Introductory Statistics: cram pack

Formula sheets, the key method steps for every topic and 12 practice problems, with full solutions at the back.

New problemsBack to the course

In the print window, choose “Save as PDF” to keep a copy on your device.

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: 39, 14, 32, 33, 51.

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

A quantity is normally distributed with mean 170 and standard deviation 10. What proportion P of values is above 182.5?

μ = 170σ = 10x = 182.5
P =
3.ProbabilityMedium
xp(x)00.2510.3520.3530.05μ = 1.2

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

x: 0, 1, 2, 3p(x): 0.25, 0.35, 0.35, 0.05
Var(X) =
4.ProbabilityHard
n = 12, p = 0.7: P(X = 2) = ?k0123456789101112

A test has 12 multiple-choice questions. A student who guesses gets each one right with probability 0.7, independently. What is the probability P of exactly 2 correct answers?

n = 12p = 0.7k = 2
P =
5.Normal distributionMedium
−3−2−10123zshaded area between −z and z = 0.9736

Drag z so that the shaded area between −z and z under the standard normal curve is 0.9736. Find z to two decimals.

A = 0.9736
z =
6.Normal distributionHard
0.03 mm0D − d?

Hole diameters D are normal with mean 25.03 mm and standard deviation 0.01 mm. Shaft diameters d are independent and normal with mean 25 mm and standard deviation 0.01 mm. A randomly chosen shaft fits when D − d > 0. Find the probability P that it fits. Round z to two decimals.

hole: μ = 25.03 mm, σ = 0.01 mmshaft: μ = 25 mm, σ = 0.01 mm
P =
7.Confidence intervalsMedium
n = 12, s = 1LUx̄ = 63.5E = ?99%

A sample of 12 items has mean 63.5 and standard deviation s = 1. The population is roughly normal. Find the margin of error E of a 99% confidence interval for the mean.

n = 12x̄ = 63.5s = 199%t(11) = 3.106
E =
8.Confidence intervalsHard
σ = 13, n = ?x̄ − Ex̄ + Ex̄E = 495%

The population standard deviation is 13. 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 = 13E = 495%
n =
9.Hypothesis tests & confidence intervalsMedium
−2.0932.093t = ?df = 19

20 employees take a typing test before and after a training course. For each person the difference d = after − before (words per minute) is recorded. The mean difference is 3.78 and the standard deviation of the differences is 8.9. Find the paired test statistic t for H₀: μd = 0.

n = 20d̄ = 3.78sd = 8.9
t =
10.Hypothesis tests & confidence intervalsHard
−1.961.96z = ?

In a trial, 30 of 100 patients on treatment A recover, and 70 of 200 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.

x₁ = 30n₁ = 100x₂ = 70n₂ = 200
z =
11.Linear regression & correlationMedium
xy165219642287251022811531128xy1520253050100

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
xy350455568667772xy46506070x = 4.5

The table gives 5 paired observations of x = advertising spend in k€ and y = sales in k€. Use the least-squares line ŷ = a + b·x. A new observation (4.5, 56) is made. Find its residual e = y − ŷ.

n = 5(4.5, 56)
e =

3Solutions

1.Descriptive statistics

Answer: s = 13.41

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 = 169/5 = 33.8

Squared deviations

Σ(x − x̄)² = 5.2² + (−19.8)² + (−1.8)² + (−0.8)² + 17.2² = 718.8

Standard deviation

s = √(718.8/4) = 13.41

Answer

s = 13.41
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: Var(X) = 0.76

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 = 0·0.25 + 1·0.35 + 2·0.35 + 3·0.05 = 1.2

Mean of the squares

E(X²) = 0·0.25 + 1·0.35 + 4·0.35 + 9·0.05 = 2.2

Variance

Var(X) = 2.2 − 1.2² = 0.76

Answer

Var(X) = 0.76
4.Probability

Answer: P = 0.000191

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 = 2) = 66·0.72·0.310 = 66·0.49·5.905 × 10⁻⁶ = 0.000191

Answer

P = 0.000191
5.Normal distribution

Answer: z = 2.22

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 + 0.9736)/2 = 0.9868

Read the z-table backwards

Find 0.9868 inside the table and read off z.

z = 2.22

Answer

z = 2.22
6.Normal distribution

Answer: P = 0.983

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 difference

μ = 25.03 − 25 = 0.03 mm, σ = 0.01414 mm

Standardise

z = (0 − 0.03)/0.01414 = −2.12

Probability

P = 1 − Φ(−2.12) = 0.983

Answer

P = 0.983
7.Confidence intervals

Answer: E = 0.8966

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.

Critical value

df = n − 1 = 11, t = 3.106

Margin of error

E = 3.106·1/√12 = 0.8966

Answer

E = 0.8966
8.Confidence intervals

Answer: n = 41

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 = (1.96·13/4)² = 40.58

Round up

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

n = 41

Answer

n = 41
9.Hypothesis tests & confidence intervals

Answer: t = 1.899

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 of the mean difference

SE = 8.9/√20 = 1.99

Test statistic

t = 3.78/1.99 = 1.899

Answer

t = 1.899

With df = 19, the two-sided 5% critical value is 2.093; |t| is smaller, so H₀ is not rejected.

10.Hypothesis tests & confidence intervals

Answer: z = −0.866

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 = 30/100 = 0.3
p̂₂ = 70/200 = 0.35
p̂ = 100/300 = 0.3333

Standard error under H₀

SE = √(0.3333·0.6667·(1/100 + 1/200)) = 0.05774

Test statistic

z = (0.3 − 0.35)/0.05774 = −0.866

Answer

z = −0.866

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

11.Linear regression & correlation

Answer: r² = 0.9905

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 = 548/6 = 91.33

Sums of squares and products

Sxx = Σ(x − x̄)² = 157.5
Sxy = Σ(x − x̄)(y − ȳ) = 822
Syy = Σ(y − ȳ)² = 4331

Coefficient of determination

r² = 822²/(157.5·4331) = 0.9905

Answer

r² = 0.9905

About 99.05% of the variation in y is explained by the line.

12.Linear regression & correlation

Answer: e = −3.6

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 = 25/5 = 5
ȳ = 312/5 = 62.4

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = 56

Regression line

b = 56/10 = 5.6
a = 62.4 − 5.6·5 = 34.4
ŷ = 34.4 + 5.6·x

Predicted value

ŷ = 34.4 + 5.6·4.5 = 59.6

Residual

e = 56 − 59.6 = −3.6

Answer

e = −3.6

The point lies below the line.

ExamProblems · examproblems.com/courses/statistics-1