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: 30, 52, 15, 60, 52.
n = 5
s =
2.Descriptive statisticsHard
A quantity is normally distributed with mean 50 and standard deviation 20. What proportion P of values is above 20?
μ = 50σ = 20x = 20
P =
3.ProbabilityMedium
For the distribution shown, find the variance Var(X) of X.
x: 10, 20, 30, 40p(x): 0.25, 0.2, 0.35, 0.2
Var(X) =
4.ProbabilityHard
X has the distribution shown, and Y = 5X + 4. Find the standard deviation σY of Y.
x: 1, 2, 3, 4, 5p(x): 0.15, 0.1, 0.3, 0.35, 0.1Y = 5X + 4
σY =
5.Normal distributionMedium
The masses of bags of cement are independent and normal with mean 25 kg and standard deviation 0.5 kg. Find the probability P that the total mass of 12 of them is below 298.3 kg. Round z to two decimals.
μ = 25 kgσ = 0.5 kgn = 12< 298.3 kg
P =
6.Normal distributionHard
Runner A's lap time X and runner B's lap time Y are independent and normal: X has mean 52 s and standard deviation 4 s, Y has mean 48 s and standard deviation 3 s. Find the probability P that X is greater than Y. Round z to two decimals.
μX = 52 s, σX = 4 sμY = 48 s, σY = 3 s
P =
7.Confidence intervalsMedium
A sample of 11 items has mean 35.5 and standard deviation s = 15.5. The population is roughly normal. Find the upper limit of the 99% confidence interval for the mean.
n = 11x̄ = 35.5s = 15.599%t(10) = 3.169
upper limit =
8.Confidence intervalsHard
The population standard deviation is 19. 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 4?
σ = 19E = 495%
n =
9.Hypothesis tests & confidence intervalsMedium
7 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 10.08 and the standard deviation of the differences is 11.6. Find the paired test statistic t for H₀: μd = 0.
n = 7d̄ = 10.08sd = 11.6
t =
10.Hypothesis tests & confidence intervalsHard
In a trial, 65 of 100 patients on treatment A recover, and 52 of 100 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.
x₁ = 65n₁ = 100x₂ = 52n₂ = 100
z =
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 6 paired observations of x = hours studied and y = exam score. Use the least-squares line ŷ = a + b·x. A new observation (10, 108) is made. Find its residual e = y − ŷ.
n = 6(10, 108)
e =
3Solutions
1.Descriptive statistics
Answer: s = 18.69
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 = 209/5 = 41.8
Squared deviations
Σ(x − x̄)² = (−11.8)² + 10.2² + (−26.8)² + 18.2² + 10.2² = 1397
Standard deviation
s = √(1397/4) = 18.69
Answer
s = 18.69
2.Descriptive statistics
Answer: P = 0.9332
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 = −1.5
Standard normal table
Φ(−1.5) = 0.06681
P = 1 − 0.06681 = 0.9332
Answer
P = 0.9332
3.Probability
Answer: Var(X) = 115
Discrete random variables: a probability distribution lists every value x a variable can take with its probability p(x); the probabilities are between 0 and 1 and add up to 1. The expected value (mean) is E(X) = Σx·p(x), the long-run average. The variance is Var(X) = Σ(x − μ)²·p(x) = E(X²) − μ², and the standard deviation σ = √Var(X). For a linear change Y = aX + b: E(Y) = a·E(X) + b, Var(Y) = a²·Var(X) and σY = |a|·σX; adding a constant does not change the spread.
Mean
μ = 10·0.25 + 20·0.2 + 30·0.35 + 40·0.2 = 25
Mean of the squares
E(X²) = 100·0.25 + 400·0.2 + 900·0.35 + 1600·0.2 = 740
Variance
Var(X) = 740 − 25² = 115
Answer
Var(X) = 115
4.Probability
Answer: σY = 5.974
Discrete random variables: a probability distribution lists every value x a variable can take with its probability p(x); the probabilities are between 0 and 1 and add up to 1. The expected value (mean) is E(X) = Σx·p(x), the long-run average. The variance is Var(X) = Σ(x − μ)²·p(x) = E(X²) − μ², and the standard deviation σ = √Var(X). For a linear change Y = aX + b: E(Y) = a·E(X) + b, Var(Y) = a²·Var(X) and σY = |a|·σX; adding a constant does not change the spread.
