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, 12, 38, 17, 29, 54, 41.

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

Exam scores are normally distributed with mean 170 and standard deviation 5. Find the z-score z of a score of 180.

μ = 170σ = 5x = 180
z =
3.ProbabilityMedium
red: 5blue: 8

A bag holds 5 red and 8 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?

5 / 8
P =
4.ProbabilityHard
xp(x)100.15200.25300.35400.25μ

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

x: 10, 20, 30, 40p(x): 0.15, 0.25, 0.35, 0.25Y = 2X + 1
σY =
5.Normal distributionMedium
105 s111.4 sX + Y?

X is normal with mean 60 s and standard deviation 5 s; Y is independent and normal with mean 45 s and standard deviation 4 s. Find the probability P that X + Y is more than 111.4 s. Round z to two decimals.

μX = 60, σX = 5μY = 45, σY = 4X + Y > 111.4 s
P =
6.Normal distributionHard
10 cm0X − Y?

The heights X of men and Y of women in a group are independent and normal: X has mean 180 cm and standard deviation 10 cm, Y has mean 170 cm and standard deviation 12 cm. Find the probability P that X is greater than Y. Round z to two decimals.

μX = 180 cm, σX = 10 cmμY = 170 cm, σY = 12 cm
P =
7.Confidence intervalsMedium
n = 13, s = 6L = ?Ux̄ = 11E95%

A sample of 13 items has mean 11 and standard deviation s = 6. The population is roughly normal. Find the lower limit of the 95% confidence interval for the mean.

n = 13x̄ = 11s = 695%t(12) = 2.179
lower limit =
8.Confidence intervalsHard
n = 11Lσ = ?s = 695%

The diameters of 11 machined shafts have a sample standard deviation of 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 = 10, 2.5% of the χ² distribution lies below 3.247 and 2.5% lies above 20.483.

n = 11s = 6 µmχ² = 3.247, 20.483
σ = µm
9.Hypothesis tests & confidence intervalsMedium
−2.06p = ?

The mean of a population is claimed to be 100; the population standard deviation is known to be 30. A random sample of 100 observations has a mean of 93.82. Test H₀: μ = 100 against H₁: μ < 100. Find the p-value p.

μ₀ = 100σ = 30n = 100x̄ = 93.82H₁: μ < μ₀
p =
10.Hypothesis tests & confidence intervalsHard
σ = 20, n = ?x̄ − Ex̄ + Ex̄E = 395%

How large must a sample be so that the 95% confidence interval for a mean has a margin of error of at most 3? The standard deviation is 20. Find the smallest sample size n.

σ = 20E = 395%
n =
11.Linear regression & correlationMedium
xy1679198722111251312814331156xy15202530100150

The table gives 6 paired observations of x = temperature in °C and y = ice creams sold. Find Pearson's correlation coefficient r.

n = 6
r =
12.Linear regression & correlationHard
xy149357562771975xy5506070ŷ = 53

The table gives 5 paired observations of x = hours studied and y = exam score. Use the least-squares line ŷ = a + b·x. For which value x does the line predict ŷ = 53?

n = 5ŷ = 53
x =

3Solutions

1.Descriptive statistics

Answer: s = 14.35

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 = 222/7 = 31.71

Squared deviations

Σ(x − x̄)² = (−0.7143)² + (−19.71)² + 6.286² + (−14.71)² + (−2.714)² + 22.29² + 9.286² = 1235

Standard deviation

s = √(1235/6) = 14.35

Answer

s = 14.35
2.Descriptive statistics

Answer: z = 2

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 = (180 − 170)/5 = 2

Answer

z = 2
3.Probability

Answer: P = 0.1282

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 = 5/13 · 4/12 = 20/156 = 0.1282

Answer

P = 0.1282
4.Probability

Answer: σY = 20.1

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 = 27, Var(X) = 101, σX = 10.05

Linear change

σY = |2|·10.05 = 20.1

Answer

σY = 20.1

The mean does move: E(Y) = 2·27 + 1 = 55.

5.Normal distribution

Answer: P = 0.1587

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 = 105 s, σ = √(5² + 4²) = 6.403 s

Standardise

z = (111.4 − 105)/6.403 = 1

Upper tail

P = 1 − Φ(1) = 1 − 0.8413 = 0.1587

Answer

P = 0.1587
6.Normal distribution

Answer: P = 0.7389

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 = 180 − 170 = 10 cm, σ = 15.62 cm

Standardise

z = (0 − 10)/15.62 = −0.64

Probability

P = 1 − Φ(−0.64) = 0.7389

Answer

P = 0.7389
7.Confidence intervals

Answer: lower limit = 7.374

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 = 12, t = 2.179

Margin of error

E = 2.179·6/√13 = 3.626

Interval

11 ± 3.626 → [7.374, 14.63]

Answer

lower limit = 7.374
8.Confidence intervals

Answer: σ = 10.53 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 = 10·6² = 360

Limits for σ²

360/20.483 = 17.58
360/3.247 = 110.9

Limit for σ

σ = √110.9 = 10.53 µm

Answer

σ = 10.53 µm

95% interval for σ: [4.192, 10.53] µm.

9.Hypothesis tests & confidence intervals

Answer: p = 0.0197

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 = 30/√100 = 3
z = (93.82 − 100)/3 = −2.06

Tail area

Φ(2.06) = 0.9803
1 − Φ(2.06) = 0.0197

p-value

p = 0.0197

Answer

p = 0.0197

p ≤ 0.05, so H₀ is rejected at the 5% level.

10.Hypothesis tests & confidence intervals

Answer: n = 171

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).

Solve for n

n = (1.96·20/3)² = 170.7

Round up

n = 171

Answer

n = 171
11.Linear regression & correlation

Answer: r = 0.9917

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 = 141/6 = 23.5
ȳ = 707/6 = 117.8

Sums of squares and products

Sxx = Σ(x − x̄)² = 157.5
Sxy = Σ(x − x̄)(y − ȳ) = 859.5
Syy = Σ(y − ȳ)² = 4769

Correlation coefficient

r = 859.5/√(157.5·4769) = 0.9917

Answer

r = 0.9917
12.Linear regression & correlation

Answer: x = 2.03

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 = 314/5 = 62.8

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 132

Regression line

b = 132/40 = 3.3
a = 62.8 − 3.3·5 = 46.3
ŷ = 46.3 + 3.3·x

Solve for x

x = (53 − 46.3)/3.3 = 2.03

Answer

x = 2.03

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