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
2025303540455055

Find the sample standard deviation s of the data: 43, 16, 55, 56, 20.

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

Exam scores are normally distributed with mean 60 and standard deviation 20. Find the z-score z of a score of 72.

μ = 60σ = 20x = 72
z =
3.ProbabilityMedium
xp(x)10.3520.3530.1540.15μ = 2.1

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

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

X has the distribution shown, and Y = 5X + 4. Find the standard deviation σY of Y.

x: 0, 1, 2p(x): 0.3, 0.35, 0.35Y = 5X + 4
σY =
5.Normal distributionMedium
−3−2−10123zshaded area to the right of z = 0.0256

Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.0256. Find z to two decimals.

A = 0.0256
z =
6.Normal distributionHard
557085100115130145pointstop 80%

The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. Drag the cut-off so the shaded top 80% is to the right of it. Find the cut-off x.

μ = 100 pointsσ = 15 points80%
x = points
7.Confidence intervalsMedium
n = 6, s = 11.5LU = ?x̄ = 118E99%

A sample of 6 items has mean 118 and standard deviation s = 11.5. The population is roughly normal. Find the upper limit of the 99% confidence interval for the mean.

n = 6x̄ = 118s = 11.599%t(5) = 4.032
upper limit =
8.Confidence intervalsHard
σ = 9, n = ?x̄ − Ex̄ + Ex̄E = 199%

The population standard deviation is 9. How large a sample n do you need so that a 99% confidence interval for the mean has a margin of error of at most 1?

σ = 9E = 199%
n =
9.Hypothesis tests & confidence intervalsMedium
t −2.01102.011

A sample of 49 students has a mean score of 196.2 and a sample standard deviation of 12. Find the test statistic t for H₀: μ = 200.

μ₀ = 200x̄ = 196.2s = 12n = 49
t =
10.Hypothesis tests & confidence intervalsHard
H₀H₁μ₀x̄cμ₀ + 2α = 0.05βσ = 18, power 90%, n = ?

A one-sided z test at α = 0.05 must detect a shift of the mean of 2 units with a power of 90%. The standard deviation is 18. Find the minimum sample size n.

α = 0.05power = 90%δ = 2σ = 18zα = 1.645zβ = 1.282
n =
11.Linear regression & correlationMedium
xy130229327426522620xy246202530

The table gives 6 paired observations of x = car age in years and y = price in k€. Find Pearson's correlation coefficient r.

n = 6
r =
12.Linear regression & correlationHard
xy2104206278361040xy51010203040x = 3

The table gives 5 paired observations of x = load in kN and y = extension in mm. Use the least-squares line ŷ = a + b·x. A new observation (3, 19) is made. Find its residual e = y − ŷ.

n = 5(3, 19)
e =

3Solutions

1.Descriptive statistics

Answer: s = 19.01

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 = 190/5 = 38

Squared deviations

Σ(x − x̄)² = 5² + (−22)² + 17² + 18² + (−18)² = 1446

Standard deviation

s = √(1446/4) = 19.01

Answer

s = 19.01
2.Descriptive statistics

Answer: z = 0.6

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 = (72 − 60)/20 = 0.6

Answer

z = 0.6
3.Probability

Answer: Var(X) = 1.09

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 = 1·0.35 + 2·0.35 + 3·0.15 + 4·0.15 = 2.1

Mean of the squares

E(X²) = 1·0.35 + 4·0.35 + 9·0.15 + 16·0.15 = 5.5

Variance

Var(X) = 5.5 − 2.1² = 1.09

Answer

Var(X) = 1.09
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 + 4 = 9.25.

5.Normal distribution

Answer: z = 1.95

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 − 0.0256 = 0.9744

Read the z-table backwards

Find 0.9744 inside the table and read off z.

z = 1.95

Answer

z = 1.95
6.Normal distribution

Answer: x = 87.4 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 = 100 − 0.84·15 = 87.4 points

Answer

x = 87.4 points
7.Confidence intervals

Answer: upper limit = 136.9

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 = 5, t = 4.032

Margin of error

E = 4.032·11.5/√6 = 18.93

Interval

118 ± 18.93 → [99.07, 136.9]

Answer

upper limit = 136.9
8.Confidence intervals

Answer: n = 538

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 = (2.576·9/1)² = 537.5

Round up

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

n = 538

Answer

n = 538
9.Hypothesis tests & confidence intervals

Answer: t = −2.217

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 = 12/√49 = 12/7 = 1.714

Test statistic

t = (196.2 − 200)/1.714 = −2.217

Answer

t = −2.217
t0.025; df = 48 = 2.011

|t| > 2.011, so H₀ is rejected at the 5% level (two-sided).

10.Hypothesis tests & confidence intervals

Answer: n = 694

Sample size for a test: to detect a true shift δ with power 1 − β in a one-sided z test at level α, the standard error must satisfy δ = (z_α + z_β)·σ/√n, so n = ((z_α + z_β)·σ/δ)², rounded up.

Solve for n

n = ((1.645 + 1.282)·18/2)² = 694

Round up

n = 694

Answer

n = 694
11.Linear regression & correlation

Answer: r = −0.9786

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 = 21/6 = 3.5
ȳ = 154/6 = 25.67

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −36
Syy = Σ(y − ȳ)² = 77.33

Correlation coefficient

r = −36/√(17.5·77.33) = −0.9786

Answer

r = −0.9786
12.Linear regression & correlation

Answer: e = 3.8

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 = 133/5 = 26.6

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 152

Regression line

b = 152/40 = 3.8
a = 26.6 − 3.8·6 = 3.8
ŷ = 3.8 + 3.8·x

Predicted value

ŷ = 3.8 + 3.8·3 = 15.2

Residual

e = 19 − 15.2 = 3.8

Answer

e = 3.8

The point lies above the line.

ExamProblems · examproblems.com/courses/statistics-1