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
15202530354045

Find the sample standard deviation s of the data: 13, 14, 21, 11, 45.

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

A quantity is normally distributed with mean 500 and standard deviation 50. What proportion P of values is above 450?

μ = 500σ = 50x = 450
P =
3.ProbabilityMedium
n = 10k = 2N = ?

In how many ways N can a committee of 2 people be chosen from a group of 10? The order does not matter.

n = 10k = 2
N =
4.ProbabilityHard
xp(x)100.35200.25300.1400.3μ

X has the distribution shown, and Y = 10X + 4. Find the standard deviation σY of Y.

x: 10, 20, 30, 40p(x): 0.35, 0.25, 0.1, 0.3Y = 10X + 4
σY =
5.Normal distributionMedium
−3−2−10123zshaded area between −z and z = 0.8415

Drag z so that the shaded area between −z and z under the standard normal curve is 0.8415. Find z to two decimals.

A = 0.8415
z =
6.Normal distributionHard
157164171178185192199cmtop 75%

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 75% is to the right of it. Find the cut-off x.

μ = 178 cmσ = 7 cm75%
x = cm
7.Confidence intervalsMedium
μ₁ − μ₂0L = ?Ux̄₁ − x̄₂E90%

The battery life of two phone models is compared. Model 1: 100 phones, mean 148 h, known standard deviation 5 h. Model 2: 40 phones, mean 152 h, known standard deviation 5 h. Find the lower limit of the 90% confidence interval for μ₁ − μ₂.

n₁ = 100x̄₁ = 148 hσ₁ = 5 hn₂ = 40x̄₂ = 152 hσ₂ = 5 h90%
lower limit = h
8.Confidence intervalsHard
σ = 34, n = ?x̄ − Ex̄ + Ex̄E = 399%

The population standard deviation is 34. 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 3?

σ = 34E = 399%
n =
9.Hypothesis tests & confidence intervalsMedium
t −2.0302.03

A sample of 36 students has a mean score of 504.8 and a sample standard deviation of 12. Find the test statistic t for H₀: μ = 500.

μ₀ = 500x̄ = 504.8s = 12n = 36
t =
10.Hypothesis tests & confidence intervalsHard
H₀H₁μ₀x̄cμ₀ + 2α = 0.05βσ = 15, power 80%, n = ?

A one-sided z test at α = 0.05 must detect a shift of the mean of 2 units with a power of 80%. The standard deviation is 15. Find the minimum sample size n.

α = 0.05power = 80%δ = 2σ = 15zα = 1.645zβ = 0.842
n =
11.Linear regression & correlationMedium
xy256467683810110107xy5106080100

The table gives 5 paired observations of x = hours studied and y = exam score. Find Pearson's correlation coefficient r.

n = 5
r =
12.Linear regression & correlationHard
xy3515607709791194xy5106080ŷ = 87

The table gives 5 paired observations of x = hours studied and y = exam score. Use the least-squares line ŷ = a + b·x. For which value x does the line predict ŷ = 87?

n = 5ŷ = 87
x =

3Solutions

1.Descriptive statistics

Answer: s = 14.04

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 = 104/5 = 20.8

Squared deviations

Σ(x − x̄)² = (−7.8)² + (−6.8)² + 0.2² + (−9.8)² + 24.2² = 788.8

Standard deviation

s = √(788.8/4) = 14.04

Answer

s = 14.04
2.Descriptive statistics

Answer: P = 0.8413

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

Standard normal table

Φ(−1) = 0.1587
P = 1 − 0.1587 = 0.8413

Answer

P = 0.8413
3.Probability

Answer: N = 45

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.

Combinations

C(10, 2) = (10·9)/(2·1) = 45

Answer

N = 45
4.Probability

Answer: σY = 123.6

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 = 23.5, Var(X) = 152.8, σX = 12.36

Linear change

σY = |10|·12.36 = 123.6

Answer

σY = 123.6

The mean does move: E(Y) = 10·23.5 + 4 = 239.

5.Normal distribution

Answer: z = 1.41

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 + 0.8415)/2 = 0.9207

Read the z-table backwards

Find 0.9207 inside the table and read off z.

z = 1.41

Answer

z = 1.41
6.Normal distribution

Answer: x = 173.3 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) = 0.25

z from the table

z ≈ −0.67

Back to the original units

x = 178 − 0.67·7 = 173.3 cm

Answer

x = 173.3 cm
7.Confidence intervals

Answer: lower limit = −5.539 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 = 148 − 152 = −4

Standard error

SE = √(5²/100 + 5²/40) = 0.9354

Margin of error

E = 1.645·0.9354 = 1.539

Interval

−4 ± 1.539 → [−5.539, −2.461]

Answer

lower limit = −5.539 h

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

8.Confidence intervals

Answer: n = 853

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 = (2.576·34/3)² = 852.3

Round up

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

n = 853

Answer

n = 853
9.Hypothesis tests & confidence intervals

Answer: t = 2.4

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/√36 = 12/6 = 2

Test statistic

t = (504.8 − 500)/2 = 2.4

Answer

t = 2.4
t0.025; df = 35 = 2.03

|t| > 2.03, so H₀ is rejected at the 5% level (two-sided).

10.Hypothesis tests & confidence intervals

Answer: n = 348

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 + 0.842)·15/2)² = 347.9

Round up

n = 348

Answer

n = 348
11.Linear regression & correlation

Answer: r = 0.9906

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 = 414/5 = 82.8

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 272
Syy = Σ(y − ȳ)² = 1885

Correlation coefficient

r = 272/√(40·1885) = 0.9906

Answer

r = 0.9906
12.Linear regression & correlation

Answer: x = 10.09

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 = 35/5 = 7
ȳ = 354/5 = 70.8

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 210

Regression line

b = 210/40 = 5.25
a = 70.8 − 5.25·7 = 34.05
ŷ = 34.05 + 5.25·x

Solve for x

x = (87 − 34.05)/5.25 = 10.09

Answer

x = 10.09

ExamProblems · examproblems.com/courses/statistics-1