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: 45, 58, 28, 51, 11, 16.
n = 6
s =
2.Descriptive statisticsHard
Exam scores are normally distributed with mean 70 and standard deviation 8. Find the z-score z of a score of 66.
μ = 70σ = 8x = 66
z =
3.ProbabilityMedium
A bag holds 3 red and 7 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?
3 / 7
P =
4.ProbabilityHard
5 components are tested, and each one fails with probability 0.3, independently. What is the probability P that at most one fails?
n = 5p = 0.3
P =
5.Normal distributionMedium
Drag z so that the shaded area between −z and z under the standard normal curve is 0.9281. Find z to two decimals.
A = 0.9281
z =
6.Normal distributionHard
The exam scores are normally distributed with mean 65 points and standard deviation 10 points. Drag the cut-off so the shaded top 10% is to the right of it. Find the cut-off x.
μ = 65 pointsσ = 10 points10%
x = points
7.Confidence intervalsMedium
A 90% 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 if the same study uses 256 observations instead?
90%n₁ = 64E₁ = 10n₂ = 256
E₂ =
8.Confidence intervalsHard
The population standard deviation is 13. 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?
σ = 13E = 295%
n =
9.Hypothesis tests & confidence intervalsMedium
A sample of 25 students has a mean score of 67.8 and a sample standard deviation of 20. Find the test statistic t for H₀: μ = 75.
μ₀ = 75x̄ = 67.8s = 20n = 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 16. Find the minimum sample size n.
α = 0.01power = 80%δ = 2σ = 16zα = 2.326zβ = 0.842
n =
11.Linear regression & correlationMedium
The table gives 6 paired observations of x = advertising spend in k€ and y = sales in k€. Find the coefficient of determination r² of the least-squares line.
n = 6
r² =
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 (3.5, 70) is made. Find its residual e = y − ŷ.
n = 6(3.5, 70)
e =
3Solutions
1.Descriptive statistics
Answer: s = 19.34
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 = 209/6 = 34.83
Squared deviations
Σ(x − x̄)² = 10.17² + 23.17² + (−6.833)² + 16.17² + (−23.83)² + (−18.83)² = 1871
Standard deviation
s = √(1871/5) = 19.34
Answer
s = 19.34
2.Descriptive statistics
Answer: z = −0.5
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 = (66 − 70)/8 = −0.5
Answer
z = −0.5
3.Probability
Answer: P = 0.06667
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 = 3/10 · 2/9 = 6/90 = 0.06667
Answer
P = 0.06667
4.Probability
Answer: P = 0.5282
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) = 0.75 = 0.1681
P(X = 1) = 5·0.3·0.74 = 0.3601
Add them
P = 0.1681 + 0.3601 = 0.5282
Answer
P = 0.5282
5.Normal distribution
Answer: z = 1.8
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 + 0.9281)/2 = 0.9641
Read the z-table backwards
Find 0.9641 inside the table and read off z.
z = 1.80
Answer
z = 1.80
6.Normal distribution
Answer: x = 77.8 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) = 0.9
z from the table
z ≈ 1.28
Back to the original units
x = 65 + 1.28·10 = 77.8 points
Answer
x = 77.8 points
7.Confidence intervals
Answer: E₂ = 5
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 = (1.645/1.645)·√(64/256) = 0.5
New margin
E₂ = 10·0.5 = 5
Answer
E₂ = 5
8.Confidence intervals
Answer: n = 163
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 = (1.96·13/2)² = 162.3
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 163
Answer
n = 163
9.Hypothesis tests & confidence intervals
Answer: t = −1.8
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 = 20/√25 = 20/5 = 4
Test statistic
t = (67.8 − 75)/4 = −1.8
Answer
t = −1.8
t0.025; df = 24 = 2.064
|t| < 2.064, so H₀ is not rejected at the 5% level (two-sided).
10.Hypothesis tests & confidence intervals
Answer: n = 643
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)·16/2)² = 642.3
Round up
n = 643
Answer
n = 643
11.Linear regression & correlation
Answer: r² = 0.9284
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 = 48/6 = 8
ȳ = 307/6 = 51.17
Sums of squares and products
Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 151
Syy = Σ(y − ȳ)² = 350.8
Coefficient of determination
r² = 151²/(70·350.8) = 0.9284
Answer
r² = 0.9284
About 92.84% of the variation in y is explained by the line.
12.Linear regression & correlation
Answer: e = 2.295
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 = 27/6 = 4.5
ȳ = 446/6 = 74.33
Sums of squares and products
Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = 116
Regression line
b = 116/17.5 = 6.629
a = 74.33 − 6.629·4.5 = 44.5
ŷ = 44.5 + 6.629·x
Predicted value
ŷ = 44.5 + 6.629·3.5 = 67.7
Residual
e = 70 − 67.7 = 2.295
Answer
e = 2.295
The point lies above the line.
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