Spread of X
μ = 3.15, Var(X) = 1.428, σX = 1.195
Linear change
σY = |5|·1.195 = 5.974
Answer
σY = 5.974
The mean does move: E(Y) = 5·3.15 + 4 = 19.75.
5.Normal distribution
Answer: P = 0.1635
Sums and differences of normal variables: if X and Y are independent and normal, X + Y and X − Y are normal too. Means add or subtract: E(X ± Y) = μX ± μY. Variances always add, also for a difference: Var(X ± Y) = σX² + σY², so σ = √(σX² + σY²). Standard deviations never add directly. The total of n independent items with the same distribution has mean n·μ and standard deviation σ·√n. Then standardise as usual: z = (x − mean)/σ and read Φ(z).
Distribution of the total
μT = 300 kg, σT = 0.5·√12 = 1.732 kg
Standardise
z = (298.3 − 300)/1.732 = −0.98
Probability
P = Φ(−0.98) = 0.1635
Answer
P = 0.1635
6.Normal distribution
Answer: P = 0.7881
Sums and differences of normal variables: if X and Y are independent and normal, X + Y and X − Y are normal too. Means add or subtract: E(X ± Y) = μX ± μY. Variances always add, also for a difference: Var(X ± Y) = σX² + σY², so σ = √(σX² + σY²). Standard deviations never add directly. The total of n independent items with the same distribution has mean n·μ and standard deviation σ·√n. Then standardise as usual: z = (x − mean)/σ and read Φ(z).
Distribution of the difference
μ = 52 − 48 = 4 s, σ = 5 s
Standardise
z = (0 − 4)/5 = −0.8
Probability
P = 1 − Φ(−0.8) = 0.7881
Answer
P = 0.7881
7.Confidence intervals
Answer: upper limit = 50.31
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.
Critical value
df = n − 1 = 10, t = 3.169
Margin of error
E = 3.169·15.5/√11 = 14.81
Interval
35.5 ± 14.81 → [20.69, 50.31]
Answer
upper limit = 50.31
8.Confidence intervals
Answer: n = 87
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 = (1.96·19/4)² = 86.68
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 87
Answer
n = 87
9.Hypothesis tests & confidence intervals
Answer: t = 2.299
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 = 11.6/√7 = 4.384
Test statistic
t = 10.08/4.384 = 2.299
Answer
t = 2.299
With df = 6, the two-sided 5% critical value is 2.447; |t| is smaller, so H₀ is not rejected.
10.Hypothesis tests & confidence intervals
Answer: z = 1.866
Hypothesis test: the test statistic = (estimate − value under H₀)/(standard error under H₀). The p-value is the probability, assuming H₀ is true, of a statistic at least as extreme as the one observed: the area in one tail for a one-sided H₁, twice the tail area for a two-sided H₁. Reject H₀ when p ≤ α.
Sample proportions
p̂₁ = 65/100 = 0.65
p̂₂ = 52/100 = 0.52
p̂ = 117/200 = 0.585
Standard error under H₀
SE = √(0.585·0.415·(1/100 + 1/100)) = 0.06968
Test statistic
z = (0.65 − 0.52)/0.06968 = 1.866
Answer
z = 1.866
|z| < 1.96, so H₀ is not rejected at the 5% level (two-sided).
11.Linear regression & correlation
Answer: r² = 0.9756
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 = 141/6 = 23.5
ȳ = 784/6 = 130.7
Sums of squares and products
Sxx = Σ(x − x̄)² = 157.5
Sxy = Σ(x − x̄)(y − ȳ) = 855
Syy = Σ(y − ȳ)² = 4757
Coefficient of determination
r² = 855²/(157.5·4757) = 0.9756
Answer
r² = 0.9756
About 97.56% of the variation in y is explained by the line.
12.Linear regression & correlation
Answer: e = 4.876
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 = 36/6 = 6
ȳ = 496/6 = 82.67
Sums of squares and products
Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 358
Regression line
b = 358/70 = 5.114
a = 82.67 − 5.114·6 = 51.98
ŷ = 51.98 + 5.114·x
Predicted value
ŷ = 51.98 + 5.114·10 = 103.1
Residual
e = 108 − 103.1 = 4.876
Answer
e = 4.876
The point lies above the line.
ExamProblems · examproblems.com/courses/statistics-1