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
Introductory Statistics: cram pack
Formula sheets, the key method steps for every topic and 12 practice problems, with full solutions at the back.
1Formulas and methods
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
- Convert the value to a z-score: z = (x − μ)/σ.
- Look up Φ(z), the area to the left of z.
- For the area to the right use 1 − Φ(z); for an interval subtract the two left areas.
- 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
- Decide between z (population σ known) and t (only the sample s, with n − 1 degrees of freedom).
- Find the critical value for the confidence level.
- Margin of error: E = critical value × standard deviation/√n.
- 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
- State H₀ and H₁ and decide whether the test is one- or two-sided.
- Calculate the test statistic: (x̄ − μ₀)/(σ/√n), or with s and t.
- Compare with the critical value, or find the p-value.
- 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
Find the sample standard deviation s of the data: 41, 27, 11, 55, 32, 53, 14.
n = 7
s =
2.Descriptive statisticsHard
Exam scores are normally distributed with mean 50 and standard deviation 12. Find the z-score z of a score of 74.
μ = 50σ = 12x = 74
z =
3.ProbabilityMedium
A bag holds 3 red and 4 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?
3 / 4
P =
4.ProbabilityHard
11 components are tested, and each one fails with probability 0.25, independently. What is the probability P that at most one fails?
n = 11p = 0.25
P =
5.Normal distributionMedium
The exam scores are normally distributed with mean 65 points and standard deviation 10 points. What fraction lies between 55 and 75.5 points? Give P as a decimal.
μ = 65σ = 10x1 = 55x2 = 75.5
P =
6.Normal distributionHard
The delivery times are normally distributed with mean 32 min and standard deviation 6 min. Find the value x below which 99% of them lie (the 99th percentile).
μ = 32σ = 6P = 99%
x = min
7.Confidence intervalsMedium
A sample of 7 items has mean 21 and standard deviation s = 7.5. The population is roughly normal. Find the upper limit of the 90% confidence interval for the mean.
n = 7x̄ = 21s = 7.590%t(6) = 1.943
upper limit =
8.Confidence intervalsHard
The population standard deviation is 38. 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 2.5?
σ = 38E = 2.595%
n =
9.Hypothesis tests & confidence intervalsMedium
A sample of 25 students has a mean score of 77.1 and a sample standard deviation of 5. Find the test statistic t for H₀: μ = 75.
μ₀ = 75x̄ = 77.1s = 5n = 25
t =
10.Hypothesis tests & confidence intervalsHard
A one-sided z test at α = 0.01 must detect a shift of the mean of 2 units with a power of 80%. The standard deviation is 6. Find the minimum sample size n.
α = 0.01power = 80%δ = 2σ = 6zα = 2.326zβ = 0.842
n =
11.Linear regression & correlationMedium
The table gives 5 paired observations of x = car age in years and y = price in k€. Their sums are given. Find Pearson's correlation coefficient r from these sums.
n = 5Σx = 15Σy = 90Σx² = 55Σy² = 1690Σxy = 245
r =
12.Linear regression & correlationHard
The table gives 6 paired observations of x = hours studied and y = exam score. Use the least-squares line ŷ = a + b·x. A new observation (1.5, 51) is made. Find its residual e = y − ŷ.
n = 6(1.5, 51)
e =
3Solutions
1.Descriptive statistics
Answer: s = 17.46
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 = 233/7 = 33.29
Squared deviations
Σ(x − x̄)² = 7.714² + (−6.286)² + (−22.29)² + 21.71² + (−1.286)² + 19.71² + (−19.29)² = 1829
Standard deviation
s = √(1829/6) = 17.46
Answer
s = 17.46
2.Descriptive statistics
Answer: z = 2
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 = (74 − 50)/12 = 2
Answer
z = 2
3.Probability
Answer: P = 0.1429
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.
First and second draw
P = 3/7 · 2/6 = 6/42 = 0.1429
Answer
P = 0.1429
4.Probability
Answer: P = 0.1971
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) = 0.7511 = 0.04224
P(X = 1) = 11·0.25·0.7510 = 0.1549
Add them
P = 0.04224 + 0.1549 = 0.1971
Answer
P = 0.1971
5.Normal distribution
Answer: P = 0.6945
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-scores
z₁ = (55 − 65)/10 = −1
z₂ = (75.5 − 65)/10 = 1.05
Area between
P = Φ(1.05) − Φ(−1) = 0.8531 − 0.1587 = 0.6945
Answer
P = 0.6945
6.Normal distribution
Answer: x = 45.96 min
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) = 0.99 → z = 2.326
Back to x
x = 32 + 2.326·6 = 45.96 min
Answer
x = 45.96 min
7.Confidence intervals
Answer: upper limit = 26.51
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 = 6, t = 1.943
Margin of error
E = 1.943·7.5/√7 = 5.508
Interval
21 ± 5.508 → [15.49, 26.51]
Answer
upper limit = 26.51
8.Confidence intervals
Answer: n = 888
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 = (1.96·38/2.5)² = 887.6
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 888
Answer
n = 888
9.Hypothesis tests & confidence intervals
Answer: t = 2.1
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 = 5/√25 = 5/5 = 1
Test statistic
t = (77.1 − 75)/1 = 2.1
Answer
t = 2.1
t0.025; df = 24 = 2.064
|t| > 2.064, so H₀ is rejected at the 5% level (two-sided).
10.Hypothesis tests & confidence intervals
Answer: n = 91
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 = ((2.326 + 0.842)·6/2)² = 90.33
Round up
n = 91
Answer
n = 91
11.Linear regression & correlation
Answer: r = −0.9449
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 = 245 − 15·90/5 = −25
Sxx = 55 − 15²/5 = 10
Syy = 1690 − 90²/5 = 70
Correlation coefficient
r = −25/√(10·70) = −0.9449
Answer
r = −0.9449
12.Linear regression & correlation
Answer: e = −5.286
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 = 372/6 = 62
Sums of squares and products
Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = 50
Regression line
b = 50/17.5 = 2.857
a = 62 − 2.857·3.5 = 52
ŷ = 52 + 2.857·x
Predicted value
ŷ = 52 + 2.857·1.5 = 56.29
Residual
e = 51 − 56.29 = −5.286
Answer
e = −5.286
The point lies below the line.
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