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
152025303540455055

Find the sample standard deviation s of the data: 15, 11, 50, 59, 19, 19.

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

A quantity is normally distributed with mean 70 and standard deviation 10. What proportion P of values is above 82.5?

μ = 70σ = 10x = 82.5
P =
3.ProbabilityMedium
n = 15k = 3N = ?

In how many ways N can a committee of 3 people be chosen from a group of 15? The order does not matter.

n = 15k = 3
N =
4.ProbabilityHard
xp(x)00.1510.120.330.2540.2μ

X has the distribution shown, and Y = 3X + 20. Find the standard deviation σY of Y.

x: 0, 1, 2, 3, 4p(x): 0.15, 0.1, 0.3, 0.25, 0.2Y = 3X + 20
σY =
5.Normal distributionMedium
−3−2−10123zshaded area between −z and z = 0.8198

Drag z so that the shaded area between −z and z under the standard normal curve is 0.8198. Find z to two decimals.

A = 0.8198
z =
6.Normal distributionHard
0.05 mm0D − d?

Hole diameters D are normal with mean 25.05 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 = 25.05 mm, σ = 0.01 mmshaft: μ = 25 mm, σ = 0.02 mm
P =
7.Confidence intervalsMedium
x̄E₁ = 11.999%, n = 25x̄E₂ = ?95%, n = 25

A 99% confidence interval for a mean, based on 25 observations and a known σ, has a margin of error of 11.9. What is the margin of error E₂ of a 95% interval from the same data?

99%n₁ = 25E₁ = 11.995%
E₂ =
8.Confidence intervalsHard
n = 16Lσ = ?s = 2.695%

The diameters of 16 machined shafts have a sample standard deviation of 2.6 µm. Assuming the diameters are normally distributed, find the upper limit σ of the 95% confidence interval for the population standard deviation. With df = 15, 2.5% of the χ² distribution lies below 6.262 and 2.5% lies above 27.488.

n = 16s = 2.6 µmχ² = 6.262, 27.488
σ = µm
9.Hypothesis tests & confidence intervalsMedium
t −2.06402.064

A sample of 25 students has a mean score of 191.2 and a sample standard deviation of 20. Find the test statistic t for H₀: μ = 200.

μ₀ = 200x̄ = 191.2s = 20n = 25
t =
10.Hypothesis tests & confidence intervalsHard
−1.961.96z = ?

In a trial, 113 of 250 patients on treatment A recover, and 64 of 200 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.

x₁ = 113n₁ = 250x₂ = 64n₂ = 200
z =
11.Linear regression & correlationMedium
xy371476577689794xy46708090

The table gives 5 paired observations of x = hours studied and y = exam score. Find the coefficient of determination r² of the least-squares line.

n = 5
r² =
12.Linear regression & correlationHard
xy17522063236926822978xy20253050607080x = 18.5

The table gives 5 paired observations of x = temperature in °C and y = ice creams sold. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 18.5.

n = 5x = 18.5
ŷ =

3Solutions

1.Descriptive statistics

Answer: s = 20.3

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 = 173/6 = 28.83

Squared deviations

Σ(x − x̄)² = (−13.83)² + (−17.83)² + 21.17² + 30.17² + (−9.833)² + (−9.833)² = 2061

Standard deviation

s = √(2061/5) = 20.3

Answer

s = 20.3
2.Descriptive statistics

Answer: P = 0.1056

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 = 1.25

Standard normal table

Φ(1.25) = 0.8944
P = 1 − 0.8944 = 0.1056

Answer

P = 0.1056
3.Probability

Answer: N = 455

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, 3) = (15·14·13)/(3·2·1) = 455

Answer

N = 455
4.Probability

Answer: σY = 3.897

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 = 2.25, Var(X) = 1.688, σX = 1.299

Linear change

σY = |3|·1.299 = 3.897

Answer

σY = 3.897

The mean does move: E(Y) = 3·2.25 + 20 = 26.75.

5.Normal distribution

Answer: z = 1.34

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 + 0.8198)/2 = 0.9099

Read the z-table backwards

Find 0.9099 inside the table and read off z.

z = 1.34

Answer

z = 1.34
6.Normal distribution

Answer: P = 0.9875

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 = 0.02236 mm

Standardise

z = (0 − 0.05)/0.02236 = −2.24

Probability

P = 1 − Φ(−2.24) = 0.9875

Answer

P = 0.9875
7.Confidence intervals

Answer: E₂ = 9.054

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 = (1.96/2.576)·√(25/25) = 0.7609

New margin

E₂ = 11.9·0.7609 = 9.054

Answer

E₂ = 9.054
8.Confidence intervals

Answer: σ = 4.024 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 = 15·2.6² = 101.4

Limits for σ²

101.4/27.488 = 3.689
101.4/6.262 = 16.19

Limit for σ

σ = √16.19 = 4.024 µm

Answer

σ = 4.024 µm

95% interval for σ: [1.921, 4.024] µm.

9.Hypothesis tests & confidence intervals

Answer: t = −2.2

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 = (191.2 − 200)/4 = −2.2

Answer

t = −2.2
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: z = 2.848

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.

Sample proportions

p̂₁ = 113/250 = 0.452
p̂₂ = 64/200 = 0.32
p̂ = 177/450 = 0.3933

Standard error under H₀

SE = √(0.3933·0.6067·(1/250 + 1/200)) = 0.04634

Test statistic

z = (0.452 − 0.32)/0.04634 = 2.848

Answer

z = 2.848

|z| > 1.96, so H₀ is rejected at the 5% level (two-sided).

11.Linear regression & correlation

Answer: r² = 0.9327

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 = 25/5 = 5
ȳ = 407/5 = 81.4

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = 59
Syy = Σ(y − ȳ)² = 373.2

Coefficient of determination

r² = 59²/(10·373.2) = 0.9327

Answer

r² = 0.9327

About 93.27% of the variation in y is explained by the line.

12.Linear regression & correlation

Answer: ŷ = 58.15

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 = 115/5 = 23
ȳ = 344/5 = 68.8

Sums of squares and products

Sxx = Σ(x − x̄)² = 90
Sxy = Σ(x − x̄)(y − ȳ) = 213

Regression line

b = 213/90 = 2.367
a = 68.8 − 2.367·23 = 14.37
ŷ = 14.37 + 2.367·x

Prediction

ŷ = 14.37 + 2.367·18.5 = 58.15

Answer

ŷ = 58.15

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