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
202530354045505560

Find the sample standard deviation s of the data: 39, 18, 18, 25, 25, 60, 20.

n = 7
s =
2.Descriptive statisticsHard
z −2−1012

Exam scores are normally distributed with mean 170 and standard deviation 8. Find the z-score z of a score of 180.

μ = 170σ = 8x = 180
z =
3.ProbabilityMedium
red: 7blue: 4

A bag holds 7 red and 4 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?

7 / 4
P =
4.ProbabilityHard
xp(x)10.220.3530.2540.2μ

X has the distribution shown, and Y = 0.5X + 5. Find the standard deviation σY of Y.

x: 1, 2, 3, 4p(x): 0.2, 0.35, 0.25, 0.2Y = 0.5X + 5
σY =
5.Normal distributionMedium
−3−2−10123zshaded area to the right of z = 0.8159

Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.8159. Find z to two decimals.

A = 0.8159
z =
6.Normal distributionHard
10 cm0X − Y?

The heights X of men and Y of women in a group are independent and normal: X has mean 180 cm and standard deviation 10 cm, Y has mean 170 cm and standard deviation 12 cm. Find the probability P that X is greater than Y. Round z to two decimals.

μX = 180 cm, σX = 10 cmμY = 170 cm, σY = 12 cm
P =
7.Confidence intervalsMedium
n = 13, s = 13LUx̄ = 40E = ?90%

A sample of 13 items has mean 40 and standard deviation s = 13. The population is roughly normal. Find the margin of error E of a 90% confidence interval for the mean.

n = 13x̄ = 40s = 1390%t(12) = 1.782
E =
8.Confidence intervalsHard
n = 21Lσ = ?s = 395%

The diameters of 21 machined shafts have a sample standard deviation of 3 µm. Assuming the diameters are normally distributed, find the upper limit σ of the 95% confidence interval for the population standard deviation. With df = 20, 2.5% of the χ² distribution lies below 9.591 and 2.5% lies above 34.17.

n = 21s = 3 µmχ² = 9.591, 34.17
σ = µm
9.Hypothesis tests & confidence intervalsMedium
t −2.0302.03

A sample of 36 students has a mean score of 72.6 and a sample standard deviation of 8. Find the test statistic t for H₀: μ = 75.

μ₀ = 75x̄ = 72.6s = 8n = 36
t =
10.Hypothesis tests & confidence intervalsHard
−1.961.96z = ?

In a trial, 40 of 80 patients on treatment A recover, and 57 of 100 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.

x₁ = 40n₁ = 80x₂ = 57n₂ = 100
z =
11.Linear regression & correlationMedium
xy128231328431539638xy246303540

The table gives 6 paired observations of x = advertising spend in k€ and y = sales in k€. Find Pearson's correlation coefficient r.

n = 6
r =
12.Linear regression & correlationHard
xy123222319413513610xy246101520ŷ = 14

The table gives 6 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 = 14?

n = 6ŷ = 14
x =

3Solutions

1.Descriptive statistics

Answer: s = 15.36

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 = 205/7 = 29.29

Squared deviations

Σ(x − x̄)² = 9.714² + (−11.29)² + (−11.29)² + (−4.286)² + (−4.286)² + 30.71² + (−9.286)² = 1415

Standard deviation

s = √(1415/6) = 15.36

Answer

s = 15.36
2.Descriptive statistics

Answer: z = 1.25

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 = (180 − 170)/8 = 1.25

Answer

z = 1.25
3.Probability

Answer: P = 0.3818

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.

First and second draw

P = 7/11 · 6/10 = 42/110 = 0.3818

Answer

P = 0.3818
4.Probability

Answer: σY = 0.5117

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 = 2.45, Var(X) = 1.047, σX = 1.023

Linear change

σY = |0.5|·1.023 = 0.5117

Answer

σY = 0.5117

The mean does move: E(Y) = 0.5·2.45 + 5 = 6.225.

5.Normal distribution

Answer: z = −0.9

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 − 0.8159 = 0.1841

Read the z-table backwards

Find 0.1841 inside the table and read off z.

z = −0.90

Answer

z = −0.90
6.Normal distribution

Answer: P = 0.7389

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 = 180 − 170 = 10 cm, σ = 15.62 cm

Standardise

z = (0 − 10)/15.62 = −0.64

Probability

P = 1 − Φ(−0.64) = 0.7389

Answer

P = 0.7389
7.Confidence intervals

Answer: E = 6.425

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 = 12, t = 1.782

Margin of error

E = 1.782·13/√13 = 6.425

Answer

E = 6.425
8.Confidence intervals

Answer: σ = 4.332 µm

Interval for a variance: (n − 1)s²/σ² has a χ² distribution with n − 1 degrees of freedom. The 95% interval for σ² runs from (n − 1)s² divided by the upper χ² value (2.5% above it) to (n − 1)s² divided by the lower χ² value (2.5% below it); the interval for σ is the square root of both limits. It is not symmetric around s.

Sum of squares

(n − 1)s² = 20·3² = 180

Limits for σ²

180/34.17 = 5.268
180/9.591 = 18.77

Limit for σ

σ = √18.77 = 4.332 µm

Answer

σ = 4.332 µm

95% interval for σ: [2.295, 4.332] µm.

9.Hypothesis tests & confidence intervals

Answer: t = −1.8

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 = 8/√36 = 8/6 = 1.333

Test statistic

t = (72.6 − 75)/1.333 = −1.8

Answer

t = −1.8
t0.025; df = 35 = 2.03

|t| < 2.03, so H₀ is not rejected at the 5% level (two-sided).

10.Hypothesis tests & confidence intervals

Answer: z = −0.9362

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 = 40/80 = 0.5
p̂₂ = 57/100 = 0.57
p̂ = 97/180 = 0.5389

Standard error under H₀

SE = √(0.5389·0.4611·(1/80 + 1/100)) = 0.07477

Test statistic

z = (0.5 − 0.57)/0.07477 = −0.9362

Answer

z = −0.9362

|z| < 1.96, so H₀ is not rejected at the 5% level (two-sided).

11.Linear regression & correlation

Answer: r = 0.849

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 = 21/6 = 3.5
ȳ = 195/6 = 32.5

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = 38.5
Syy = Σ(y − ȳ)² = 117.5

Correlation coefficient

r = 38.5/√(17.5·117.5) = 0.849

Answer

r = 0.849
12.Linear regression & correlation

Answer: x = 4.452

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 = 21/6 = 3.5
ȳ = 100/6 = 16.67

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −49

Regression line

b = −49/17.5 = −2.8
a = 16.67 − (−2.8)·3.5 = 26.47
ŷ = 26.47 − 2.8·x

Solve for x

x = (14 − 26.47)/(−2.8) = 4.452

Answer

x = 4.452

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