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
202530354045505560

Find the sample standard deviation s of the data: 39, 42, 39, 60, 16, 29.

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

Exam scores are normally distributed with mean 50 and standard deviation 5. Find the z-score z of a score of 45.

μ = 50σ = 5x = 45
z =
3.ProbabilityMedium
xp(x)00.3510.320.230.15μ = 1.15

For the distribution shown, find the standard deviation σ of X.

x: 0, 1, 2, 3p(x): 0.35, 0.3, 0.2, 0.15
σ =
4.ProbabilityHard
xp(x)00.2510.0520.330.340.1μ

X has the distribution shown, and Y = 10X + 5. Find the standard deviation σY of Y.

x: 0, 1, 2, 3, 4p(x): 0.25, 0.05, 0.3, 0.3, 0.1Y = 10X + 5
σY =
5.Normal distributionMedium
48 kg45.6 kgT = X₁ + … + X₄?

The masses of boxes are independent and normal with mean 12 kg and standard deviation 0.8 kg. Find the probability P that the total mass of 4 of them is below 45.6 kg. Round z to two decimals.

μ = 12 kgσ = 0.8 kgn = 4< 45.6 kg
P =
6.Normal distributionHard
μ = 250x = ?99%

The bag weights are normally distributed with mean 250 g and standard deviation 5 g. Find the value x below which 99% of them lie (the 99th percentile).

μ = 250σ = 5P = 99%
x = g
7.Confidence intervalsMedium
μ₁ − μ₂0L = ?Ux̄₁ − x̄₂E95%

The battery life of two phone models is compared. Model 1: 80 phones, mean 144.5 h, known standard deviation 14 h. Model 2: 40 phones, mean 144 h, known standard deviation 15 h. Find the lower limit of the 95% confidence interval for μ₁ − μ₂.

n₁ = 80x̄₁ = 144.5 hσ₁ = 14 hn₂ = 40x̄₂ = 144 hσ₂ = 15 h95%
lower limit = h
8.Confidence intervalsHard
σ = 10, n = ?x̄ − Ex̄ + Ex̄E = 399%

The population standard deviation is 10. How large a sample n do you need so that a 99% confidence interval for the mean has a margin of error of at most 3?

σ = 10E = 399%
n =
9.Hypothesis tests & confidence intervalsMedium
t −2.06402.064

A sample of 25 students has a mean score of 501.4 and a sample standard deviation of 10. Find the test statistic t for H₀: μ = 500.

μ₀ = 500x̄ = 501.4s = 10n = 25
t =
10.Hypothesis tests & confidence intervalsHard
H₀H₁μ₀x̄cμ₀ + 2α = 0.05βσ = 17, power 95%, n = ?

A one-sided z test at α = 0.05 must detect a shift of the mean of 2 units with a power of 95%. The standard deviation is 17. Find the minimum sample size n.

α = 0.05power = 95%δ = 2σ = 17zα = 1.645zβ = 1.645
n =
11.Linear regression & correlationMedium
xy18722083221022410726120xy202580100120

The table gives 5 paired observations of x = temperature in °C and y = ice creams sold. Their sums are given. Find Pearson's correlation coefficient r from these sums.

n = 5Σx = 110Σy = 484Σx² = 2460Σy² = 48326Σxy = 10888
r =
12.Linear regression & correlationHard
xy128222324419522617xy2462025x = 4.5

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. A new observation (4.5, 18) is made. Find its residual e = y − ŷ.

n = 6(4.5, 18)
e =

3Solutions

1.Descriptive statistics

Answer: s = 14.6

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 = 225/6 = 37.5

Squared deviations

Σ(x − x̄)² = 1.5² + 4.5² + 1.5² + 22.5² + (−21.5)² + (−8.5)² = 1066

Standard deviation

s = √(1066/5) = 14.6

Answer

s = 14.6
2.Descriptive statistics

Answer: z = −1

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 = (45 − 50)/5 = −1

Answer

z = −1
3.Probability

Answer: σ = 1.062

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 = 0·0.35 + 1·0.3 + 2·0.2 + 3·0.15 = 1.15

Mean of the squares

E(X²) = 0·0.35 + 1·0.3 + 4·0.2 + 9·0.15 = 2.45

Variance

Var(X) = 2.45 − 1.15² = 1.128
σ = √1.128 = 1.062

Answer

σ = 1.062
4.Probability

Answer: σY = 13.22

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 = 1.95, Var(X) = 1.748, σX = 1.322

Linear change

σY = |10|·1.322 = 13.22

Answer

σY = 13.22

The mean does move: E(Y) = 10·1.95 + 5 = 24.5.

5.Normal distribution

Answer: P = 0.06681

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 total

μT = 48 kg, σT = 0.8·√4 = 1.6 kg

Standardise

z = (45.6 − 48)/1.6 = −1.5

Probability

P = Φ(−1.5) = 0.06681

Answer

P = 0.06681
6.Normal distribution

Answer: x = 261.6 g

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) = 0.99 → z = 2.326

Back to x

x = 250 + 2.326·5 = 261.63 g

Answer

x = 261.63 g
7.Confidence intervals

Answer: lower limit = −5.07 h

Interval for a difference: estimate ± critical value × standard error. For two independent means with known σ: SE = √(σ₁²/n₁ + σ₂²/n₂) and z. With equal unknown variances: pooled s_p and t with n₁ + n₂ − 2 degrees of freedom. For paired data, work with the differences: d̄ ± t·s_d/√n.

Point estimate

x̄₁ − x̄₂ = 144.5 − 144 = 0.5

Standard error

SE = √(14²/80 + 15²/40) = 2.842

Margin of error

E = 1.96·2.842 = 5.57

Interval

0.5 ± 5.57 → [−5.07, 6.07]

Answer

lower limit = −5.07 h

The interval contains 0, so the data do not show a significant difference.

8.Confidence intervals

Answer: n = 74

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 = (2.576·10/3)² = 73.73

Round up

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

n = 74

Answer

n = 74
9.Hypothesis tests & confidence intervals

Answer: t = 0.7

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 = 10/√25 = 10/5 = 2

Test statistic

t = (501.4 − 500)/2 = 0.7

Answer

t = 0.7
t0.025; df = 24 = 2.064

|t| < 2.064, so H₀ is not rejected at the 5% level (two-sided).

10.Hypothesis tests & confidence intervals

Answer: n = 783

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 = ((1.645 + 1.645)·17/2)² = 782

Round up

n = 783

Answer

n = 783
11.Linear regression & correlation

Answer: r = 0.9881

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 = 10888 − 110·484/5 = 240
Sxx = 2460 − 110²/5 = 40
Syy = 48326 − 484²/5 = 1475

Correlation coefficient

r = 240/√(40·1475) = 0.9881

Answer

r = 0.9881
12.Linear regression & correlation

Answer: e = −2.286

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 = 132/6 = 22

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −30

Regression line

b = −30/17.5 = −1.714
a = 22 − (−1.714)·3.5 = 28
ŷ = 28 − 1.714·x

Predicted value

ŷ = 28 + (−1.714)·4.5 = 20.29

Residual

e = 18 − 20.29 = −2.286

Answer

e = −2.286

The point lies below the line.

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