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: 57, 59, 42, 53, 48, 56.
n = 6
s =
2.Descriptive statisticsHard
A quantity is normally distributed with mean 60 and standard deviation 12. What proportion P of values is above 67.2?
μ = 60σ = 12x = 67.2
P =
3.ProbabilityMedium
A fair die is rolled 8 times. What is the probability P of getting at least one six?
n = 8
P =
4.ProbabilityHard
X has the distribution shown, and Y = 3X + 20. Find the standard deviation σY of Y.
x: 1, 2, 3, 4p(x): 0.35, 0.3, 0.1, 0.25Y = 3X + 20
σY =
5.Normal distributionMedium
Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.0384. Find z to two decimals.
A = 0.0384
z =
6.Normal distributionHard
The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. Drag the cut-off so the shaded bottom 10% is to the left of it. Find the cut-off x.
μ = 100 pointsσ = 15 points10%
x = points
7.Confidence intervalsMedium
A 95% confidence interval for a mean, based on 64 observations and a known σ, has a margin of error of 10. What is the margin of error E₂ of a 90% interval from the same data?
95%n₁ = 64E₁ = 1090%
E₂ =
8.Confidence intervalsHard
The population standard deviation is 7. 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 0.5?
σ = 7E = 0.595%
n =
9.Hypothesis tests & confidence intervalsMedium
A sample of 100 students has a mean score of 76.2 and a sample standard deviation of 10. Find the test statistic t for H₀: μ = 75.
μ₀ = 75x̄ = 76.2s = 10n = 100
t =
10.Hypothesis tests & confidence intervalsHard
A sample of 36 measurements has mean 498.3. The population standard deviation is 8. Find the upper limit U of the 99% confidence interval for the mean.
x̄ = 498.3σ = 8n = 3699%
U =
11.Linear regression & correlationMedium
The table gives 6 paired observations of x = hours studied and y = exam score. Their sums are given. Find Pearson's correlation coefficient r from these sums.
n = 6Σx = 42Σy = 498Σx² = 364Σy² = 43896Σxy = 3904
r =
12.Linear regression & correlationHard
The table gives 6 paired observations of x = advertising spend in k€ and y = sales in k€. Use the least-squares line ŷ = a + b·x. For which value x does the line predict ŷ = 57?
n = 6ŷ = 57
x =
3Solutions
1.Descriptive statistics
Answer: s = 6.411
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 = 315/6 = 52.5
Squared deviations
Σ(x − x̄)² = 4.5² + 6.5² + (−10.5)² + 0.5² + (−4.5)² + 3.5² = 205.5
Standard deviation
s = √(205.5/5) = 6.411
Answer
s = 6.411
2.Descriptive statistics
Answer: P = 0.2743
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 = 0.6
Standard normal table
Φ(0.6) = 0.7257
P = 1 − 0.7257 = 0.2743
Answer
P = 0.2743
3.Probability
Answer: P = 0.7674
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.
Complement
P(no six) = (5/6)8 = 0.2326
P = 1 − 0.2326 = 0.7674
Answer
P = 0.7674
4.Probability
Answer: σY = 3.534
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 = 2.25, Var(X) = 1.387, σX = 1.178
Linear change
σY = |3|·1.178 = 3.534
Answer
σY = 3.534
The mean does move: E(Y) = 3·2.25 + 20 = 26.75.
5.Normal distribution
Answer: z = 1.77
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 − 0.0384 = 0.9616
Read the z-table backwards
Find 0.9616 inside the table and read off z.
z = 1.77
Answer
z = 1.77
6.Normal distribution
Answer: x = 80.8 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) = 0.1
z from the table
z ≈ −1.28
Back to the original units
x = 100 − 1.28·15 = 80.8 points
Answer
x = 80.8 points
7.Confidence intervals
Answer: E₂ = 8.393
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.
Ratio of margins
E₂/E₁ = (1.645/1.96)·√(64/64) = 0.8393
New margin
E₂ = 10·0.8393 = 8.393
Answer
E₂ = 8.393
8.Confidence intervals
Answer: n = 753
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 = (1.96·7/0.5)² = 753
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 753
Answer
n = 753
9.Hypothesis tests & confidence intervals
Answer: t = 1.2
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 = 10/√100 = 10/10 = 1
Test statistic
t = (76.2 − 75)/1 = 1.2
Answer
t = 1.2
t0.025; df = 99 = 1.984
|t| < 1.984, so H₀ is not rejected at the 5% level (two-sided).
10.Hypothesis tests & confidence intervals
Answer: U = 501.7
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 = 2.576·8/√36 = 3.435
Interval limit
U = 498.3 + 3.435 = 501.7
Answer
U = 501.7
11.Linear regression & correlation
Answer: r = 0.987
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 = 3904 − 42·498/6 = 418
Sxx = 364 − 42²/6 = 70
Syy = 43896 − 498²/6 = 2562
Correlation coefficient
r = 418/√(70·2562) = 0.987
Answer
r = 0.987
12.Linear regression & correlation
Answer: x = 3.556
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 = 33/6 = 5.5
ȳ = 397/6 = 66.17
Sums of squares and products
Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = 82.5
Regression line
b = 82.5/17.5 = 4.714
a = 66.17 − 4.714·5.5 = 40.24
ŷ = 40.24 + 4.714·x
Solve for x
x = (57 − 40.24)/4.714 = 3.556
Answer
x = 3.556
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