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
15202530354045505560

Find the sample standard deviation s of the data: 37, 32, 39, 60, 13.

n = 5
s =
2.Descriptive statisticsHard
z −2−1012

A quantity is normally distributed with mean 60 and standard deviation 5. What proportion P of values is above 54?

μ = 60σ = 5x = 54
P =
3.ProbabilityMedium
xp(x)10.1520.330.340.25μ = 2.65

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

x: 1, 2, 3, 4p(x): 0.15, 0.3, 0.3, 0.25
Var(X) =
4.ProbabilityHard
n = 10, p = 0.6: P(X = 5) = ?k012345678910

A test has 10 multiple-choice questions. A student who guesses gets each one right with probability 0.6, independently. What is the probability P of exactly 5 correct answers?

n = 10p = 0.6k = 5
P =
5.Normal distributionMedium
−3−2−10123zshaded area between −z and z = 0.3688

Drag z so that the shaded area between −z and z under the standard normal curve is 0.3688. Find z to two decimals.

A = 0.3688
z =
6.Normal distributionHard
380420460500540580620hbottom 5%

The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. Drag the cut-off so the shaded bottom 5% is to the left of it. Find the cut-off x.

μ = 500 hσ = 40 h5%
x = h
7.Confidence intervalsMedium
μ₁ − μ₂0LUx̄₁ − x̄₂E = ?90%

The battery life of two phone models is compared. Model 1: 60 phones, mean 44.5 h, known standard deviation 7 h. Model 2: 30 phones, mean 50.5 h, known standard deviation 14 h. Find the margin of error E of the 90% confidence interval for μ₁ − μ₂.

n₁ = 60x̄₁ = 44.5 hσ₁ = 7 hn₂ = 30x̄₂ = 50.5 hσ₂ = 14 h90%
E = h
8.Confidence intervalsHard
σ = 32, n = ?x̄ − Ex̄ + Ex̄E = 495%

The population standard deviation is 32. 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 4?

σ = 32E = 495%
n =
9.Hypothesis tests & confidence intervalsMedium
t −1.99801.998

A sample of 64 students has a mean score of 48.9 and a sample standard deviation of 4. Find the test statistic t for H₀: μ = 50.

μ₀ = 50x̄ = 48.9s = 4n = 64
t =
10.Hypothesis tests & confidence intervalsHard
−1.961.96z = ?

In a trial, 180 of 300 patients on treatment A recover, and 168 of 300 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.

x₁ = 180n₁ = 300x₂ = 168n₂ = 300
z =
11.Linear regression & correlationMedium
xy1655196722772585289931119xy152025306080100120

The table gives 6 paired observations of x = temperature in °C and y = ice creams sold. Find the coefficient of determination r² of the least-squares line.

n = 6
r² =
12.Linear regression & correlationHard
xy126223325419520xy242025x = 3.5

The table gives 5 paired observations of x = car age in years and y = price in k€. Use the least-squares line ŷ = a + b·x. A new observation (3.5, 26) is made. Find its residual e = y − ŷ.

n = 5(3.5, 26)
e =

3Solutions

1.Descriptive statistics

Answer: s = 16.81

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 = 181/5 = 36.2

Squared deviations

Σ(x − x̄)² = 0.8² + (−4.2)² + 2.8² + 23.8² + (−23.2)² = 1131

Standard deviation

s = √(1131/4) = 16.81

Answer

s = 16.81
2.Descriptive statistics

Answer: P = 0.8849

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 = −1.2

Standard normal table

Φ(−1.2) = 0.1151
P = 1 − 0.1151 = 0.8849

Answer

P = 0.8849
3.Probability

Answer: Var(X) = 1.028

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 = 1·0.15 + 2·0.3 + 3·0.3 + 4·0.25 = 2.65

Mean of the squares

E(X²) = 1·0.15 + 4·0.3 + 9·0.3 + 16·0.25 = 8.05

Variance

Var(X) = 8.05 − 2.65² = 1.028

Answer

Var(X) = 1.028
4.Probability

Answer: P = 0.2007

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.

Binomial formula

P(X = 5) = 252·0.65·0.45 = 252·0.07776·0.01024 = 0.2007

Answer

P = 0.2007
5.Normal distribution

Answer: z = 0.48

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 + 0.3688)/2 = 0.6844

Read the z-table backwards

Find 0.6844 inside the table and read off z.

z = 0.48

Answer

z = 0.48
6.Normal distribution

Answer: x = 434 h

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) = 0.05

z from the table

z ≈ −1.64

Back to the original units

x = 500 − 1.64·40 = 434 h

Answer

x = 434 h
7.Confidence intervals

Answer: E = 4.46 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 = 44.5 − 50.5 = −6

Standard error

SE = √(7²/60 + 14²/30) = 2.711

Margin of error

E = 1.645·2.711 = 4.46

Answer

E = 4.46 h

The interval does not contain 0, so the means differ significantly.

8.Confidence intervals

Answer: n = 246

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 = (1.96·32/4)² = 245.9

Round up

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

n = 246

Answer

n = 246
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 = 4/√64 = 4/8 = 0.5

Test statistic

t = (48.9 − 50)/0.5 = −2.2

Answer

t = −2.2
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: z = 0.9926

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 = 180/300 = 0.6
p̂₂ = 168/300 = 0.56
p̂ = 348/600 = 0.58

Standard error under H₀

SE = √(0.58·0.42·(1/300 + 1/300)) = 0.0403

Test statistic

z = (0.6 − 0.56)/0.0403 = 0.9926

Answer

z = 0.9926

|z| < 1.96, so H₀ is not rejected at the 5% level (two-sided).

11.Linear regression & correlation

Answer: r² = 0.9768

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 = 141/6 = 23.5
ȳ = 502/6 = 83.67

Sums of squares and products

Sxx = Σ(x − x̄)² = 157.5
Sxy = Σ(x − x̄)(y − ȳ) = 636
Syy = Σ(y − ȳ)² = 2629

Coefficient of determination

r² = 636²/(157.5·2629) = 0.9768

Answer

r² = 0.9768

About 97.68% of the variation in y is explained by the line.

12.Linear regression & correlation

Answer: e = 4.2

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 = 113/5 = 22.6

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = −16

Regression line

b = −16/10 = −1.6
a = 22.6 − (−1.6)·3 = 27.4
ŷ = 27.4 − 1.6·x

Predicted value

ŷ = 27.4 + (−1.6)·3.5 = 21.8

Residual

e = 26 − 21.8 = 4.2

Answer

e = 4.2

The point lies above the line.

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