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
1520253035404550

Find the sample standard deviation s of the data: 31, 46, 13, 16, 52, 35.

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

A quantity is normally distributed with mean 60 and standard deviation 10. What proportion P of values is below 64?

μ = 60σ = 10x = 64
P =
3.ProbabilityMedium
red: 5blue: 9

A bag holds 5 red and 9 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?

5 / 9
P =
4.ProbabilityHard
xp(x)00.310.320.230.2μ

X has the distribution shown, and Y = −2X − 3. Find the standard deviation σY of Y.

x: 0, 1, 2, 3p(x): 0.3, 0.3, 0.2, 0.2Y = −2X − 3
σY =
5.Normal distributionMedium
490542P = ?

The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. What fraction lies between 490 and 542 h? Give P as a decimal.

μ = 500σ = 40x1 = 490x2 = 542
P =
6.Normal distributionHard
0.02 mm0D − d?

Hole diameters D are normal with mean 25.02 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 = 25.02 mm, σ = 0.015 mmshaft: μ = 25 mm, σ = 0.01 mm
P =
7.Confidence intervalsMedium
n = 9, s = 8.5LUx̄ = 51E = ?90%

A sample of 9 items has mean 51 and standard deviation s = 8.5. The population is roughly normal. Find the margin of error E of a 90% confidence interval for the mean.

n = 9x̄ = 51s = 8.590%t(8) = 1.86
E =
8.Confidence intervalsHard
n = 30σ = ?Us = 2.695%

The diameters of 30 machined shafts have a sample standard deviation of 2.6 µm. Assuming the diameters are normally distributed, find the lower limit σ of the 95% confidence interval for the population standard deviation. With df = 29, 2.5% of the χ² distribution lies below 16.047 and 2.5% lies above 45.722.

n = 30s = 2.6 µmχ² = 16.047, 45.722
σ = µm
9.Hypothesis tests & confidence intervalsMedium
t −2.06402.064

A sample of 25 students has a mean score of 121.7 and a sample standard deviation of 12. Find the test statistic t for H₀: μ = 120.

μ₀ = 120x̄ = 121.7s = 12n = 25
t =
10.Hypothesis tests & confidence intervalsHard
z −1.9601.96

A sample of 49 measurements has mean 118.9. The population standard deviation is 10. Find the upper limit U of the 95% confidence interval for the mean.

x̄ = 118.9σ = 10n = 4995%
U =
11.Linear regression & correlationMedium
xy264136208231029xy510102030

The table gives 5 paired observations of x = load in kN and y = extension in mm. Find the coefficient of determination r² of the least-squares line.

n = 5
r² =
12.Linear regression & correlationHard
xy1373415477549581168xy51040506070ŷ = 45

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. For which value x does the line predict ŷ = 45?

n = 6ŷ = 45
x =

3Solutions

1.Descriptive statistics

Answer: s = 15.64

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 = 193/6 = 32.17

Squared deviations

Σ(x − x̄)² = (−1.167)² + 13.83² + (−19.17)² + (−16.17)² + 19.83² + 2.833² = 1223

Standard deviation

s = √(1223/5) = 15.64

Answer

s = 15.64
2.Descriptive statistics

Answer: P = 0.6554

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 = 0.4

Standard normal table

Φ(0.4) = 0.6554
P = 0.6554

Answer

P = 0.6554
3.Probability

Answer: P = 0.1099

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 = 5/14 · 4/13 = 20/182 = 0.1099

Answer

P = 0.1099
4.Probability

Answer: σY = 2.2

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.

Spread of X

μ = 1.3, Var(X) = 1.21, σX = 1.1

Linear change

σY = |−2|·1.1 = 2.2

Answer

σY = 2.2

The mean does move: E(Y) = −2·1.3 − 3 = −5.6.

5.Normal distribution

Answer: P = 0.4518

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-scores

z₁ = (490 − 500)/40 = −0.25
z₂ = (542 − 500)/40 = 1.05

Area between

P = Φ(1.05) − Φ(−0.25) = 0.8531 − 0.4013 = 0.4518

Answer

P = 0.4518
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: E = 5.27

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 = 1.86·8.5/√9 = 5.27

Answer

E = 5.27
8.Confidence intervals

Answer: σ = 2.071 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 = 29·2.6² = 196

Limits for σ²

196/45.722 = 4.288
196/16.047 = 12.22

Limit for σ

σ = √4.288 = 2.071 µm

Answer

σ = 2.071 µm

95% interval for σ: [2.071, 3.495] µm.

9.Hypothesis tests & confidence intervals

Answer: t = 0.7083

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 = 12/√25 = 12/5 = 2.4

Test statistic

t = (121.7 − 120)/2.4 = 0.7083

Answer

t = 0.7083
t0.025; df = 24 = 2.064

|t| < 2.064, so H₀ is not rejected at the 5% level (two-sided).

10.Hypothesis tests & confidence intervals

Answer: U = 121.7

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 = 1.96·10/√49 = 2.8

Interval limit

U = 118.9 + 2.8 = 121.7

Answer

U = 121.7
11.Linear regression & correlation

Answer: r² = 0.9837

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 = 91/5 = 18.2

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 112
Syy = Σ(y − ȳ)² = 318.8

Coefficient of determination

r² = 112²/(40·318.8) = 0.9837

Answer

r² = 0.9837

About 98.37% of the variation in y is explained by the line.

12.Linear regression & correlation

Answer: x = 4.083

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 = 36/6 = 6
ȳ = 305/6 = 50.83

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 213

Regression line

b = 213/70 = 3.043
a = 50.83 − 3.043·6 = 32.58
ŷ = 32.58 + 3.043·x

Solve for x

x = (45 − 32.58)/3.043 = 4.083

Answer

x = 4.083

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