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: 19, 47, 31, 41, 43, 25.

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

A quantity is normally distributed with mean 50 and standard deviation 8. What proportion P of values is above 46?

μ = 50σ = 8x = 46
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)00.3520.1540.260.3μ

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

x: 0, 2, 4, 6p(x): 0.35, 0.15, 0.2, 0.3Y = 2X − 3
σY =
5.Normal distributionMedium
−3−2−10123zshaded area between −z and z = 0.9342

Drag z so that the shaded area between −z and z under the standard normal curve is 0.9342. Find z to two decimals.

A = 0.9342
z =
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 0.015 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 = 0.015 mm
P =
7.Confidence intervalsMedium
n = 16, s = 5LUx̄ = 68E = ?95%

A sample of 16 items has mean 68 and standard deviation s = 5. The population is roughly normal. Find the margin of error E of a 95% confidence interval for the mean.

n = 16x̄ = 68s = 595%t(15) = 2.131
E =
8.Confidence intervalsHard
σ = 23, n = ?x̄ − Ex̄ + Ex̄E = 390%

The population standard deviation is 23. How large a sample n do you need so that a 90% confidence interval for the mean has a margin of error of at most 3?

σ = 23E = 390%
n =
9.Hypothesis tests & confidence intervalsMedium
t −2.13102.131

A sample of 16 students has a mean score of 103.4 and a sample standard deviation of 8. Find the test statistic t for H₀: μ = 100.

μ₀ = 100x̄ = 103.4s = 8n = 16
t =
10.Hypothesis tests & confidence intervalsHard
μ₀x̄c

A one-sided z test of H₀: μ = 80 against H₁: μ < 80 is done at α = 0.05 with a sample of 25; the standard deviation is known to be 15. In reality the mean is 76.1. Find the power 1 − β of the test in percent.

μ₀ = 80μ = 76.1σ = 15n = 25α = 0.05
1 − β = %
11.Linear regression & correlationMedium
xy37058278491021110913121xy51080100120

The table gives 6 paired observations of x = hours studied and y = exam score. Their sums are given. Find Pearson's correlation coefficient r from these sums.

n = 6Σx = 48Σy = 568Σx² = 454Σy² = 55606Σxy = 4898
r =
12.Linear regression & correlationHard
xy1403485547649701183xy510406080x = 10

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. A new observation (10, 74) is made. Find its residual e = y − ŷ.

n = 6(10, 74)
e =

3Solutions

1.Descriptive statistics

Answer: s = 11.08

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 = 206/6 = 34.33

Squared deviations

Σ(x − x̄)² = (−15.33)² + 12.67² + (−3.333)² + 6.667² + 8.667² + (−9.333)² = 613.3

Standard deviation

s = √(613.3/5) = 11.08

Answer

s = 11.08
2.Descriptive statistics

Answer: P = 0.6915

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 = −0.5

Standard normal table

Φ(−0.5) = 0.3085
P = 1 − 0.3085 = 0.6915

Answer

P = 0.6915
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 = 4.976

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 = 2.9, Var(X) = 6.19, σX = 2.488

Linear change

σY = |2|·2.488 = 4.976

Answer

σY = 4.976

The mean does move: E(Y) = 2·2.9 − 3 = 2.8.

5.Normal distribution

Answer: z = 1.84

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 + 0.9342)/2 = 0.9671

Read the z-table backwards

Find 0.9671 inside the table and read off z.

z = 1.84

Answer

z = 1.84
6.Normal distribution

Answer: P = 0.8264

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 = 0.02121 mm

Standardise

z = (0 − 0.02)/0.02121 = −0.94

Probability

P = 1 − Φ(−0.94) = 0.8264

Answer

P = 0.8264
7.Confidence intervals

Answer: E = 2.664

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 = 15, t = 2.131

Margin of error

E = 2.131·5/√16 = 2.664

Answer

E = 2.664
8.Confidence intervals

Answer: n = 160

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.

Solve for n

n = (1.645·23/3)² = 159.1

Round up

A sample size is a whole number, and rounding down would make the margin too large.

n = 160

Answer

n = 160
9.Hypothesis tests & confidence intervals

Answer: t = 1.7

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).

Standard error

SE = 8/√16 = 8/4 = 2

Test statistic

t = (103.4 − 100)/2 = 1.7

Answer

t = 1.7
t0.025; df = 15 = 2.131

|t| < 2.131, so H₀ is not rejected at the 5% level (two-sided).

10.Hypothesis tests & confidence intervals

Answer: 1 − β = 36.5 %

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/√25 = 3
x̄c = 80 − 1.645·3 = 75.06

z under the true mean

z = (75.06 − 76.1)/3 = −0.345

Power

power = P(Z < −0.345) = 36.5%
1 − β = 36.5%

Answer

1 − β = 36.5%
11.Linear regression & correlation

Answer: r = 0.9876

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 = 4898 − 48·568/6 = 354
Sxx = 454 − 48²/6 = 70
Syy = 55606 − 568²/6 = 1835

Correlation coefficient

r = 354/√(70·1835) = 0.9876

Answer

r = 0.9876
12.Linear regression & correlation

Answer: e = −2.462

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 = 359/6 = 59.83

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 291

Regression line

b = 291/70 = 4.157
a = 59.83 − 4.157·6 = 34.89
ŷ = 34.89 + 4.157·x

Predicted value

ŷ = 34.89 + 4.157·10 = 76.46

Residual

e = 74 − 76.46 = −2.462

Answer

e = −2.462

The point lies below the line.

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