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
303540455055

Find the sample standard deviation s of the data: 52, 59, 50, 29, 39, 36.

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

A quantity is normally distributed with mean 60 and standard deviation 15. What proportion P of values is above 78.75?

μ = 60σ = 15x = 78.75
P =
3.ProbabilityMedium
red: 7blue: 6

A bag holds 7 red and 6 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?

7 / 6
P =
4.ProbabilityHard
n = 12, p = 0.1: P(X ≤ 1) = ?k0123456789101112

12 components are tested, and each one fails with probability 0.1, independently. What is the probability P that at most one fails?

n = 12p = 0.1
P =
5.Normal distributionMedium
640 g644.8 gT = X₁ + … + X₁₆?

The masses of bolts are independent and normal with mean 40 g and standard deviation 1.2 g. Find the probability P that the total mass of 16 of them exceeds 644.8 g. Round z to two decimals.

μ = 40 gσ = 1.2 gn = 16> 644.8 g
P =
6.Normal distributionHard
μ = 100x = ?5%

The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. Find the value x below which 5% of them lie (the 5th percentile).

μ = 100σ = 15P = 5%
x = points
7.Confidence intervalsMedium
n = 25, s = 14LUx̄ = 51.5E = ?99%

A sample of 25 items has mean 51.5 and standard deviation s = 14. The population is roughly normal. Find the margin of error E of a 99% confidence interval for the mean.

n = 25x̄ = 51.5s = 1499%t(24) = 2.797
E =
8.Confidence intervalsHard
μ₁ − μ₂, t(15) = 1.7530LU = ?x̄₁ − x̄₂E90%

Two fertilisers are tested. Plot group 1: 8 plots, mean yield 90 kg, standard deviation 6.5 kg. Plot group 2: 9 plots, mean 79.5 kg, standard deviation 8 kg. Assume equal variances (pooled t, df = n₁ + n₂ − 2). Find the upper limit of the 90% confidence interval for μ₁ − μ₂.

n₁ = 8x̄₁ = 90s₁ = 6.5n₂ = 9x̄₂ = 79.5s₂ = 890%t(15) = 1.753
upper limit = kg
9.Hypothesis tests & confidence intervalsMedium
t −1.9901.99

A sample of 81 students has a mean score of 97 and a sample standard deviation of 15. Find the test statistic t for H₀: μ = 100.

μ₀ = 100x̄ = 97s = 15n = 81
t =
10.Hypothesis tests & confidence intervalsHard
μ₀x̄c

A one-sided z test of H₀: μ = 100 against H₁: μ < 100 is done at α = 0.05 with a sample of 36; the standard deviation is known to be 15. In reality the mean is 92. Find the power 1 − β of the test in percent.

μ₀ = 100μ = 92σ = 15n = 36α = 0.05
1 − β = %
11.Linear regression & correlationMedium
xy130234340447547651xy246304050

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

n = 6Σx = 21Σy = 249Σx² = 91Σy² = 10675Σxy = 947
r =
12.Linear regression & correlationHard
xy21341962983310421249xy5102040ŷ = 38

The table gives 6 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 = 38?

n = 6ŷ = 38
x =

3Solutions

1.Descriptive statistics

Answer: s = 11.3

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 = 265/6 = 44.17

Squared deviations

Σ(x − x̄)² = 7.833² + 14.83² + 5.833² + (−15.17)² + (−5.167)² + (−8.167)² = 638.8

Standard deviation

s = √(638.8/5) = 11.3

Answer

s = 11.3
2.Descriptive statistics

Answer: P = 0.1056

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 = 1.25

Standard normal table

Φ(1.25) = 0.8944
P = 1 − 0.8944 = 0.1056

Answer

P = 0.1056
3.Probability

Answer: P = 0.2692

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 = 7/13 · 6/12 = 42/156 = 0.2692

Answer

P = 0.2692
4.Probability

Answer: P = 0.659

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) = 0.912 = 0.2824
P(X = 1) = 12·0.1·0.911 = 0.3766

Add them

P = 0.2824 + 0.3766 = 0.659

Answer

P = 0.659
5.Normal distribution

Answer: P = 0.1587

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 = 640 g, σT = 1.2·√16 = 4.8 g

Standardise

z = (644.8 − 640)/4.8 = 1

Probability

P = 1 − Φ(1) = 0.1587

Answer

P = 0.1587
6.Normal distribution

Answer: x = 75.33 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) = 0.05 → z = −1.645

Back to x

x = 100 + −1.645·15 = 75.33 points

Answer

x = 75.33 points
7.Confidence intervals

Answer: E = 7.832

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 = 24, t = 2.797

Margin of error

E = 2.797·14/√25 = 7.832

Answer

E = 7.832
8.Confidence intervals

Answer: upper limit = 16.75 kg

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.

Pooled standard deviation

sp² = (7·6.5² + 8·8²)/15 = 53.85
sp = 7.338

Margin of error

E = 1.753·7.338·√(1/8 + 1/9) = 6.251

Interval

10.5 ± 6.251 → [4.249, 16.75]

Answer

upper limit = 16.75 kg
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 = 15/√81 = 15/9 = 1.667

Test statistic

t = (97 − 100)/1.667 = −1.8

Answer

t = −1.8
t0.025; df = 80 = 1.99

|t| < 1.99, so H₀ is not rejected at the 5% level (two-sided).

10.Hypothesis tests & confidence intervals

Answer: 1 − β = 94 %

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 = 15/√36 = 2.5
x̄c = 100 − 1.645·2.5 = 95.89

z under the true mean

z = (95.89 − 92)/2.5 = 1.555

Power

power = P(Z < 1.555) = 94%
1 − β = 94%

Answer

1 − β = 94%
11.Linear regression & correlation

Answer: r = 0.9766

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 = 947 − 21·249/6 = 75.5
Sxx = 91 − 21²/6 = 17.5
Syy = 10675 − 249²/6 = 341.5

Correlation coefficient

r = 75.5/√(17.5·341.5) = 0.9766

Answer

r = 0.9766
12.Linear regression & correlation

Answer: x = 8.983

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 = 42/6 = 7
ȳ = 185/6 = 30.83

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 253

Regression line

b = 253/70 = 3.614
a = 30.83 − 3.614·7 = 5.533
ŷ = 5.533 + 3.614·x

Solve for x

x = (38 − 5.533)/3.614 = 8.983

Answer

x = 8.983

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