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
1820222426283032

Find the sample standard deviation s of the data: 29, 18, 33, 22, 31.

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

Exam scores are normally distributed with mean 170 and standard deviation 5. Find the z-score z of a score of 175.

μ = 170σ = 5x = 175
z =
3.ProbabilityMedium
xp(x)10.1520.3530.0540.1550.3μ = 3.1

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

x: 1, 2, 3, 4, 5p(x): 0.15, 0.35, 0.05, 0.15, 0.3
Var(X) =
4.ProbabilityHard
n = 6, p = 0.1: P(X ≤ 1) = ?k0123456

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

n = 6p = 0.1
P =
5.Normal distributionMedium
−3−2−10123zshaded area to the right of z = 0.1611

Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.1611. Find z to two decimals.

A = 0.1611
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 0.02 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 = 0.02 mm
P =
7.Confidence intervalsMedium
n = 8, s = 3LUx̄ = 38E = ?95%

A sample of 8 items has mean 38 and standard deviation s = 3. The population is roughly normal. Find the margin of error E of a 95% confidence interval for the mean.

n = 8x̄ = 38s = 395%t(7) = 2.365
E =
8.Confidence intervalsHard
n = 25σ = ?Us = 4.195%

The diameters of 25 machined shafts have a sample standard deviation of 4.1 µm. Assuming the diameters are normally distributed, find the lower limit σ of the 95% confidence interval for the population standard deviation. With df = 24, 2.5% of the χ² distribution lies below 12.401 and 2.5% lies above 39.364.

n = 25s = 4.1 µmχ² = 12.401, 39.364
σ = µm
9.Hypothesis tests & confidence intervalsMedium
t −2.06402.064

A sample of 25 students has a mean score of 58.4 and a sample standard deviation of 20. Find the test statistic t for H₀: μ = 50.

μ₀ = 50x̄ = 58.4s = 20n = 25
t =
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 64; the standard deviation is known to be 16. In reality the mean is 85. Find the power 1 − β of the test in percent.

μ₀ = 80μ = 85σ = 16n = 64α = 0.01
1 − β = %
11.Linear regression & correlationMedium
xy234334436541645xy246354045

The table gives 5 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 = 5Σx = 20Σy = 190Σx² = 90Σy² = 7314Σxy = 789
r =
12.Linear regression & correlationHard
xy126225323419519617xy2462025x = 5.5

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. A new observation (5.5, 22) is made. Find its residual e = y − ŷ.

n = 6(5.5, 22)
e =

3Solutions

1.Descriptive statistics

Answer: s = 6.348

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 = 133/5 = 26.6

Squared deviations

Σ(x − x̄)² = 2.4² + (−8.6)² + 6.4² + (−4.6)² + 4.4² = 161.2

Standard deviation

s = √(161.2/4) = 6.348

Answer

s = 6.348
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 = (175 − 170)/5 = 1

Answer

z = 1
3.Probability

Answer: Var(X) = 2.29

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 = 1·0.15 + 2·0.35 + 3·0.05 + 4·0.15 + 5·0.3 = 3.1

Mean of the squares

E(X²) = 1·0.15 + 4·0.35 + 9·0.05 + 16·0.15 + 25·0.3 = 11.9

Variance

Var(X) = 11.9 − 3.1² = 2.29

Answer

Var(X) = 2.29
4.Probability

Answer: P = 0.8857

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) = 0.96 = 0.5314
P(X = 1) = 6·0.1·0.95 = 0.3543

Add them

P = 0.5314 + 0.3543 = 0.8857

Answer

P = 0.8857
5.Normal distribution

Answer: z = 0.99

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 − 0.1611 = 0.8389

Read the z-table backwards

Find 0.8389 inside the table and read off z.

z = 0.99

Answer

z = 0.99
6.Normal distribution

Answer: P = 0.8133

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 = 0.02236 mm

Standardise

z = (0 − 0.02)/0.02236 = −0.89

Probability

P = 1 − Φ(−0.89) = 0.8133

Answer

P = 0.8133
7.Confidence intervals

Answer: E = 2.508

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 = 7, t = 2.365

Margin of error

E = 2.365·3/√8 = 2.508

Answer

E = 2.508
8.Confidence intervals

Answer: σ = 3.201 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 = 24·4.1² = 403.4

Limits for σ²

403.4/39.364 = 10.25
403.4/12.401 = 32.53

Limit for σ

σ = √10.25 = 3.201 µm

Answer

σ = 3.201 µm

95% interval for σ: [3.201, 5.704] µm.

9.Hypothesis tests & confidence intervals

Answer: t = 2.1

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 = 20/√25 = 20/5 = 4

Test statistic

t = (58.4 − 50)/4 = 2.1

Answer

t = 2.1
t0.025; df = 24 = 2.064

|t| > 2.064, so H₀ is rejected at the 5% level (two-sided).

10.Hypothesis tests & confidence intervals

Answer: 1 − β = 56.91 %

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 = 16/√64 = 2
x̄c = 80 + 2.326·2 = 84.65

z under the true mean

z = (84.65 − 85)/2 = −0.174

Power

power = P(Z > −0.174) = 56.91%
1 − β = 56.91%

Answer

1 − β = 56.91%
11.Linear regression & correlation

Answer: r = 0.9459

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 = 789 − 20·190/5 = 29
Sxx = 90 − 20²/5 = 10
Syy = 7314 − 190²/5 = 94

Correlation coefficient

r = 29/√(10·94) = 0.9459

Answer

r = 0.9459
12.Linear regression & correlation

Answer: e = 4.329

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 = 129/6 = 21.5

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −33.5

Regression line

b = −33.5/17.5 = −1.914
a = 21.5 − (−1.914)·3.5 = 28.2
ŷ = 28.2 − 1.914·x

Predicted value

ŷ = 28.2 + (−1.914)·5.5 = 17.67

Residual

e = 22 − 17.67 = 4.329

Answer

e = 4.329

The point lies above the line.

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