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
10152025303540455055

Find the sample standard deviation s of the data: 34, 10, 49, 52, 13, 58.

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

Exam scores are normally distributed with mean 50 and standard deviation 8. Find the z-score z of a score of 62.

μ = 50σ = 8x = 62
z =
3.ProbabilityMedium
red: 7blue: 4

A bag holds 7 red and 4 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?

7 / 4
P =
4.ProbabilityHard
xp(x)00.310.3520.35μ

X has the distribution shown, and Y = 0.5X + 1. Find the standard deviation σY of Y.

x: 0, 1, 2p(x): 0.3, 0.35, 0.35Y = 0.5X + 1
σY =
5.Normal distributionMedium
248.75254.25P = ?

The bag weights are normally distributed with mean 250 g and standard deviation 5 g. What fraction lies between 248.75 and 254.25 g? Give P as a decimal.

μ = 250σ = 5x1 = 248.75x2 = 254.25
P =
6.Normal distributionHard
35455565758595pointsbottom 40%

The exam scores are normally distributed with mean 65 points and standard deviation 10 points. Drag the cut-off so the shaded bottom 40% is to the left of it. Find the cut-off x.

μ = 65 pointsσ = 10 points40%
x = points
7.Confidence intervalsMedium
μ₁ − μ₂0LU = ?x̄₁ − x̄₂E99%

The battery life of two phone models is compared. Model 1: 80 phones, mean 73.5 h, known standard deviation 8 h. Model 2: 80 phones, mean 67.5 h, known standard deviation 16 h. Find the upper limit of the 99% confidence interval for μ₁ − μ₂.

n₁ = 80x̄₁ = 73.5 hσ₁ = 8 hn₂ = 80x̄₂ = 67.5 hσ₂ = 16 h99%
upper limit = h
8.Confidence intervalsHard
σ = 23, n = ?x̄ − Ex̄ + Ex̄E = 195%

The population standard deviation is 23. How large a sample n do you need so that a 95% confidence interval for the mean has a margin of error of at most 1?

σ = 23E = 195%
n =
9.Hypothesis tests & confidence intervalsMedium
−2.1792.179t = ?df = 12

13 employees take a typing test before and after a training course. For each person the difference d = after − before (words per minute) is recorded. The mean difference is 4.44 and the standard deviation of the differences is 8. Find the paired test statistic t for H₀: μd = 0.

n = 13d̄ = 4.44sd = 8
t =
10.Hypothesis tests & confidence intervalsHard
−1.961.96z = ?

In a trial, 65 of 100 patients on treatment A recover, and 44 of 80 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.

x₁ = 65n₁ = 100x₂ = 44n₂ = 80
z =
11.Linear regression & correlationMedium
xy3515507579701172xy510506070

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
xy145317672070239026105xy1520256080100ŷ = 83

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. For which value x does the line predict ŷ = 83?

n = 5ŷ = 83
x =

3Solutions

1.Descriptive statistics

Answer: s = 20.58

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 = 216/6 = 36

Squared deviations

Σ(x − x̄)² = (−2)² + (−26)² + 13² + 16² + (−23)² + 22² = 2118

Standard deviation

s = √(2118/5) = 20.58

Answer

s = 20.58
2.Descriptive statistics

Answer: z = 1.5

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 = (62 − 50)/8 = 1.5

Answer

z = 1.5
3.Probability

Answer: P = 0.3818

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 = 7/11 · 6/10 = 42/110 = 0.3818

Answer

P = 0.3818
4.Probability

Answer: σY = 0.4023

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 = 1.05, Var(X) = 0.6475, σX = 0.8047

Linear change

σY = |0.5|·0.8047 = 0.4023

Answer

σY = 0.4023

The mean does move: E(Y) = 0.5·1.05 + 1 = 1.525.

5.Normal distribution

Answer: P = 0.401

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-scores

z₁ = (248.75 − 250)/5 = −0.25
z₂ = (254.25 − 250)/5 = 0.85

Area between

P = Φ(0.85) − Φ(−0.25) = 0.8023 − 0.4013 = 0.401

Answer

P = 0.401
6.Normal distribution

Answer: x = 62.5 points

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

Φ(z) = 0.4

z from the table

z ≈ −0.25

Back to the original units

x = 65 − 0.25·10 = 62.5 points

Answer

x = 62.5 points
7.Confidence intervals

Answer: upper limit = 11.15 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 = 73.5 − 67.5 = 6

Standard error

SE = √(8²/80 + 16²/80) = 2

Margin of error

E = 2.576·2 = 5.152

Interval

6 ± 5.152 → [0.848, 11.15]

Answer

upper limit = 11.15 h

The interval does not contain 0, so the means differ significantly.

8.Confidence intervals

Answer: n = 2033

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 = (1.96·23/1)² = 2032

Round up

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

n = 2033

Answer

n = 2033
9.Hypothesis tests & confidence intervals

Answer: t = 2.001

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 of the mean difference

SE = 8/√13 = 2.219

Test statistic

t = 4.44/2.219 = 2.001

Answer

t = 2.001

With df = 12, the two-sided 5% critical value is 2.179; |t| is smaller, so H₀ is not rejected.

10.Hypothesis tests & confidence intervals

Answer: z = 1.364

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 = 65/100 = 0.65
p̂₂ = 44/80 = 0.55
p̂ = 109/180 = 0.6056

Standard error under H₀

SE = √(0.6056·0.3944·(1/100 + 1/80)) = 0.07331

Test statistic

z = (0.65 − 0.55)/0.07331 = 1.364

Answer

z = 1.364

|z| < 1.96, so H₀ is not rejected at the 5% level (two-sided).

11.Linear regression & correlation

Answer: r² = 0.8857

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 = 35/5 = 7
ȳ = 300/5 = 60

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 124
Syy = Σ(y − ȳ)² = 434

Coefficient of determination

r² = 124²/(40·434) = 0.8857

Answer

r² = 0.8857

About 88.57% of the variation in y is explained by the line.

12.Linear regression & correlation

Answer: x = 21.42

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 = 100/5 = 20
ȳ = 385/5 = 77

Sums of squares and products

Sxx = Σ(x − x̄)² = 90
Sxy = Σ(x − x̄)(y − ȳ) = 381

Regression line

b = 381/90 = 4.233
a = 77 − 4.233·20 = −7.667
ŷ = −7.667 + 4.233·x

Solve for x

x = (83 − (−7.667))/4.233 = 21.42

Answer

x = 21.42

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