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
2530354045

Find the sample standard deviation s of the data: 45, 38, 29, 22, 48.

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

Exam scores are normally distributed with mean 100 and standard deviation 10. Find the z-score z of a score of 106.

μ = 100σ = 10x = 106
z =
3.ProbabilityMedium
n = 15k = 4N = ?

In how many ways N can a committee of 4 people be chosen from a group of 15? The order does not matter.

n = 15k = 4
N =
4.ProbabilityHard
n = 9, p = 0.4: P(X ≤ 1) = ?k0123456789

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

n = 9p = 0.4
P =
5.Normal distributionMedium
425 kg398.2 kgT = X₁ + … + X₅?

The masses of passengers with luggage are independent and normal with mean 85 kg and standard deviation 12 kg. Find the probability P that the total mass of 5 of them is below 398.2 kg. Round z to two decimals.

μ = 85 kgσ = 12 kgn = 5< 398.2 kg
P =
6.Normal distributionHard
0.04 mm0D − d?

Hole diameters D are normal with mean 25.04 mm and standard deviation 0.015 mm. Shaft diameters d are independent and normal with mean 25 mm and standard deviation 0.01 mm. A randomly chosen shaft fits when D − d > 0. Find the probability P that it fits. Round z to two decimals.

hole: μ = 25.04 mm, σ = 0.015 mmshaft: μ = 25 mm, σ = 0.01 mm
P =
7.Confidence intervalsMedium
n = 9, s = 6L = ?Ux̄ = 106E90%

A sample of 9 items has mean 106 and standard deviation s = 6. The population is roughly normal. Find the lower limit of the 90% confidence interval for the mean.

n = 9x̄ = 106s = 690%t(8) = 1.86
lower limit =
8.Confidence intervalsHard
n = 10Lσ = ?s = 10.695%

The diameters of 10 machined shafts have a sample standard deviation of 10.6 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 = 10.6 µmχ² = 2.7, 19.023
σ = µm
9.Hypothesis tests & confidence intervalsMedium
t −2.06402.064

A sample of 25 students has a mean score of 57.2 and a sample standard deviation of 15. Find the test statistic t for H₀: μ = 50.

μ₀ = 50x̄ = 57.2s = 15n = 25
t =
10.Hypothesis tests & confidence intervalsHard
μ₀x̄c

A one-sided z test of H₀: μ = 100 against H₁: μ < 100 is done at α = 0.10 with a sample of 100; the standard deviation is known to be 15. In reality the mean is 97.6. Find the probability of a Type II error β in percent.

μ₀ = 100μ = 97.6σ = 15n = 100α = 0.10
β = %
11.Linear regression & correlationMedium
xy340437541649748xy46404550

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 = 25Σy = 215Σx² = 135Σy² = 9355Σxy = 1103
r =
12.Linear regression & correlationHard
xy12121831841351069xy246101520x = 1.5

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 (1.5, 23) is made. Find its residual e = y − ŷ.

n = 6(1.5, 23)
e =

3Solutions

1.Descriptive statistics

Answer: s = 10.88

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 = 182/5 = 36.4

Squared deviations

Σ(x − x̄)² = 8.6² + 1.6² + (−7.4)² + (−14.4)² + 11.6² = 473.2

Standard deviation

s = √(473.2/4) = 10.88

Answer

s = 10.88
2.Descriptive statistics

Answer: z = 0.6

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 = (106 − 100)/10 = 0.6

Answer

z = 0.6
3.Probability

Answer: N = 1365

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.

Combinations

C(15, 4) = (15·14·13·12)/(4·3·2·1) = 1365

Answer

N = 1365
4.Probability

Answer: P = 0.07054

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) = 0.69 = 0.01008
P(X = 1) = 9·0.4·0.68 = 0.06047

Add them

P = 0.01008 + 0.06047 = 0.07054

Answer

P = 0.07054
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 = 425 kg, σT = 12·√5 = 26.83 kg

Standardise

z = (398.2 − 425)/26.83 = −1

Probability

P = Φ(−1) = 0.1587

Answer

P = 0.1587
6.Normal distribution

Answer: P = 0.9868

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 = 25.04 − 25 = 0.04 mm, σ = 0.01803 mm

Standardise

z = (0 − 0.04)/0.01803 = −2.22

Probability

P = 1 − Φ(−2.22) = 0.9868

Answer

P = 0.9868
7.Confidence intervals

Answer: lower limit = 102.3

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 = 8, t = 1.86

Margin of error

E = 1.86·6/√9 = 3.72

Interval

106 ± 3.72 → [102.3, 109.7]

Answer

lower limit = 102.3
8.Confidence intervals

Answer: σ = 19.35 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 = 9·10.6² = 1011

Limits for σ²

1011/19.023 = 53.16
1011/2.7 = 374.5

Limit for σ

σ = √374.5 = 19.35 µm

Answer

σ = 19.35 µm

95% interval for σ: [7.291, 19.35] µm.

9.Hypothesis tests & confidence intervals

Answer: t = 2.4

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/√25 = 15/5 = 3

Test statistic

t = (57.2 − 50)/3 = 2.4

Answer

t = 2.4
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: β = 37.52 %

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/√100 = 1.5
x̄c = 100 − 1.282·1.5 = 98.08

z under the true mean

z = (98.08 − 97.6)/1.5 = 0.318

Type II error

power = P(Z < 0.318) = 62.48%
β = 100% − 62.48% = 37.52%

Answer

β = 37.52%
11.Linear regression & correlation

Answer: r = 0.8442

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 = 1103 − 25·215/5 = 28
Sxx = 135 − 25²/5 = 10
Syy = 9355 − 215²/5 = 110

Correlation coefficient

r = 28/√(10·110) = 0.8442

Answer

r = 0.8442
12.Linear regression & correlation

Answer: e = 3.081

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 = 89/6 = 14.83

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −44.5

Regression line

b = −44.5/17.5 = −2.543
a = 14.83 − (−2.543)·3.5 = 23.73
ŷ = 23.73 − 2.543·x

Predicted value

ŷ = 23.73 + (−2.543)·1.5 = 19.92

Residual

e = 23 − 19.92 = 3.081

Answer

e = 3.081

The point lies above the line.

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