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
152025303540455055

Find the sample standard deviation s of the data: 32, 17, 13, 57, 21, 22, 18.

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

A quantity is normally distributed with mean 70 and standard deviation 12. What proportion P of values is above 88?

μ = 70σ = 12x = 88
P =
3.ProbabilityMedium
xp(x)10.120.2530.3540.3μ = 2.85

For the distribution shown, find the standard deviation σ of X.

x: 1, 2, 3, 4p(x): 0.1, 0.25, 0.35, 0.3
σ =
4.ProbabilityHard
n = 8, p = 0.3: P(X ≤ 1) = ?k012345678

8 components are tested, and each one fails with probability 0.3, independently. What is the probability P that at most one fails?

n = 8p = 0.3
P =
5.Normal distributionMedium
−3−2−10123zshaded area to the right of z = 0.9495

Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.9495. Find z to two decimals.

A = 0.9495
z =
6.Normal distributionHard
μ = 250x = ?75%

The bag weights are normally distributed with mean 250 g and standard deviation 5 g. Find the value x below which 75% of them lie (the 75th percentile).

μ = 250σ = 5P = 75%
x = g
7.Confidence intervalsMedium
μ₁ − μ₂0LUx̄₁ − x̄₂E = ?95%

The battery life of two phone models is compared. Model 1: 60 phones, mean 54 h, known standard deviation 11 h. Model 2: 40 phones, mean 44 h, known standard deviation 19 h. Find the margin of error E of the 95% confidence interval for μ₁ − μ₂.

n₁ = 60x̄₁ = 54 hσ₁ = 11 hn₂ = 40x̄₂ = 44 hσ₂ = 19 h95%
E = h
8.Confidence intervalsHard
σ = 10, n = ?x̄ − Ex̄ + Ex̄E = 1.590%

The population standard deviation is 10. How large a sample n do you need so that a 90% confidence interval for the mean has a margin of error of at most 1.5?

σ = 10E = 1.590%
n =
9.Hypothesis tests & confidence intervalsMedium
t −1.98401.984

A sample of 100 students has a mean score of 73.24 and a sample standard deviation of 8. Find the test statistic t for H₀: μ = 75.

μ₀ = 75x̄ = 73.24s = 8n = 100
t =
10.Hypothesis tests & confidence intervalsHard
z −1.9601.96

A sample of 100 measurements has mean 118.6. The population standard deviation is 12. Find the upper limit U of the 95% confidence interval for the mean.

x̄ = 118.6σ = 12n = 10095%
U =
11.Linear regression & correlationMedium
xy234961381910201226xy5101020

The table gives 6 paired observations of x = load in kN and y = extension in mm. Find the coefficient of determination r² of the least-squares line.

n = 6
r² =
12.Linear regression & correlationHard
xy155018592184248527104xy1520256080100ŷ = 97

The table gives 5 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 = 97?

n = 5ŷ = 97
x =

3Solutions

1.Descriptive statistics

Answer: s = 15.01

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 = 180/7 = 25.71

Squared deviations

Σ(x − x̄)² = 6.286² + (−8.714)² + (−12.71)² + 31.29² + (−4.714)² + (−3.714)² + (−7.714)² = 1351

Standard deviation

s = √(1351/6) = 15.01

Answer

s = 15.01
2.Descriptive statistics

Answer: P = 0.06681

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 = 1.5

Standard normal table

Φ(1.5) = 0.9332
P = 1 − 0.9332 = 0.06681

Answer

P = 0.06681
3.Probability

Answer: σ = 0.9631

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 = 1·0.1 + 2·0.25 + 3·0.35 + 4·0.3 = 2.85

Mean of the squares

E(X²) = 1·0.1 + 4·0.25 + 9·0.35 + 16·0.3 = 9.05

Variance

Var(X) = 9.05 − 2.85² = 0.9275
σ = √0.9275 = 0.9631

Answer

σ = 0.9631
4.Probability

Answer: P = 0.2553

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.

Zero and one failure

P(X = 0) = 0.78 = 0.05765
P(X = 1) = 8·0.3·0.77 = 0.1977

Add them

P = 0.05765 + 0.1977 = 0.2553

Answer

P = 0.2553
5.Normal distribution

Answer: z = −1.64

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 − 0.9495 = 0.0505

Read the z-table backwards

Find 0.0505 inside the table and read off z.

z = −1.64

Answer

z = −1.64
6.Normal distribution

Answer: x = 253.4 g

Normal distribution: convert a value x to a z-score, z = (x − μ)/σ, which counts standard deviations from the mean. A z-table (or Φ(z)) gives the area to the left of z, which is P(X < x). The area to the right is 1 − Φ(z), and the area between two values is the difference of their two areas.

z for the percentile

Φ(z) = 0.75 → z = 0.6745

Back to x

x = 250 + 0.6745·5 = 253.37 g

Answer

x = 253.37 g
7.Confidence intervals

Answer: E = 6.513 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 = 54 − 44 = 10

Standard error

SE = √(11²/60 + 19²/40) = 3.323

Margin of error

E = 1.96·3.323 = 6.513

Answer

E = 6.513 h

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

8.Confidence intervals

Answer: n = 121

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 = (1.645·10/1.5)² = 120.3

Round up

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

n = 121

Answer

n = 121
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 = 8/√100 = 8/10 = 0.8

Test statistic

t = (73.24 − 75)/0.8 = −2.2

Answer

t = −2.2
t0.025; df = 99 = 1.984

|t| > 1.984, so H₀ is rejected at the 5% level (two-sided).

10.Hypothesis tests & confidence intervals

Answer: U = 121

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).

Margin of error

E = 1.96·12/√100 = 2.352

Interval limit

U = 118.6 + 2.352 = 121

Answer

U = 121
11.Linear regression & correlation

Answer: r² = 0.9792

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 = 42/6 = 7
ȳ = 90/6 = 15

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 154
Syy = Σ(y − ȳ)² = 346

Coefficient of determination

r² = 154²/(70·346) = 0.9792

Answer

r² = 0.9792

About 97.92% of the variation in y is explained by the line.

12.Linear regression & correlation

Answer: x = 25.61

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 = 105/5 = 21
ȳ = 382/5 = 76.4

Sums of squares and products

Sxx = Σ(x − x̄)² = 90
Sxy = Σ(x − x̄)(y − ȳ) = 402

Regression line

b = 402/90 = 4.467
a = 76.4 − 4.467·21 = −17.4
ŷ = −17.4 + 4.467·x

Solve for x

x = (97 − (−17.4))/4.467 = 25.61

Answer

x = 25.61

ExamProblems · examproblems.com/courses/statistics-1