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
202530354045

Find the sample standard deviation s of the data: 30, 16, 44, 47, 31.

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

A quantity is normally distributed with mean 70 and standard deviation 15. What proportion P of values is below 55?

μ = 70σ = 15x = 55
P =
3.ProbabilityMedium
xp(x)10.3520.1530.1540.35μ = 2.5

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

x: 1, 2, 3, 4p(x): 0.35, 0.15, 0.15, 0.35
Var(X) =
4.ProbabilityHard
xp(x)00.3510.3520.0530.0540.2μ

X has the distribution shown, and Y = −2X + 1. Find the standard deviation σY of Y.

x: 0, 1, 2, 3, 4p(x): 0.35, 0.35, 0.05, 0.05, 0.2Y = −2X + 1
σY =
5.Normal distributionMedium
76.75130P = ?

The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. What fraction lies between 76.75 and 130 points? Give P as a decimal.

μ = 100σ = 15x1 = 76.75x2 = 130
P =
6.Normal distributionHard
0.05 mm0D − d?

Hole diameters D are normal with mean 25.05 mm and standard deviation 0.02 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 = 25.05 mm, σ = 0.02 mmshaft: μ = 25 mm, σ = 0.02 mm
P =
7.Confidence intervalsMedium
x̄E₁ = 4.899%, n = 36x̄E₂ = ?90%, n = 144

A 99% confidence interval for a mean, based on 36 observations and a known σ, has a margin of error of 4.8. What is the margin of error E₂ of a 90% interval based on 144 observations?

99%n₁ = 36E₁ = 4.890%n₂ = 144
E₂ =
8.Confidence intervalsHard
n = 20Lσ = ?s = 1.395%

The diameters of 20 machined shafts have a sample standard deviation of 1.3 µm. Assuming the diameters are normally distributed, find the upper limit σ of the 95% confidence interval for the population standard deviation. With df = 19, 2.5% of the χ² distribution lies below 8.907 and 2.5% lies above 32.852.

n = 20s = 1.3 µmχ² = 8.907, 32.852
σ = µm
9.Hypothesis tests & confidence intervalsMedium
t −1.9901.99

A sample of 81 students has a mean score of 101.5 and a sample standard deviation of 8. Find the test statistic t for H₀: μ = 100.

μ₀ = 100x̄ = 101.5s = 8n = 81
t =
10.Hypothesis tests & confidence intervalsHard
z −1.64501.645

A sample of 16 measurements has mean 44. The population standard deviation is 12. Find the lower limit L of the 90% confidence interval for the mean.

x̄ = 44σ = 12n = 1690%
L =
11.Linear regression & correlationMedium
xy260478689810110114xy5106080100120

The table gives 5 paired observations of x = hours studied and y = exam score. Their sums are given. Find Pearson's correlation coefficient r from these sums.

n = 5Σx = 30Σy = 442Σx² = 220Σy² = 40802Σxy = 2914
r =
12.Linear regression & correlationHard
xy36958078799411113xy51080100x = 4

The table gives 5 paired observations of x = hours studied and y = exam score. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 4.

n = 5x = 4
ŷ =

3Solutions

1.Descriptive statistics

Answer: s = 12.42

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 = 168/5 = 33.6

Squared deviations

Σ(x − x̄)² = (−3.6)² + (−17.6)² + 10.4² + 13.4² + (−2.6)² = 617.2

Standard deviation

s = √(617.2/4) = 12.42

Answer

s = 12.42
2.Descriptive statistics

Answer: P = 0.1587

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

Standard normal table

Φ(−1) = 0.1587
P = 0.1587

Answer

P = 0.1587
3.Probability

Answer: Var(X) = 1.65

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 = 1·0.35 + 2·0.15 + 3·0.15 + 4·0.35 = 2.5

Mean of the squares

E(X²) = 1·0.35 + 4·0.15 + 9·0.15 + 16·0.35 = 7.9

Variance

Var(X) = 7.9 − 2.5² = 1.65

Answer

Var(X) = 1.65
4.Probability

Answer: σY = 2.993

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 = 1.4, Var(X) = 2.24, σX = 1.497

Linear change

σY = |−2|·1.497 = 2.993

Answer

σY = 2.993

The mean does move: E(Y) = −2·1.4 + 1 = −1.8.

5.Normal distribution

Answer: P = 0.9167

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 = (76.75 − 100)/15 = −1.55
z₂ = (130 − 100)/15 = 2

Area between

P = Φ(2) − Φ(−1.55) = 0.9772 − 0.06057 = 0.9167

Answer

P = 0.9167
6.Normal distribution

Answer: P = 0.9616

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 = 25.05 − 25 = 0.05 mm, σ = 0.02828 mm

Standardise

z = (0 − 0.05)/0.02828 = −1.77

Probability

P = 1 − Φ(−1.77) = 0.9616

Answer

P = 0.9616
7.Confidence intervals

Answer: E₂ = 1.533

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.

Ratio of margins

E₂/E₁ = (1.645/2.576)·√(36/144) = 0.3193

New margin

E₂ = 4.8·0.3193 = 1.533

Answer

E₂ = 1.533
8.Confidence intervals

Answer: σ = 1.899 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 = 19·1.3² = 32.11

Limits for σ²

32.11/32.852 = 0.9774
32.11/8.907 = 3.605

Limit for σ

σ = √3.605 = 1.899 µm

Answer

σ = 1.899 µm

95% interval for σ: [0.9886, 1.899] µm.

9.Hypothesis tests & confidence intervals

Answer: t = 1.688

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 = 8/√81 = 8/9 = 0.8889

Test statistic

t = (101.5 − 100)/0.8889 = 1.688

Answer

t = 1.688
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: L = 39.06

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

L = 44 − 4.935 = 39.06

Answer

L = 39.06
11.Linear regression & correlation

Answer: r = 0.9962

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 = 2914 − 30·442/5 = 262
Sxx = 220 − 30²/5 = 40
Syy = 40802 − 442²/5 = 1729

Correlation coefficient

r = 262/√(40·1729) = 0.9962

Answer

r = 0.9962
12.Linear regression & correlation

Answer: ŷ = 73.3

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 = 35/5 = 7
ȳ = 443/5 = 88.6

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 204

Regression line

b = 204/40 = 5.1
a = 88.6 − 5.1·7 = 52.9
ŷ = 52.9 + 5.1·x

Prediction

ŷ = 52.9 + 5.1·4 = 73.3

Answer

ŷ = 73.3

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