−10123zshaded area to the right of z = 0.6331

Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.6331. Find z to two decimals.

A = 0.6331
z =
6.Normal distributionHard
35455565758595pointstop 80%

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

μ = 65 pointsσ = 10 points80%
x = points
7.Confidence intervalsMedium
n = 8, sd​ = 0.70LUd̄ = 6.9E = ?90%

The blood pressure of 8 patients is measured before and after a treatment. The differences d = before − after (mmHg) have mean 6.9 and standard deviation 0.7. Find the margin of error E of the 90% confidence interval for the mean difference.

n = 8d̄ = 6.9sd = 0.790%t(7) = 1.895
E = mmHg
8.Confidence intervalsHard
σ = 23, n = ?x̄ − Ex̄ + Ex̄E = 190%

The population standard deviation is 23. 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 1?

σ = 23E = 190%
n =
9.Hypothesis tests & confidence intervalsMedium
t −1.99801.998

A sample of 64 students has a mean score of 196.1 and a sample standard deviation of 10. Find the test statistic t for H₀: μ = 200.

μ₀ = 200x̄ = 196.1s = 10n = 64
t =
10.Hypothesis tests & confidence intervalsHard
σ = 4, n = ?x̄ − Ex̄ + Ex̄E = 199%

How large must a sample be so that the 99% confidence interval for a mean has a margin of error of at most 1? The standard deviation is 4. Find the smallest sample size n.

σ = 4E = 199%
n =
11.Linear regression & correlationMedium
xy2941061582110251226xy5101020

The table gives 6 paired observations of x = load in kN and y = extension in mm. Their sums are given. Find Pearson's correlation coefficient r from these sums.

n = 6Σx = 42Σy = 106Σx² = 364Σy² = 2148Σxy = 878
r =
12.Linear regression & correlationHard
xy181032012422128241432615628174xy2025100150ŷ = 151

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 = 151?

n = 6ŷ = 151
x =

3Solutions

1.Descriptive statistics

Answer: s = 12.56

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 = 256/6 = 42.67

Squared deviations

Σ(x − x̄)² = 16.33² + 1.333² + (−11.67)² + 4.333² + (−17.67)² + 7.333² = 789.3

Standard deviation

s = √(789.3/5) = 12.56

Answer

s = 12.56
2.Descriptive statistics

Answer: P = 0.1056

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 = 1.25

Standard normal table

Φ(1.25) = 0.8944
P = 1 − 0.8944 = 0.1056

Answer

P = 0.1056
3.Probability

Answer: σ = 1.071

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 = 0·0.2 + 1·0.35 + 2·0.3 + 3·0.1 + 4·0.05 = 1.45

Mean of the squares

E(X²) = 0·0.2 + 1·0.35 + 4·0.3 + 9·0.1 + 16·0.05 = 3.25

Variance

Var(X) = 3.25 − 1.45² = 1.147
σ = √1.147 = 1.071

Answer

σ = 1.071
4.Probability

Answer: E(G) = 2.5 €

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: 5 (p = 0.25), 95 (p = 0.05), −5 (p = 0.7)

Expected value

E(G) = 5·0.25 + 95·0.05 − 5·0.7 = 2.5 €

Answer

E(G) = 2.5 €

A positive value means you win money on average.

5.Normal distribution

Answer: z = −0.34

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 − 0.6331 = 0.3669

Read the z-table backwards

Find 0.3669 inside the table and read off z.

z = −0.34

Answer

z = −0.34
6.Normal distribution

Answer: x = 56.6 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) = 0.2

z from the table

z ≈ −0.84

Back to the original units

x = 65 − 0.84·10 = 56.6 points

Answer

x = 56.6 points
7.Confidence intervals

Answer: E = 0.469 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 = 7, t = 1.895

Margin of error

E = 1.895·0.7/√8 = 0.469

Answer

E = 0.469 mmHg
8.Confidence intervals

Answer: n = 1432

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 = (1.645·23/1)² = 1431

Round up

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

n = 1432

Answer

n = 1432
9.Hypothesis tests & confidence intervals

Answer: t = −3.12

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 = 10/√64 = 10/8 = 1.25

Test statistic

t = (196.1 − 200)/1.25 = −3.12

Answer

t = −3.12
t0.025; df = 63 = 1.998

|t| > 1.998, so H₀ is rejected at the 5% level (two-sided).

10.Hypothesis tests & confidence intervals

Answer: n = 107

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 = (2.576·4/1)² = 106.2

Round up

n = 107

Answer

n = 107
11.Linear regression & correlation

Answer: r = 0.9796

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.

Centred sums

Sxy = 878 − 42·106/6 = 136
Sxx = 364 − 42²/6 = 70
Syy = 2148 − 106²/6 = 275.3

Correlation coefficient

r = 136/√(70·275.3) = 0.9796

Answer

r = 0.9796
12.Linear regression & correlation

Answer: x = 24.95

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 = 138/6 = 23
ȳ = 828/6 = 138

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 466

Regression line

b = 466/70 = 6.657
a = 138 − 6.657·23 = −15.11
ŷ = −15.11 + 6.657·x

Solve for x

x = (151 − (−15.11))/6.657 = 24.95

Answer

x = 24.95

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