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: 29, 16, 45, 51, 20, 56, 25.
n = 7
s =
2.Descriptive statisticsHard
Exam scores are normally distributed with mean 500 and standard deviation 60. Find the z-score z of a score of 560.
μ = 500σ = 60x = 560
z =
3.ProbabilityMedium
In how many ways N can a committee of 4 people be chosen from a group of 13? The order does not matter.
n = 13k = 4
N =
4.ProbabilityHard
A test has 11 multiple-choice questions. A student who guesses gets each one right with probability 0.2, independently. What is the probability P of exactly 2 correct answers?
n = 11p = 0.2k = 2
P =
5.Normal distributionMedium
The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. What fraction lies between 92.5 and 123.25 points? Give P as a decimal.
μ = 100σ = 15x1 = 92.5x2 = 123.25
P =
6.Normal distributionHard
The bag weights are normally distributed with mean 250 g and standard deviation 5 g. Find the value x below which 90% of them lie (the 90th percentile).
μ = 250σ = 5P = 90%
x = g
7.Confidence intervalsMedium
The blood pressure of 10 patients is measured before and after a treatment. The differences d = before − after (mmHg) have mean 9 and standard deviation 0.6. Find the upper limit of the 90% confidence interval for the mean difference.
n = 10d̄ = 9sd = 0.690%t(9) = 1.833
upper limit = mmHg
8.Confidence intervalsHard
The population standard deviation is 40. 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 = 40E = 2.595%
n =
9.Hypothesis tests & confidence intervalsMedium
A sample of 16 students has a mean score of 73.88 and a sample standard deviation of 5. Find the test statistic t for H₀: μ = 75.
μ₀ = 75x̄ = 73.88s = 5n = 16
t =
10.Hypothesis tests & confidence intervalsHard
How large must a sample be so that the 99% confidence interval for a mean has a margin of error of at most 0.5? The standard deviation is 12. Find the smallest sample size n.
σ = 12E = 0.599%
n =
11.Linear regression & correlationMedium
The table gives 6 paired observations of x = temperature in °C and y = ice creams sold. Find the coefficient of determination r² of the least-squares line.
n = 6
r² =
12.Linear regression & correlationHard
The table gives 5 paired observations of x = car age in years and y = price in k€. Use the least-squares line ŷ = a + b·x. For which value x does the line predict ŷ = 21?
n = 5ŷ = 21
x =
3Solutions
1.Descriptive statistics
Answer: s = 15.9
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 = 242/7 = 34.57
Squared deviations
Σ(x − x̄)² = (−5.571)² + (−18.57)² + 10.43² + 16.43² + (−14.57)² + 21.43² + (−9.571)² = 1518
Standard deviation
s = √(1518/6) = 15.9
Answer
s = 15.9
2.Descriptive statistics
Answer: z = 1
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 = (560 − 500)/60 = 1
Answer
z = 1
3.Probability
Answer: N = 715
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(13, 4) = (13·12·11·10)/(4·3·2·1) = 715
Answer
N = 715
4.Probability
Answer: P = 0.2953
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) = 55·0.22·0.89 = 55·0.04·0.1342 = 0.2953
Answer
P = 0.2953
5.Normal distribution
Answer: P = 0.6309
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 = (92.5 − 100)/15 = −0.5
z₂ = (123.25 − 100)/15 = 1.55
Area between
P = Φ(1.55) − Φ(−0.5) = 0.9394 − 0.3085 = 0.6309
Answer
P = 0.6309
6.Normal distribution
Answer: x = 256.4 g
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) = 0.9 → z = 1.282
Back to x
x = 250 + 1.282·5 = 256.41 g
Answer
x = 256.41 g
7.Confidence intervals
Answer: upper limit = 9.348 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 = 9, t = 1.833
Margin of error
E = 1.833·0.6/√10 = 0.3478
Interval
9 ± 0.3478 → [8.652, 9.348]
Answer
upper limit = 9.348 mmHg
8.Confidence intervals
Answer: n = 984
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 = (1.96·40/2.5)² = 983.4
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 984
Answer
n = 984
9.Hypothesis tests & confidence intervals
Answer: t = −0.896
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/√16 = 5/4 = 1.25
Test statistic
t = (73.88 − 75)/1.25 = −0.896
Answer
t = −0.896
t0.025; df = 15 = 2.131
|t| < 2.131, so H₀ is not rejected at the 5% level (two-sided).
10.Hypothesis tests & confidence intervals
Answer: n = 3823
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 = (2.576·12/0.5)² = 3822
Round up
n = 3823
Answer
n = 3823
11.Linear regression & correlation
Answer: r² = 0.9818
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 = 129/6 = 21.5
ȳ = 670/6 = 111.7
Sums of squares and products
Sxx = Σ(x − x̄)² = 157.5
Sxy = Σ(x − x̄)(y − ȳ) = 945
Syy = Σ(y − ȳ)² = 5775
Coefficient of determination
r² = 945²/(157.5·5775) = 0.9818
Answer
r² = 0.9818
About 98.18% of the variation in y is explained by the line.
12.Linear regression & correlation
Answer: x = 2.364
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 = 98/5 = 19.6
Sums of squares and products
Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = −22
Regression line
b = −22/10 = −2.2
a = 19.6 − (−2.2)·3 = 26.2
ŷ = 26.2 − 2.2·x
Solve for x
x = (21 − 26.2)/(−2.2) = 2.364
Answer
x = 2.364
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