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: 55, 20, 35, 28, 14, 23.
n = 6
s =
2.Descriptive statisticsHard
A quantity is normally distributed with mean 60 and standard deviation 8. What proportion P of values is above 52?
μ = 60σ = 8x = 52
P =
3.ProbabilityMedium
In how many ways N can a committee of 3 people be chosen from a group of 7? The order does not matter.
n = 7k = 3
N =
4.ProbabilityHard
11 components are tested, and each one fails with probability 0.1, independently. What is the probability P that at most one fails?
n = 11p = 0.1
P =
5.Normal distributionMedium
The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. What fraction lies between 480 and 540 h? Give P as a decimal.
μ = 500σ = 40x1 = 480x2 = 540
P =
6.Normal distributionHard
The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. Find the value x below which 75% of them lie (the 75th percentile).
μ = 178σ = 7P = 75%
x = cm
7.Confidence intervalsMedium
The blood pressure of 11 patients is measured before and after a treatment. The differences d = before − after (mmHg) have mean 5.9 and standard deviation 2.8. Find the lower limit of the 95% confidence interval for the mean difference.
n = 11d̄ = 5.9sd = 2.895%t(10) = 2.228
lower limit = mmHg
8.Confidence intervalsHard
The population standard deviation is 17. How large a sample n do you need so that a 99% confidence interval for the mean has a margin of error of at most 0.5?
σ = 17E = 0.599%
n =
9.Hypothesis tests & confidence intervalsMedium
A sample of 25 students has a mean score of 100.6 and a sample standard deviation of 4. Find the test statistic t for H₀: μ = 100.
μ₀ = 100x̄ = 100.6s = 4n = 25
t =
10.Hypothesis tests & confidence intervalsHard
A sample of 16 measurements has mean 116.8. The population standard deviation is 5. Find the upper limit U of the 95% confidence interval for the mean.
x̄ = 116.8σ = 5n = 1695%
U =
11.Linear regression & correlationMedium
The table gives 6 paired observations of x = load in kN and y = extension in mm. Find Pearson's correlation coefficient r.
n = 6
r =
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 = 22?
n = 5ŷ = 22
x =
3Solutions
1.Descriptive statistics
Answer: s = 14.52
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 = 175/6 = 29.17
Squared deviations
Σ(x − x̄)² = 25.83² + (−9.167)² + 5.833² + (−1.167)² + (−15.17)² + (−6.167)² = 1055
Standard deviation
s = √(1055/5) = 14.52
Answer
s = 14.52
2.Descriptive statistics
Answer: P = 0.8413
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 = −1
Standard normal table
Φ(−1) = 0.1587
P = 1 − 0.1587 = 0.8413
Answer
P = 0.8413
3.Probability
Answer: N = 35
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(7, 3) = (7·6·5)/(3·2·1) = 35
Answer
N = 35
4.Probability
Answer: P = 0.6974
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) = 0.911 = 0.3138
P(X = 1) = 11·0.1·0.910 = 0.3835
Add them
P = 0.3138 + 0.3835 = 0.6974
Answer
P = 0.6974
5.Normal distribution
Answer: P = 0.5328
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 = (480 − 500)/40 = −0.5
z₂ = (540 − 500)/40 = 1
Area between
P = Φ(1) − Φ(−0.5) = 0.8413 − 0.3085 = 0.5328
Answer
P = 0.5328
6.Normal distribution
Answer: x = 182.7 cm
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) = 0.75 → z = 0.6745
Back to x
x = 178 + 0.6745·7 = 182.72 cm
Answer
x = 182.72 cm
7.Confidence intervals
Answer: lower limit = 4.019 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 = 10, t = 2.228
Margin of error
E = 2.228·2.8/√11 = 1.881
Interval
5.9 ± 1.881 → [4.019, 7.781]
Answer
lower limit = 4.019 mmHg
8.Confidence intervals
Answer: n = 7671
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 = (2.576·17/0.5)² = 7671
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 7671
Answer
n = 7671
9.Hypothesis tests & confidence intervals
Answer: t = 0.75
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 = 4/√25 = 4/5 = 0.8
Test statistic
t = (100.6 − 100)/0.8 = 0.75
Answer
t = 0.75
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: U = 119.3
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 = 1.96·5/√16 = 2.45
Interval limit
U = 116.8 + 2.45 = 119.3
Answer
U = 119.3
11.Linear regression & correlation
Answer: r = 0.9863
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 = 42/6 = 7
ȳ = 107/6 = 17.83
Sums of squares and products
Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 155
Syy = Σ(y − ȳ)² = 352.8
Correlation coefficient
r = 155/√(70·352.8) = 0.9863
Answer
r = 0.9863
12.Linear regression & correlation
Answer: x = 3.348
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 = 114/5 = 22.8
Sums of squares and products
Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = −23
Regression line
b = −23/10 = −2.3
a = 22.8 − (−2.3)·3 = 29.7
ŷ = 29.7 − 2.3·x
Solve for x
x = (22 − 29.7)/(−2.3) = 3.348
Answer
x = 3.348
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