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
1015202530354045

Find the sample standard deviation s of the data: 17, 46, 49, 10, 27, 44.

n = 6
s =
2.Descriptive statisticsHard
z −2−1012

Exam scores are normally distributed with mean 70 and standard deviation 20. Find the z-score z of a score of 82.

μ = 70σ = 20x = 82
z =
3.ProbabilityMedium
n = 6k = 3N = ?

In how many ways N can a committee of 3 people be chosen from a group of 6? The order does not matter.

n = 6k = 3
N =
4.ProbabilityHard
xp(x)10.320.230.0540.3550.1μ

X has the distribution shown, and Y = 2X + 5. Find the standard deviation σY of Y.

x: 1, 2, 3, 4, 5p(x): 0.3, 0.2, 0.05, 0.35, 0.1Y = 2X + 5
σY =
5.Normal distributionMedium
72 kg73 kgT = X₁ + … + X₆?

The masses of boxes are independent and normal with mean 12 kg and standard deviation 0.8 kg. Find the probability P that the total mass of 6 of them exceeds 73 kg. Round z to two decimals.

μ = 12 kgσ = 0.8 kgn = 6> 73 kg
P =
6.Normal distributionHard
μ = 250x = ?20%

The bag weights are normally distributed with mean 250 g and standard deviation 5 g. Find the value x below which 20% of them lie (the 20th percentile).

μ = 250σ = 5P = 20%
x = g
7.Confidence intervalsMedium
n = 30, s = 16.5LUx̄ = 53E = ?99%

A sample of 30 items has mean 53 and standard deviation s = 16.5. The population is roughly normal. Find the margin of error E of a 99% confidence interval for the mean.

n = 30x̄ = 53s = 16.599%t(29) = 2.756
E =
8.Confidence intervalsHard
n = 20σ = ?Us = 4.895%

The diameters of 20 machined shafts have a sample standard deviation of 4.8 µm. Assuming the diameters are normally distributed, find the lower limit σ of the 95% confidence interval for the population standard deviation. With df = 19, 2.5% of the χ² distribution lies below 8.907 and 2.5% lies above 32.852.

n = 20s = 4.8 µmχ² = 8.907, 32.852
σ = µm
9.Hypothesis tests & confidence intervalsMedium
t −2.01102.011

A sample of 49 students has a mean score of 199.5 and a sample standard deviation of 4. Find the test statistic t for H₀: μ = 200.

μ₀ = 200x̄ = 199.5s = 4n = 49
t =
10.Hypothesis tests & confidence intervalsHard
H₀H₁μ₀x̄cμ₀ + 5α = 0.05βσ = 30, power 90%, n = ?

A one-sided z test at α = 0.05 must detect a shift of the mean of 5 units with a power of 90%. The standard deviation is 30. Find the minimum sample size n.

α = 0.05power = 90%δ = 5σ = 30zα = 1.645zβ = 1.282
n =
11.Linear regression & correlationMedium
xy1780199821104231172511627138xy202580100120140

The table gives 6 paired observations of x = temperature in °C and y = ice creams sold. Their sums are given. Find Pearson's correlation coefficient r from these sums.

n = 6Σx = 132Σy = 653Σx² = 2974Σy² = 73009Σxy = 14723
r =
12.Linear regression & correlationHard
xy17531968217723832596xy20256080100x = 18

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. Predict ŷ for x = 18.

n = 5x = 18
ŷ =

3Solutions

1.Descriptive statistics

Answer: s = 16.51

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 = 193/6 = 32.17

Squared deviations

Σ(x − x̄)² = (−15.17)² + 13.83² + 16.83² + (−22.17)² + (−5.167)² + 11.83² = 1363

Standard deviation

s = √(1363/5) = 16.51

Answer

s = 16.51
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 = (82 − 70)/20 = 0.6

Answer

z = 0.6
3.Probability

Answer: N = 20

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(6, 3) = (6·5·4)/(3·2·1) = 20

Answer

N = 20
4.Probability

Answer: σY = 2.89

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 = 2.75, Var(X) = 2.087, σX = 1.445

Linear change

σY = |2|·1.445 = 2.89

Answer

σY = 2.89

The mean does move: E(Y) = 2·2.75 + 5 = 10.5.

5.Normal distribution

Answer: P = 0.305

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 total

μT = 72 kg, σT = 0.8·√6 = 1.96 kg

Standardise

z = (73 − 72)/1.96 = 0.51

Probability

P = 1 − Φ(0.51) = 0.305

Answer

P = 0.305
6.Normal distribution

Answer: x = 245.8 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) = 0.2 → z = −0.8416

Back to x

x = 250 + −0.8416·5 = 245.79 g

Answer

x = 245.79 g
7.Confidence intervals

Answer: E = 8.302

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 = 29, t = 2.756

Margin of error

E = 2.756·16.5/√30 = 8.302

Answer

E = 8.302
8.Confidence intervals

Answer: σ = 3.65 µm

Interval for a variance: (n − 1)s²/σ² has a χ² distribution with n − 1 degrees of freedom. The 95% interval for σ² runs from (n − 1)s² divided by the upper χ² value (2.5% above it) to (n − 1)s² divided by the lower χ² value (2.5% below it); the interval for σ is the square root of both limits. It is not symmetric around s.

Sum of squares

(n − 1)s² = 19·4.8² = 437.8

Limits for σ²

437.8/32.852 = 13.33
437.8/8.907 = 49.15

Limit for σ

σ = √13.33 = 3.65 µm

Answer

σ = 3.65 µm

95% interval for σ: [3.65, 7.011] µm.

9.Hypothesis tests & confidence intervals

Answer: t = −0.875

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/√49 = 4/7 = 0.5714

Test statistic

t = (199.5 − 200)/0.5714 = −0.875

Answer

t = −0.875
t0.025; df = 48 = 2.011

|t| < 2.011, so H₀ is not rejected at the 5% level (two-sided).

10.Hypothesis tests & confidence intervals

Answer: n = 309

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)·30/5)² = 308.4

Round up

n = 309

Answer

n = 309
11.Linear regression & correlation

Answer: r = 0.9686

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.

Centred sums

Sxy = 14723 − 132·653/6 = 357
Sxx = 2974 − 132²/6 = 70
Syy = 73009 − 653²/6 = 1941

Correlation coefficient

r = 357/√(70·1941) = 0.9686

Answer

r = 0.9686
12.Linear regression & correlation

Answer: ŷ = 60.25

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 = 377/5 = 75.4

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 202

Regression line

b = 202/40 = 5.05
a = 75.4 − 5.05·21 = −30.65
ŷ = −30.65 + 5.05·x

Prediction

ŷ = −30.65 + 5.05·18 = 60.25

Answer

ŷ = 60.25

ExamProblems · examproblems.com/courses/statistics-1