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
10152025303540455055

Find the sample standard deviation s of the data: 10, 23, 31, 55, 34, 15, 35.

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

A quantity is normally distributed with mean 100 and standard deviation 15. What proportion P of values is above 70?

μ = 100σ = 15x = 70
P =
3.ProbabilityMedium
xp(x)00.110.2520.1530.3540.15μ = 2.2

For the distribution shown, find the variance Var(X) of X.

x: 0, 1, 2, 3, 4p(x): 0.1, 0.25, 0.15, 0.35, 0.15
Var(X) =
4.ProbabilityHard
xp(x)00.310.3520.35μ

X has the distribution shown, and Y = 5X + 1. Find the standard deviation σY of Y.

x: 0, 1, 2p(x): 0.3, 0.35, 0.35Y = 5X + 1
σY =
5.Normal distributionMedium
750 g745.1 gX + Y?

X is normal with mean 250 g and standard deviation 4 g; Y is independent and normal with mean 500 g and standard deviation 9 g. Find the probability P that X + Y is more than 745.1 g. Round z to two decimals.

μX = 250, σX = 4μY = 500, σY = 9X + Y > 745.1 g
P =
6.Normal distributionHard
157164171178185192199cmtop 20%

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 20% is to the right of it. Find the cut-off x.

μ = 178 cmσ = 7 cm20%
x = cm
7.Confidence intervalsMedium
n = 13, sd​ = 3.40LU = ?d̄ = 5.7E90%

The blood pressure of 13 patients is measured before and after a treatment. The differences d = before − after (mmHg) have mean 5.7 and standard deviation 3.4. Find the upper limit of the 90% confidence interval for the mean difference.

n = 13d̄ = 5.7sd = 3.490%t(12) = 1.782
upper limit = mmHg
8.Confidence intervalsHard
σ = 11, n = ?x̄ − Ex̄ + Ex̄E = 0.595%

The population standard deviation is 11. How large a sample n do you need so that a 95% confidence interval for the mean has a margin of error of at most 0.5?

σ = 11E = 0.595%
n =
9.Hypothesis tests & confidence intervalsMedium
−1.081.08p = ?

The mean of a population is claimed to be 100; the population standard deviation is known to be 15. A random sample of 36 observations has a mean of 97.3. Test H₀: μ = 100 against H₁: μ ≠ 100. Find the p-value p.

μ₀ = 100σ = 15n = 36x̄ = 97.3H₁: μ ≠ μ₀
p =
10.Hypothesis tests & confidence intervalsHard
−1.961.96z = ?

In a trial, 110 of 200 patients on treatment A recover, and 129 of 300 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.

x₁ = 110n₁ = 200x₂ = 129n₂ = 300
z =
11.Linear regression & correlationMedium
xy274156238301042xy5102040

The table gives 5 paired observations of x = load in kN and y = extension in mm. Find Pearson's correlation coefficient r.

n = 5
r =
12.Linear regression & correlationHard
xy1649186320722280249426102xy20256080100ŷ = 56

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 = 56?

n = 6ŷ = 56
x =

3Solutions

1.Descriptive statistics

Answer: s = 14.91

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 = 203/7 = 29

Squared deviations

Σ(x − x̄)² = (−19)² + (−6)² + 2² + 26² + 5² + (−14)² + 6² = 1334

Standard deviation

s = √(1334/6) = 14.91

Answer

s = 14.91
2.Descriptive statistics

Answer: P = 0.9772

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 = 1 − 0.02275 = 0.9772

Answer

P = 0.9772
3.Probability

Answer: Var(X) = 1.56

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 = 0·0.1 + 1·0.25 + 2·0.15 + 3·0.35 + 4·0.15 = 2.2

Mean of the squares

E(X²) = 0·0.1 + 1·0.25 + 4·0.15 + 9·0.35 + 16·0.15 = 6.4

Variance

Var(X) = 6.4 − 2.2² = 1.56

Answer

Var(X) = 1.56
4.Probability

Answer: σY = 4.023

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 = 1.05, Var(X) = 0.6475, σX = 0.8047

Linear change

σY = |5|·0.8047 = 4.023

Answer

σY = 4.023

The mean does move: E(Y) = 5·1.05 + 1 = 6.25.

5.Normal distribution

Answer: P = 0.6915

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 = 750 g, σ = √(4² + 9²) = 9.849 g

Standardise

z = (745.1 − 750)/9.849 = −0.5

Upper tail

P = 1 − Φ(−0.5) = 1 − 0.3085 = 0.6915

Answer

P = 0.6915
6.Normal distribution

Answer: x = 183.9 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) = 0.8

z from the table

z ≈ 0.84

Back to the original units

x = 178 + 0.84·7 = 183.9 cm

Answer

x = 183.9 cm
7.Confidence intervals

Answer: upper limit = 7.38 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 = 1.782

Margin of error

E = 1.782·3.4/√13 = 1.68

Interval

5.7 ± 1.68 → [4.02, 7.38]

Answer

upper limit = 7.38 mmHg
8.Confidence intervals

Answer: n = 1860

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 = (1.96·11/0.5)² = 1859

Round up

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

n = 1860

Answer

n = 1860
9.Hypothesis tests & confidence intervals

Answer: p = 0.2801

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.

Test statistic

SE = σ/√n = 15/√36 = 2.5
z = (97.3 − 100)/2.5 = −1.08

Tail area

Φ(1.08) = 0.8599
1 − Φ(1.08) = 0.1401

p-value

p = 2·0.1401 = 0.2801

Answer

p = 0.2801

p > 0.05, so H₀ is not rejected at the 5% level.

10.Hypothesis tests & confidence intervals

Answer: z = 2.632

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.

Sample proportions

p̂₁ = 110/200 = 0.55
p̂₂ = 129/300 = 0.43
p̂ = 239/500 = 0.478

Standard error under H₀

SE = √(0.478·0.522·(1/200 + 1/300)) = 0.0456

Test statistic

z = (0.55 − 0.43)/0.0456 = 2.632

Answer

z = 2.632

|z| > 1.96, so H₀ is rejected at the 5% level (two-sided).

11.Linear regression & correlation

Answer: r = 0.9954

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 = 30/5 = 6
ȳ = 117/5 = 23.4

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 170
Syy = Σ(y − ȳ)² = 729.2

Correlation coefficient

r = 170/√(40·729.2) = 0.9954

Answer

r = 0.9954
12.Linear regression & correlation

Answer: x = 17.05

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 = 460/6 = 76.67

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 366

Regression line

b = 366/70 = 5.229
a = 76.67 − 5.229·21 = −33.13
ŷ = −33.13 + 5.229·x

Solve for x

x = (56 − (−33.13))/5.229 = 17.05

Answer

x = 17.05

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