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
152025303540

Find the sample standard deviation s of the data: 12, 42, 26, 31, 35.

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

A quantity is normally distributed with mean 50 and standard deviation 5. What proportion P of values is below 60?

μ = 50σ = 5x = 60
P =
3.ProbabilityMedium
xp(x)100.3200.35300.3400.05μ = 21

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

x: 10, 20, 30, 40p(x): 0.3, 0.35, 0.3, 0.05
Var(X) =
4.ProbabilityHard
n = 6, p = 0.6: P(X ≤ 1) = ?k0123456

6 components are tested, and each one fails with probability 0.6, independently. What is the probability P that at most one fails?

n = 6p = 0.6
P =
5.Normal distributionMedium
−3−2−10123zshaded area between −z and z = 0.8473

Drag z so that the shaded area between −z and z under the standard normal curve is 0.8473. Find z to two decimals.

A = 0.8473
z =
6.Normal distributionHard
557085100115130145pointsbottom 75%

The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. Drag the cut-off so the shaded bottom 75% is to the left of it. Find the cut-off x.

μ = 100 pointsσ = 15 points75%
x = points
7.Confidence intervalsMedium
n = 10, s = 10L = ?Ux̄ = 73E95%

A sample of 10 items has mean 73 and standard deviation s = 10. The population is roughly normal. Find the lower limit of the 95% confidence interval for the mean.

n = 10x̄ = 73s = 1095%t(9) = 2.262
lower limit =
8.Confidence intervalsHard
n = 10Lσ = ?s = 295%

The diameters of 10 machined shafts have a sample standard deviation of 2 µm. Assuming the diameters are normally distributed, find the upper limit σ of the 95% confidence interval for the population standard deviation. With df = 9, 2.5% of the χ² distribution lies below 2.7 and 2.5% lies above 19.023.

n = 10s = 2 µmχ² = 2.7, 19.023
σ = µm
9.Hypothesis tests & confidence intervalsMedium
−2.1792.179t = ?df = 12

Two teaching methods are compared. Group 1: 9 students, mean score 96, standard deviation 11. Group 2: 5 students, mean score 109.3, standard deviation 9. 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 = 9x̄₁ = 96s₁ = 11n₂ = 5x̄₂ = 109.3s₂ = 9
t =
10.Hypothesis tests & confidence intervalsHard
z −1.64501.645

A sample of 16 measurements has mean 202.1. The population standard deviation is 12. Find the upper limit U of the 90% confidence interval for the mean.

x̄ = 202.1σ = 12n = 1690%
U =
11.Linear regression & correlationMedium
xy2841262382710351241xy5102040

The table gives 6 paired observations of x = load in kN and y = extension in mm. Their sums are given. Find Pearson's correlation coefficient r from these sums.

n = 6Σx = 42Σy = 146Σx² = 364Σy² = 4372Σxy = 1260
r =
12.Linear regression & correlationHard
xy22843664284610481249xy510304050x = 7

The table gives 6 paired observations of x = advertising spend in k€ and y = sales in k€. Use the least-squares line ŷ = a + b·x. A new observation (7, 46) is made. Find its residual e = y − ŷ.

n = 6(7, 46)
e =

3Solutions

1.Descriptive statistics

Answer: s = 11.26

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 = 146/5 = 29.2

Squared deviations

Σ(x − x̄)² = (−17.2)² + 12.8² + (−3.2)² + 1.8² + 5.8² = 506.8

Standard deviation

s = √(506.8/4) = 11.26

Answer

s = 11.26
2.Descriptive statistics

Answer: P = 0.9772

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 = 2

Standard normal table

Φ(2) = 0.9772
P = 0.9772

Answer

P = 0.9772
3.Probability

Answer: Var(X) = 79

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

Answer

Var(X) = 79
4.Probability

Answer: P = 0.04096

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) = 0.46 = 0.004096
P(X = 1) = 6·0.6·0.45 = 0.03686

Add them

P = 0.004096 + 0.03686 = 0.04096

Answer

P = 0.04096
5.Normal distribution

Answer: z = 1.43

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 + 0.8473)/2 = 0.9236

Read the z-table backwards

Find 0.9236 inside the table and read off z.

z = 1.43

Answer

z = 1.43
6.Normal distribution

Answer: x = 110.1 points

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) = 0.75

z from the table

z ≈ 0.67

Back to the original units

x = 100 + 0.67·15 = 110.1 points

Answer

x = 110.1 points
7.Confidence intervals

Answer: lower limit = 65.85

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 = 9, t = 2.262

Margin of error

E = 2.262·10/√10 = 7.153

Interval

73 ± 7.153 → [65.85, 80.15]

Answer

lower limit = 65.85
8.Confidence intervals

Answer: σ = 3.651 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 = 9·2² = 36

Limits for σ²

36/19.023 = 1.892
36/2.7 = 13.33

Limit for σ

σ = √13.33 = 3.651 µm

Answer

σ = 3.651 µm

95% interval for σ: [1.376, 3.651] µm.

9.Hypothesis tests & confidence intervals

Answer: t = −2.298

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 = (8·11² + 4·9²)/12 = 107.7
sp = 10.38

Standard error

SE = 10.38·√(1/9 + 1/5) = 5.788

Test statistic

t = (96 − 109.3)/5.788 = −2.298

Answer

t = −2.298

With df = 12, the two-sided 5% critical value is 2.179; |t| is larger, so H₀ is rejected.

10.Hypothesis tests & confidence intervals

Answer: U = 207

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 = 1.645·12/√16 = 4.935

Interval limit

U = 202.1 + 4.935 = 207

Answer

U = 207
11.Linear regression & correlation

Answer: r = 0.9938

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 = 1260 − 42·146/6 = 238
Sxx = 364 − 42²/6 = 70
Syy = 4372 − 146²/6 = 819.3

Correlation coefficient

r = 238/√(70·819.3) = 0.9938

Answer

r = 0.9938
12.Linear regression & correlation

Answer: e = 4.5

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 = 249/6 = 41.5

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 145

Regression line

b = 145/70 = 2.071
a = 41.5 − 2.071·7 = 27
ŷ = 27 + 2.071·x

Predicted value

ŷ = 27 + 2.071·7 = 41.5

Residual

e = 46 − 41.5 = 4.5

Answer

e = 4.5

The point lies above the line.

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