Introductory Statistics: cram pack

Formula sheets, the key method steps for every topic and 12 practice problems, with full solutions at the back.

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1Formulas and methods

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

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

  1. Convert the value to a z-score: z = (x − μ)/σ.
  2. Look up Φ(z), the area to the left of z.
  3. For the area to the right use 1 − Φ(z); for an interval subtract the two left areas.
  4. 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

  1. Decide between z (population σ known) and t (only the sample s, with n − 1 degrees of freedom).
  2. Find the critical value for the confidence level.
  3. Margin of error: E = critical value × standard deviation/√n.
  4. 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

  1. State H₀ and H₁ and decide whether the test is one- or two-sided.
  2. Calculate the test statistic: (x̄ − μ₀)/(σ/√n), or with s and t.
  3. Compare with the critical value, or find the p-value.
  4. 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
2025303540455055

Find the sample standard deviation s of the data: 59, 22, 57, 19, 35.

n = 5
s =
2.Descriptive statisticsHard
z −2−1012

A quantity is normally distributed with mean 70 and standard deviation 12. What proportion P of values is above 88?

μ = 70σ = 12x = 88
P =
3.ProbabilityMedium
xp(x)100.3200.35300.3400.05μ = 21

For the distribution shown, find the standard deviation σ of X.

x: 10, 20, 30, 40p(x): 0.3, 0.35, 0.3, 0.05
σ =
4.ProbabilityHard
xp(x)00.3510.0520.2530.35μ

X has the distribution shown, and Y = 10X − 3. Find the standard deviation σY of Y.

x: 0, 1, 2, 3p(x): 0.35, 0.05, 0.25, 0.35Y = 10X − 3
σY =
5.Normal distributionMedium
100 kg100.5 kgT = X₁ + … + X₄?

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 4 of them is below 100.5 kg. Round z to two decimals.

μ = 25 kgσ = 0.5 kgn = 4< 100.5 kg
P =
6.Normal distributionHard
μ = 65x = ?25%

The exam scores are normally distributed with mean 65 points and standard deviation 10 points. Find the value x below which 25% of them lie (the 25th percentile).

μ = 65σ = 10P = 25%
x = points
7.Confidence intervalsMedium
n = 21, s = 14.5LUx̄ = 42.5E = ?95%

A sample of 21 items has mean 42.5 and standard deviation s = 14.5. The population is roughly normal. Find the margin of error E of a 95% confidence interval for the mean.

n = 21x̄ = 42.5s = 14.595%t(20) = 2.086
E =
8.Confidence intervalsHard
σ = 10, n = ?x̄ − Ex̄ + Ex̄E = 499%

The population standard deviation is 10. 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 = 10E = 499%
n =
9.Hypothesis tests & confidence intervalsMedium
−2.0932.093t = ?df = 19

Two teaching methods are compared. Group 1: 15 students, mean score 108, standard deviation 11.5. Group 2: 6 students, mean score 121, standard deviation 3. Assume equal population variances and use the pooled two-sample t test. Find the test statistic t for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (group 1 minus group 2).

n₁ = 15x̄₁ = 108s₁ = 11.5n₂ = 6x̄₂ = 121s₂ = 3
t =
10.Hypothesis tests & confidence intervalsHard
z −2.57602.576

A sample of 100 measurements has mean 74. The population standard deviation is 5. Find the upper limit U of the 99% confidence interval for the mean.

x̄ = 74σ = 5n = 10099%
U =
11.Linear regression & correlationMedium
xy150353568771980xy550607080

The table gives 5 paired observations of x = hours studied and y = exam score. Their sums are given. Find Pearson's correlation coefficient r from these sums.

n = 5Σx = 25Σy = 322Σx² = 165Σy² = 21374Σxy = 1766
r =
12.Linear regression & correlationHard
xy274136198261032xy510102030ŷ = 16

The table gives 5 paired observations of x = load in kN and y = extension in mm. Use the least-squares line ŷ = a + b·x. For which value x does the line predict ŷ = 16?

n = 5ŷ = 16
x =

3Solutions

1.Descriptive statistics

Answer: s = 18.89

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 = 192/5 = 38.4

Squared deviations

Σ(x − x̄)² = 20.6² + (−16.4)² + 18.6² + (−19.4)² + (−3.4)² = 1427

Standard deviation

s = √(1427/4) = 18.89

Answer

s = 18.89
2.Descriptive statistics

Answer: P = 0.06681

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) = 0.9332
P = 1 − 0.9332 = 0.06681

Answer

P = 0.06681
3.Probability

Answer: σ = 8.888

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 = 10·0.3 + 20·0.35 + 30·0.3 + 40·0.05 = 21

Mean of the squares

E(X²) = 100·0.3 + 400·0.35 + 900·0.3 + 1600·0.05 = 520

Variance

Var(X) = 520 − 21² = 79
σ = √79 = 8.888

Answer

σ = 8.888
4.Probability

Answer: σY = 12.81

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 = 1.6, Var(X) = 1.64, σX = 1.281

Linear change

σY = |10|·1.281 = 12.81

Answer

σY = 12.81

The mean does move: E(Y) = 10·1.6 − 3 = 13.

5.Normal distribution

Answer: P = 0.6915

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 = 100 kg, σT = 0.5·√4 = 1 kg

Standardise

z = (100.5 − 100)/1 = 0.5

Probability

P = Φ(0.5) = 0.6915

Answer

P = 0.6915
6.Normal distribution

Answer: x = 58.26 points

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) = 0.25 → z = −0.6745

Back to x

x = 65 + −0.6745·10 = 58.26 points

Answer

x = 58.26 points
7.Confidence intervals

Answer: E = 6.6

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 = 20, t = 2.086

Margin of error

E = 2.086·14.5/√21 = 6.6

Answer

E = 6.6
8.Confidence intervals

Answer: n = 42

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 = (2.576·10/4)² = 41.47

Round up

A sample size is a whole number, and rounding down would make the margin too large.

n = 42

Answer

n = 42
9.Hypothesis tests & confidence intervals

Answer: t = −2.694

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.

Pooled variance

sp² = (14·11.5² + 5·3²)/19 = 99.82
sp = 9.991

Standard error

SE = 9.991·√(1/15 + 1/6) = 4.826

Test statistic

t = (108 − 121)/4.826 = −2.694

Answer

t = −2.694

With df = 19, the two-sided 5% critical value is 2.093; |t| is larger, so H₀ is rejected.

10.Hypothesis tests & confidence intervals

Answer: U = 75.29

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 = 2.576·5/√100 = 1.288

Interval limit

U = 74 + 1.288 = 75.29

Answer

U = 75.29
11.Linear regression & correlation

Answer: r = 0.9771

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 = 1766 − 25·322/5 = 156
Sxx = 165 − 25²/5 = 40
Syy = 21374 − 322²/5 = 637.2

Correlation coefficient

r = 156/√(40·637.2) = 0.9771

Answer

r = 0.9771
12.Linear regression & correlation

Answer: x = 4.921

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 = 30/5 = 6
ȳ = 97/5 = 19.4

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 126

Regression line

b = 126/40 = 3.15
a = 19.4 − 3.15·6 = 0.5
ŷ = 0.5 + 3.15·x

Solve for x

x = (16 − 0.5)/3.15 = 4.921

Answer

x = 4.921

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