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: 26, 19, 33, 48, 37, 51, 39.
n = 7
s =
2.Descriptive statisticsHard
A quantity is normally distributed with mean 60 and standard deviation 10. What proportion P of values is above 70?
μ = 60σ = 10x = 70
P =
3.ProbabilityMedium
A fair die is rolled 4 times. What is the probability P of getting at least one six?
n = 4
P =
4.ProbabilityHard
12 components are tested, and each one fails with probability 0.25, independently. What is the probability P that at most one fails?
n = 12p = 0.25
P =
5.Normal distributionMedium
Drag z so that the shaded area between −z and z under the standard normal curve is 0.6047. Find z to two decimals.
A = 0.6047
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 70% is to the right of it. Find the cut-off x.
μ = 65 pointsσ = 10 points70%
x = points
7.Confidence intervalsMedium
The blood pressure of 8 patients is measured before and after a treatment. The differences d = before − after (mmHg) have mean 7.8 and standard deviation 2.2. Find the upper limit of the 99% confidence interval for the mean difference.
n = 8d̄ = 7.8sd = 2.299%t(7) = 3.499
upper limit = mmHg
8.Confidence intervalsHard
The population standard deviation is 33. 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 4?
σ = 33E = 499%
n =
9.Hypothesis tests & confidence intervalsMedium
A sample of 49 students has a mean score of 48 and a sample standard deviation of 10. Find the test statistic t for H₀: μ = 50.
μ₀ = 50x̄ = 48s = 10n = 49
t =
10.Hypothesis tests & confidence intervalsHard
A one-sided z test of H₀: μ = 80 against H₁: μ > 80 is done at α = 0.05 with a sample of 25; the standard deviation is known to be 15. In reality the mean is 87.8. Find the power 1 − β of the test in percent.
μ₀ = 80μ = 87.8σ = 15n = 25α = 0.05
1 − β = %
11.Linear regression & correlationMedium
The table gives 5 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 = 5Σx = 35Σy = 253Σx² = 285Σy² = 13239Σxy = 1899
r =
12.Linear regression & correlationHard
The table gives 6 paired observations of x = load in kN and y = extension in mm. Use the least-squares line ŷ = a + b·x. A new observation (11, 44) is made. Find its residual e = y − ŷ.
n = 6(11, 44)
e =
3Solutions
1.Descriptive statistics
Answer: s = 11.38
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 = 253/7 = 36.14
Squared deviations
Σ(x − x̄)² = (−10.14)² + (−17.14)² + (−3.143)² + 11.86² + 0.8571² + 14.86² + 2.857² = 776.9
Standard deviation
s = √(776.9/6) = 11.38
Answer
s = 11.38
2.Descriptive statistics
Answer: P = 0.1587
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) = 0.8413
P = 1 − 0.8413 = 0.1587
Answer
P = 0.1587
3.Probability
Answer: P = 0.5177
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)4 = 0.4823
P = 1 − 0.4823 = 0.5177
Answer
P = 0.5177
4.Probability
Answer: P = 0.1584
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) = 0.7512 = 0.03168
P(X = 1) = 12·0.25·0.7511 = 0.1267
Add them
P = 0.03168 + 0.1267 = 0.1584
Answer
P = 0.1584
5.Normal distribution
Answer: z = 0.85
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 + 0.6047)/2 = 0.8023
Read the z-table backwards
Find 0.8023 inside the table and read off z.
z = 0.85
Answer
z = 0.85
6.Normal distribution
Answer: x = 59.8 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) = 0.3
z from the table
z ≈ −0.52
Back to the original units
x = 65 − 0.52·10 = 59.8 points
Answer
x = 59.8 points
7.Confidence intervals
Answer: upper limit = 10.52 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 = 7, t = 3.499
Margin of error
E = 3.499·2.2/√8 = 2.722
Interval
7.8 ± 2.722 → [5.078, 10.52]
Answer
upper limit = 10.52 mmHg
8.Confidence intervals
Answer: n = 452
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 = (2.576·33/4)² = 451.6
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 452
Answer
n = 452
9.Hypothesis tests & confidence intervals
Answer: t = −1.4
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 = 10/√49 = 10/7 = 1.429
Test statistic
t = (48 − 50)/1.429 = −1.4
Answer
t = −1.4
t0.025; df = 48 = 2.011
|t| < 2.011, so H₀ is not rejected at the 5% level (two-sided).
10.Hypothesis tests & confidence intervals
Answer: 1 − β = 83.02 %
Power of a test: power = 1 − β is the probability of rejecting H₀ when it is false. Find the cutoff x̄c from μ₀ and z_α, then compute the probability of landing beyond x̄c using the true mean: z = (x̄c − μ)/(σ/√n).
Critical sample mean
SE = 15/√25 = 3
x̄c = 80 + 1.645·3 = 84.94
z under the true mean
z = (84.94 − 87.8)/3 = −0.955
Power
power = P(Z > −0.955) = 83.02%
1 − β = 83.02%
Answer
1 − β = 83.02%
11.Linear regression & correlation
Answer: r = 0.9679
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 = 1899 − 35·253/5 = 128
Sxx = 285 − 35²/5 = 40
Syy = 13239 − 253²/5 = 437.2
Correlation coefficient
r = 128/√(40·437.2) = 0.9679
Answer
r = 0.9679
12.Linear regression & correlation
Answer: e = 4.2
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 = 162/6 = 27
Sums of squares and products
Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 224
Regression line
b = 224/70 = 3.2
a = 27 − 3.2·7 = 4.6
ŷ = 4.6 + 3.2·x
Predicted value
ŷ = 4.6 + 3.2·11 = 39.8
Residual
e = 44 − 39.8 = 4.2
Answer
e = 4.2
The point lies above the line.
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