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
3540455055

Find the sample standard deviation s of the data: 38, 57, 47, 35, 47, 53.

n = 6
s =
2.Descriptive statisticsHard
z −2−1012

Exam scores are normally distributed with mean 170 and standard deviation 15. Find the z-score z of a score of 155.

μ = 170σ = 15x = 155
z =
3.ProbabilityMedium
P(at least one six) = ??1?2?3?4?5?6?7P(six) = 1/6 on each roll

A fair die is rolled 7 times. What is the probability P of getting at least one six?

n = 7
P =
4.ProbabilityHard
xp(x)10.2520.330.2540.2μ

X has the distribution shown, and Y = −2X + 4. Find the standard deviation σY of Y.

x: 1, 2, 3, 4p(x): 0.25, 0.3, 0.25, 0.2Y = −2X + 4
σY =
5.Normal distributionMedium
53.573.5P = ?

The exam scores are normally distributed with mean 65 points and standard deviation 10 points. What fraction lies between 53.5 and 73.5 points? Give P as a decimal.

μ = 65σ = 10x1 = 53.5x2 = 73.5
P =
6.Normal distributionHard
235240245250255260265gtop 80%

The bag weights are normally distributed with mean 250 g and standard deviation 5 g. Drag the cut-off so the shaded top 80% is to the right of it. Find the cut-off x.

μ = 250 gσ = 5 g80%
x = g
7.Confidence intervalsMedium
n = 7, s = 10.5LUx̄ = 101.5E = ?99%

A sample of 7 items has mean 101.5 and standard deviation s = 10.5. The population is roughly normal. Find the margin of error E of a 99% confidence interval for the mean.

n = 7x̄ = 101.5s = 10.599%t(6) = 3.707
E =
8.Confidence intervalsHard
p = 0.5, n = ?p̂ − Ep̂ + Ep̂E = 0.0599%

Nothing is known in advance about the proportion of defective products. How many products n must be inspected so that a 99% confidence interval for the proportion has a margin of error of at most 0.05?

p = 0.5E = 0.0599%
n =
9.Hypothesis tests & confidence intervalsMedium
−2.0642.064t = ?df = 24

Two teaching methods are compared. Group 1: 17 students, mean score 104, standard deviation 11. Group 2: 9 students, mean score 94.8, standard deviation 11.5. 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 = 17x̄₁ = 104s₁ = 11n₂ = 9x̄₂ = 94.8s₂ = 11.5
t =
10.Hypothesis tests & confidence intervalsHard
z −2.57602.576

A sample of 16 measurements has mean 85.9. The population standard deviation is 15. Find the lower limit L of the 99% confidence interval for the mean.

x̄ = 85.9σ = 15n = 1699%
L =
11.Linear regression & correlationMedium
xy274136208271033xy510102030

The table gives 5 paired observations of x = load in kN and y = extension in mm. Find the coefficient of determination r² of the least-squares line.

n = 5
r² =
12.Linear regression & correlationHard
xy2514536568671076xy510506070x = 9

The table gives 5 paired observations of x = hours studied and y = exam score. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 9.

n = 5x = 9
ŷ =

3Solutions

1.Descriptive statistics

Answer: s = 8.448

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 = 277/6 = 46.17

Squared deviations

Σ(x − x̄)² = (−8.167)² + 10.83² + 0.8333² + (−11.17)² + 0.8333² + 6.833² = 356.8

Standard deviation

s = √(356.8/5) = 8.448

Answer

s = 8.448
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 = (155 − 170)/15 = −1

Answer

z = −1
3.Probability

Answer: P = 0.7209

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)7 = 0.2791
P = 1 − 0.2791 = 0.7209

Answer

P = 0.7209
4.Probability

Answer: σY = 2.135

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 = 2.4, Var(X) = 1.14, σX = 1.068

Linear change

σY = |−2|·1.068 = 2.135

Answer

σY = 2.135

The mean does move: E(Y) = −2·2.4 + 4 = −0.8.

5.Normal distribution

Answer: P = 0.6773

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 = (53.5 − 65)/10 = −1.15
z₂ = (73.5 − 65)/10 = 0.85

Area between

P = Φ(0.85) − Φ(−1.15) = 0.8023 − 0.1251 = 0.6773

Answer

P = 0.6773
6.Normal distribution

Answer: x = 245.8 g

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) = 0.2

z from the table

z ≈ −0.84

Back to the original units

x = 250 − 0.84·5 = 245.8 g

Answer

x = 245.8 g
7.Confidence intervals

Answer: E = 14.71

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 = 6, t = 3.707

Margin of error

E = 3.707·10.5/√7 = 14.71

Answer

E = 14.71
8.Confidence intervals

Answer: n = 664

Confidence interval for a proportion: p̂ ± z·√(p̂(1 − p̂)/n), where p̂ = x/n. To reach a margin E you need n ≥ z²·p(1 − p)/E², rounded up; without a prior estimate use p = 0.5, which gives the largest n.

Solve for n

n = 2.576²·0.5·0.5/0.05² = 663.6

Round up

n = 664

Answer

n = 664
9.Hypothesis tests & confidence intervals

Answer: t = 1.998

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 = (16·11² + 8·11.5²)/24 = 124.8
sp = 11.17

Standard error

SE = 11.17·√(1/17 + 1/9) = 4.604

Test statistic

t = (104 − 94.8)/4.604 = 1.998

Answer

t = 1.998

With df = 24, the two-sided 5% critical value is 2.064; |t| is smaller, so H₀ is not rejected.

10.Hypothesis tests & confidence intervals

Answer: L = 76.24

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 = 2.576·15/√16 = 9.66

Interval limit

L = 85.9 − 9.66 = 76.24

Answer

L = 76.24
11.Linear regression & correlation

Answer: r² = 0.9991

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 = 100/5 = 20

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 132
Syy = Σ(y − ȳ)² = 436

Coefficient of determination

r² = 132²/(40·436) = 0.9991

Answer

r² = 0.9991

About 99.91% of the variation in y is explained by the line.

12.Linear regression & correlation

Answer: ŷ = 70.2

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 = 30/5 = 6
ȳ = 303/5 = 60.6

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 128

Regression line

b = 128/40 = 3.2
a = 60.6 − 3.2·6 = 41.4
ŷ = 41.4 + 3.2·x

Prediction

ŷ = 41.4 + 3.2·9 = 70.2

Answer

ŷ = 70.2

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