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: 31, 35, 11, 23, 13.
n = 5
s =
2.Descriptive statisticsHard
Exam scores are normally distributed with mean 170 and standard deviation 8. Find the z-score z of a score of 154.
μ = 170σ = 8x = 154
z =
3.ProbabilityMedium
In how many ways N can a committee of 4 people be chosen from a group of 14? The order does not matter.
n = 14k = 4
N =
4.ProbabilityHard
A test has 8 multiple-choice questions. A student who guesses gets each one right with probability 0.25, independently. What is the probability P of exactly 2 correct answers?
n = 8p = 0.25k = 2
P =
5.Normal distributionMedium
Drag z so that the shaded area between −z and z under the standard normal curve is 0.4177. Find z to two decimals.
A = 0.4177
z =
6.Normal distributionHard
The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. Drag the cut-off so the shaded top 30% is to the right of it. Find the cut-off x.
μ = 178 cmσ = 7 cm30%
x = cm
7.Confidence intervalsMedium
The blood pressure of 13 patients is measured before and after a treatment. The differences d = before − after (mmHg) have mean 8.5 and standard deviation 5.7. Find the margin of error E of the 99% confidence interval for the mean difference.
n = 13d̄ = 8.5sd = 5.799%t(12) = 3.055
E = mmHg
8.Confidence intervalsHard
The population standard deviation is 33. 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 2?
σ = 33E = 290%
n =
9.Hypothesis tests & confidence intervalsMedium
A sample of 64 students has a mean score of 101.8 and a sample standard deviation of 12. Find the test statistic t for H₀: μ = 100.
μ₀ = 100x̄ = 101.8s = 12n = 64
t =
10.Hypothesis tests & confidence intervalsHard
A one-sided z test at α = 0.05 must detect a shift of the mean of 2.5 units with a power of 90%. The standard deviation is 17. Find the minimum sample size n.
α = 0.05power = 90%δ = 2.5σ = 17zα = 1.645zβ = 1.282
n =
11.Linear regression & correlationMedium
The table gives 5 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 = 5
r² =
12.Linear regression & correlationHard
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. A new observation (24.5, 100) is made. Find its residual e = y − ŷ.
n = 5(24.5, 100)
e =
3Solutions
1.Descriptive statistics
Answer: s = 10.62
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 = 113/5 = 22.6
Squared deviations
Σ(x − x̄)² = 8.4² + 12.4² + (−11.6)² + 0.4² + (−9.6)² = 451.2
Standard deviation
s = √(451.2/4) = 10.62
Answer
s = 10.62
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 = (154 − 170)/8 = −2
Answer
z = −2
3.Probability
Answer: N = 1001
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(14, 4) = (14·13·12·11)/(4·3·2·1) = 1001
Answer
N = 1001
4.Probability
Answer: P = 0.3115
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.
Binomial formula
P(X = 2) = 28·0.252·0.756 = 28·0.0625·0.178 = 0.3115
Answer
P = 0.3115
5.Normal distribution
Answer: z = 0.55
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 + 0.4177)/2 = 0.7088
Read the z-table backwards
Find 0.7088 inside the table and read off z.
z = 0.55
Answer
z = 0.55
6.Normal distribution
Answer: x = 181.6 cm
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) = 0.7
z from the table
z ≈ 0.52
Back to the original units
x = 178 + 0.52·7 = 181.6 cm
Answer
x = 181.6 cm
7.Confidence intervals
Answer: E = 4.83 mmHg
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.
Critical value
df = 12, t = 3.055
Margin of error
E = 3.055·5.7/√13 = 4.83
Answer
E = 4.83 mmHg
8.Confidence intervals
Answer: n = 737
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 = (1.645·33/2)² = 736.7
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 737
Answer
n = 737
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 = 12/√64 = 12/8 = 1.5
Test statistic
t = (101.8 − 100)/1.5 = 1.2
Answer
t = 1.2
t0.025; df = 63 = 1.998
|t| < 1.998, so H₀ is not rejected at the 5% level (two-sided).
10.Hypothesis tests & confidence intervals
Answer: n = 397
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)·17/2.5)² = 396.2
Round up
n = 397
Answer
n = 397
11.Linear regression & correlation
Answer: r² = 0.7438
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 = 35/5 = 7
ȳ = 202/5 = 40.4
Sums of squares and products
Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 62
Syy = Σ(y − ȳ)² = 129.2
Coefficient of determination
r² = 62²/(40·129.2) = 0.7438
Answer
r² = 0.7438
About 74.38% of the variation in y is explained by the line.
12.Linear regression & correlation
Answer: e = −3.25
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 = 115/5 = 23
ȳ = 481/5 = 96.2
Sums of squares and products
Sxx = Σ(x − x̄)² = 90
Sxy = Σ(x − x̄)(y − ȳ) = 423
Regression line
b = 423/90 = 4.7
a = 96.2 − 4.7·23 = −11.9
ŷ = −11.9 + 4.7·x
Predicted value
ŷ = −11.9 + 4.7·24.5 = 103.3
Residual
e = 100 − 103.3 = −3.25
Answer
e = −3.25
The point lies below the line.
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