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
15202530354045505560

Find the sample standard deviation s of the data: 45, 53, 36, 53, 13, 32, 60.

n = 7
s =
2.Descriptive statisticsHard
z −2−1012

A quantity is normally distributed with mean 500 and standard deviation 50. What proportion P of values is above 562.5?

μ = 500σ = 50x = 562.5
P =
3.ProbabilityMedium
xp(x)00.2510.1520.330.3μ = 1.65

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

x: 0, 1, 2, 3p(x): 0.25, 0.15, 0.3, 0.3
Var(X) =
4.ProbabilityHard
xp(x)00.2510.1520.130.2540.25μ

X has the distribution shown, and Y = 5X + 1. Find the standard deviation σY of Y.

x: 0, 1, 2, 3, 4p(x): 0.25, 0.15, 0.1, 0.25, 0.25Y = 5X + 1
σY =
5.Normal distributionMedium
200 mm199.8 mmX + Y?

X is normal with mean 120 mm and standard deviation 0.3 mm; Y is independent and normal with mean 80 mm and standard deviation 0.4 mm. Find the probability P that X + Y is more than 199.8 mm. Round z to two decimals.

μX = 120, σX = 0.3μY = 80, σY = 0.4X + Y > 199.8 mm
P =
6.Normal distributionHard
μ = 178x = ?20%

The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. Find the value x below which 20% of them lie (the 20th percentile).

μ = 178σ = 7P = 20%
x = cm
7.Confidence intervalsMedium
n = 10, s = 20L = ?Ux̄ = 52.5E95%

A sample of 10 items has mean 52.5 and standard deviation s = 20. The population is roughly normal. Find the lower limit of the 95% confidence interval for the mean.

n = 10x̄ = 52.5s = 2095%t(9) = 2.262
lower limit =
8.Confidence intervalsHard
σ = 15, n = ?x̄ − Ex̄ + Ex̄E = 2.590%

The population standard deviation is 15. 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 2.5?

σ = 15E = 2.590%
n =
9.Hypothesis tests & confidence intervalsMedium
−1.661.66p = ?

The mean of a population is claimed to be 50; the population standard deviation is known to be 16. A random sample of 64 observations has a mean of 53.32. Test H₀: μ = 50 against H₁: μ ≠ 50. Find the p-value p.

μ₀ = 50σ = 16n = 64x̄ = 53.32H₁: μ ≠ μ₀
p =
10.Hypothesis tests & confidence intervalsHard
H₀H₁μ₀x̄cμ₀ + 4α = 0.01βσ = 19, power 80%, n = ?

A one-sided z test at α = 0.01 must detect a shift of the mean of 4 units with a power of 80%. The standard deviation is 19. Find the minimum sample size n.

α = 0.01power = 80%δ = 4σ = 19zα = 2.326zβ = 0.842
n =
11.Linear regression & correlationMedium
xy158356568767980xy5607080

The table gives 5 paired observations of x = hours studied and y = exam score. Find Pearson's correlation coefficient r.

n = 5
r =
12.Linear regression & correlationHard
xy125221319416515612xy246152025ŷ = 16

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. For which value x does the line predict ŷ = 16?

n = 6ŷ = 16
x =

3Solutions

1.Descriptive statistics

Answer: s = 16.08

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 = 292/7 = 41.71

Squared deviations

Σ(x − x̄)² = 3.286² + 11.29² + (−5.714)² + 11.29² + (−28.71)² + (−9.714)² + 18.29² = 1551

Standard deviation

s = √(1551/6) = 16.08

Answer

s = 16.08
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: Var(X) = 1.328

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 = 0·0.25 + 1·0.15 + 2·0.3 + 3·0.3 = 1.65

Mean of the squares

E(X²) = 0·0.25 + 1·0.15 + 4·0.3 + 9·0.3 = 4.05

Variance

Var(X) = 4.05 − 1.65² = 1.328

Answer

Var(X) = 1.328
4.Probability

Answer: σY = 7.73

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 = 2.1, Var(X) = 2.39, σX = 1.546

Linear change

σY = |5|·1.546 = 7.73

Answer

σY = 7.73

The mean does move: E(Y) = 5·2.1 + 1 = 11.5.

5.Normal distribution

Answer: P = 0.6554

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 sum

μ = 200 mm, σ = √(0.3² + 0.4²) = 0.5 mm

Standardise

z = (199.8 − 200)/0.5 = −0.4

Upper tail

P = 1 − Φ(−0.4) = 1 − 0.3446 = 0.6554

Answer

P = 0.6554
6.Normal distribution

Answer: x = 172.1 cm

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 for the percentile

Φ(z) = 0.2 → z = −0.8416

Back to x

x = 178 + −0.8416·7 = 172.11 cm

Answer

x = 172.11 cm
7.Confidence intervals

Answer: lower limit = 38.19

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 = 9, t = 2.262

Margin of error

E = 2.262·20/√10 = 14.31

Interval

52.5 ± 14.31 → [38.19, 66.81]

Answer

lower limit = 38.19
8.Confidence intervals

Answer: n = 98

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 = (1.645·15/2.5)² = 97.42

Round up

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

n = 98

Answer

n = 98
9.Hypothesis tests & confidence intervals

Answer: p = 0.09691

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.

Test statistic

SE = σ/√n = 16/√64 = 2
z = (53.32 − 50)/2 = 1.66

Tail area

Φ(1.66) = 0.9515
1 − Φ(1.66) = 0.0485

p-value

p = 2·0.0485 = 0.09691

Answer

p = 0.09691

p > 0.05, so H₀ is not rejected at the 5% level.

10.Hypothesis tests & confidence intervals

Answer: n = 227

Sample size for a test: to detect a true shift δ with power 1 − β in a one-sided z test at level α, the standard error must satisfy δ = (z_α + z_β)·σ/√n, so n = ((z_α + z_β)·σ/δ)², rounded up.

Solve for n

n = ((2.326 + 0.842)·19/4)² = 226.4

Round up

n = 227

Answer

n = 227
11.Linear regression & correlation

Answer: r = 0.9106

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 = 329/5 = 65.8

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 110
Syy = Σ(y − ȳ)² = 364.8

Correlation coefficient

r = 110/√(40·364.8) = 0.9106

Answer

r = 0.9106
12.Linear regression & correlation

Answer: x = 4.314

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 = 108/6 = 18

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −43

Regression line

b = −43/17.5 = −2.457
a = 18 − (−2.457)·3.5 = 26.6
ŷ = 26.6 − 2.457·x

Solve for x

x = (16 − 26.6)/(−2.457) = 4.314

Answer

x = 4.314

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