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
3035404550

Find the sample standard deviation s of the data: 26, 50, 42, 38, 44, 39.

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

A quantity is normally distributed with mean 100 and standard deviation 10. What proportion P of values is above 85?

μ = 100σ = 10x = 85
P =
3.ProbabilityMedium
xp(x)00.320.240.3560.15μ = 2.7

For the distribution shown, find the variance Var(X) of X.

x: 0, 2, 4, 6p(x): 0.3, 0.2, 0.35, 0.15
Var(X) =
4.ProbabilityHard
n = 7, p = 0.4: P(X ≤ 1) = ?k01234567

7 components are tested, and each one fails with probability 0.4, independently. What is the probability P that at most one fails?

n = 7p = 0.4
P =
5.Normal distributionMedium
−3−2−10123zshaded area to the right of z = 0.9678

Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.9678. Find z to two decimals.

A = 0.9678
z =
6.Normal distributionHard
0.02 mm0D − d?

Hole diameters D are normal with mean 25.02 mm and standard deviation 0.01 mm. Shaft diameters d are independent and normal with mean 25 mm and standard deviation 0.015 mm. A randomly chosen shaft fits when D − d > 0. Find the probability P that it fits. Round z to two decimals.

hole: μ = 25.02 mm, σ = 0.01 mmshaft: μ = 25 mm, σ = 0.015 mm
P =
7.Confidence intervalsMedium
μ₁ − μ₂0L = ?Ux̄₁ − x̄₂E95%

The battery life of two phone models is compared. Model 1: 100 phones, mean 45.5 h, known standard deviation 7 h. Model 2: 80 phones, mean 41 h, known standard deviation 6 h. Find the lower limit of the 95% confidence interval for μ₁ − μ₂.

n₁ = 100x̄₁ = 45.5 hσ₁ = 7 hn₂ = 80x̄₂ = 41 hσ₂ = 6 h95%
lower limit = h
8.Confidence intervalsHard
σ = 18, n = ?x̄ − Ex̄ + Ex̄E = 290%

The population standard deviation is 18. How large a sample n do you need so that a 90% confidence interval for the mean has a margin of error of at most 2?

σ = 18E = 290%
n =
9.Hypothesis tests & confidence intervalsMedium
−1.74p = ?

The mean of a population is claimed to be 500; the population standard deviation is known to be 32. A random sample of 64 observations has a mean of 493.04. Test H₀: μ = 500 against H₁: μ < 500. Find the p-value p.

μ₀ = 500σ = 32n = 64x̄ = 493.04H₁: μ < μ₀
p =
10.Hypothesis tests & confidence intervalsHard
μ₀x̄c

A one-sided z test of H₀: μ = 80 against H₁: μ > 80 is done at α = 0.01 with a sample of 25; the standard deviation is known to be 5. In reality the mean is 83.9. Find the probability of a Type II error β in percent.

μ₀ = 80μ = 83.9σ = 5n = 25α = 0.01
β = %
11.Linear regression & correlationMedium
xy171022011023131261572917632187xy202530100150

The table gives 6 paired observations of x = temperature in °C and y = ice creams sold. Their sums are given. Find Pearson's correlation coefficient r from these sums.

n = 6Σx = 147Σy = 863Σx² = 3759Σy² = 130259Σxy = 22117
r =
12.Linear regression & correlationHard
xy25486118141019xy5105101520x = 3

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. A new observation (3, 11) is made. Find its residual e = y − ŷ.

n = 5(3, 11)
e =

3Solutions

1.Descriptive statistics

Answer: s = 8.01

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 = 239/6 = 39.83

Squared deviations

Σ(x − x̄)² = (−13.83)² + 10.17² + 2.167² + (−1.833)² + 4.167² + (−0.8333)² = 320.8

Standard deviation

s = √(320.8/5) = 8.01

Answer

s = 8.01
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) = 4.51

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 = 0·0.3 + 2·0.2 + 4·0.35 + 6·0.15 = 2.7

Mean of the squares

E(X²) = 0·0.3 + 4·0.2 + 16·0.35 + 36·0.15 = 11.8

Variance

Var(X) = 11.8 − 2.7² = 4.51

Answer

Var(X) = 4.51
4.Probability

Answer: P = 0.1586

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) = 0.67 = 0.02799
P(X = 1) = 7·0.4·0.66 = 0.1306

Add them

P = 0.02799 + 0.1306 = 0.1586

Answer

P = 0.1586
5.Normal distribution

Answer: z = −1.85

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 − 0.9678 = 0.0322

Read the z-table backwards

Find 0.0322 inside the table and read off z.

z = −1.85

Answer

z = −1.85
6.Normal distribution

Answer: P = 0.8665

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 = 25.02 − 25 = 0.02 mm, σ = 0.01803 mm

Standardise

z = (0 − 0.02)/0.01803 = −1.11

Probability

P = 1 − Φ(−1.11) = 0.8665

Answer

P = 0.8665
7.Confidence intervals

Answer: lower limit = 2.6 h

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.

Point estimate

x̄₁ − x̄₂ = 45.5 − 41 = 4.5

Standard error

SE = √(7²/100 + 6²/80) = 0.9695

Margin of error

E = 1.96·0.9695 = 1.9

Interval

4.5 ± 1.9 → [2.6, 6.4]

Answer

lower limit = 2.6 h

The interval does not contain 0, so the means differ significantly.

8.Confidence intervals

Answer: n = 220

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 = (1.645·18/2)² = 219.2

Round up

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

n = 220

Answer

n = 220
9.Hypothesis tests & confidence intervals

Answer: p = 0.04093

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.

Test statistic

SE = σ/√n = 32/√64 = 4
z = (493.04 − 500)/4 = −1.74

Tail area

Φ(1.74) = 0.9591
1 − Φ(1.74) = 0.0409

p-value

p = 0.04093

Answer

p = 0.04093

p ≤ 0.05, so H₀ is rejected at the 5% level.

10.Hypothesis tests & confidence intervals

Answer: β = 5.774 %

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 = 5/√25 = 1
x̄c = 80 + 2.326·1 = 82.33

z under the true mean

z = (82.33 − 83.9)/1 = −1.574

Type II error

power = P(Z > −1.574) = 94.23%
β = 100% − 94.23% = 5.774%

Answer

β = 5.774%
11.Linear regression & correlation

Answer: r = 0.9907

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 = 22117 − 147·863/6 = 973.5
Sxx = 3759 − 147²/6 = 157.5
Syy = 130259 − 863²/6 = 6131

Correlation coefficient

r = 973.5/√(157.5·6131) = 0.9907

Answer

r = 0.9907
12.Linear regression & correlation

Answer: e = 4.7

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 = 57/5 = 11.4

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 68

Regression line

b = 68/40 = 1.7
a = 11.4 − 1.7·6 = 1.2
ŷ = 1.2 + 1.7·x

Predicted value

ŷ = 1.2 + 1.7·3 = 6.3

Residual

e = 11 − 6.3 = 4.7

Answer

e = 4.7

The point lies above the line.

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