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: 33, 52, 25, 45, 27, 43, 43.
n = 7
s =
2.Descriptive statisticsHard
Exam scores are normally distributed with mean 170 and standard deviation 5. Find the z-score z of a score of 160.
μ = 170σ = 5x = 160
z =
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
9 components are tested, and each one fails with probability 0.3, independently. What is the probability P that at most one fails?
n = 9p = 0.3
P =
5.Normal distributionMedium
Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.0256. Find z to two decimals.
A = 0.0256
z =
6.Normal distributionHard
The bag weights are normally distributed with mean 250 g and standard deviation 5 g. Drag the cut-off so the shaded top 75% is to the right of it. Find the cut-off x.
μ = 250 gσ = 5 g75%
x = g
7.Confidence intervalsMedium
The blood pressure of 6 patients is measured before and after a treatment. The differences d = before − after (mmHg) have mean 2 and standard deviation 3. Find the margin of error E of the 95% confidence interval for the mean difference.
n = 6d̄ = 2sd = 395%t(5) = 2.571
E = mmHg
8.Confidence intervalsHard
Nothing is known in advance about the proportion of defective products. How many products n must be inspected so that a 99% confidence interval for the proportion has a margin of error of at most 0.04?
p = 0.5E = 0.0499%
n =
9.Hypothesis tests & confidence intervalsMedium
10 employees take a typing test before and after a training course. For each person the difference d = after − before (words per minute) is recorded. The mean difference is −2.37 and the standard deviation of the differences is 3. Find the paired test statistic t for H₀: μd = 0.
n = 10d̄ = −2.37sd = 3
t =
10.Hypothesis tests & confidence intervalsHard
How large must a sample be so that the 95% confidence interval for a mean has a margin of error of at most 1? The standard deviation is 8. Find the smallest sample size n.
σ = 8E = 195%
n =
11.Linear regression & correlationMedium
The table gives 6 paired observations of x = advertising spend in k€ and y = sales in k€. Their sums are given. Find Pearson's correlation coefficient r from these sums.
n = 6Σx = 21Σy = 175Σx² = 91Σy² = 5335Σxy = 675
r =
12.Linear regression & correlationHard
The table gives 5 paired observations of x = hours studied and y = exam score. Use the least-squares line ŷ = a + b·x. A new observation (3.5, 71) is made. Find its residual e = y − ŷ.
n = 5(3.5, 71)
e =
3Solutions
1.Descriptive statistics
Answer: s = 10.08
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 = 268/7 = 38.29
Squared deviations
Σ(x − x̄)² = (−5.286)² + 13.71² + (−13.29)² + 6.714² + (−11.29)² + 4.714² + 4.714² = 609.4
Standard deviation
s = √(609.4/6) = 10.08
Answer
s = 10.08
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 = (160 − 170)/5 = −2
Answer
z = −2
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: P = 0.196
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) = 0.79 = 0.04035
P(X = 1) = 9·0.3·0.78 = 0.1556
Add them
P = 0.04035 + 0.1556 = 0.196
Answer
P = 0.196
5.Normal distribution
Answer: z = 1.95
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 − 0.0256 = 0.9744
Read the z-table backwards
Find 0.9744 inside the table and read off z.
z = 1.95
Answer
z = 1.95
6.Normal distribution
Answer: x = 246.7 g
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) = 0.25
z from the table
z ≈ −0.67
Back to the original units
x = 250 − 0.67·5 = 246.7 g
Answer
x = 246.7 g
7.Confidence intervals
Answer: E = 3.149 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 = 5, t = 2.571
Margin of error
E = 2.571·3/√6 = 3.149
Answer
E = 3.149 mmHg
8.Confidence intervals
Answer: n = 1037
Confidence interval for a proportion: p̂ ± z·√(p̂(1 − p̂)/n), where p̂ = x/n. To reach a margin E you need n ≥ z²·p(1 − p)/E², rounded up; without a prior estimate use p = 0.5, which gives the largest n.
Solve for n
n = 2.576²·0.5·0.5/0.04² = 1037
Round up
n = 1037
Answer
n = 1037
9.Hypothesis tests & confidence intervals
Answer: t = −2.498
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 of the mean difference
SE = 3/√10 = 0.9487
Test statistic
t = −2.37/0.9487 = −2.498
Answer
t = −2.498
With df = 9, the two-sided 5% critical value is 2.262; |t| is larger, so H₀ is rejected.
10.Hypothesis tests & confidence intervals
Answer: n = 246
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).
Solve for n
n = (1.96·8/1)² = 245.9
Round up
n = 246
Answer
n = 246
11.Linear regression & correlation
Answer: r = 0.9834
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 = 675 − 21·175/6 = 62.5
Sxx = 91 − 21²/6 = 17.5
Syy = 5335 − 175²/6 = 230.8
Correlation coefficient
r = 62.5/√(17.5·230.8) = 0.9834
Answer
r = 0.9834
12.Linear regression & correlation
Answer: e = 5.35
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 = 15/5 = 3
ȳ = 319/5 = 63.8
Sums of squares and products
Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = 37
Regression line
b = 37/10 = 3.7
a = 63.8 − 3.7·3 = 52.7
ŷ = 52.7 + 3.7·x
Predicted value
ŷ = 52.7 + 3.7·3.5 = 65.65
Residual
e = 71 − 65.65 = 5.35
Answer
e = 5.35
The point lies above the line.
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