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
2025303540455055

Find the sample standard deviation s of the data: 50, 39, 18, 42, 24, 58, 37.

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

Exam scores are normally distributed with mean 500 and standard deviation 50. Find the z-score z of a score of 575.

μ = 500σ = 50x = 575
z =
3.ProbabilityMedium
xp(x)10.320.0530.2540.0550.35μ = 3.1

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

x: 1, 2, 3, 4, 5p(x): 0.3, 0.05, 0.25, 0.05, 0.35
Var(X) =
4.ProbabilityHard
n = 10, p = 0.25: P(X ≤ 1) = ?k012345678910

10 components are tested, and each one fails with probability 0.25, independently. What is the probability P that at most one fails?

n = 10p = 0.25
P =
5.Normal distributionMedium
160 g158.8 gT = X₁ + … + X₄?

The masses of bolts are independent and normal with mean 40 g and standard deviation 1.2 g. Find the probability P that the total mass of 4 of them exceeds 158.8 g. Round z to two decimals.

μ = 40 gσ = 1.2 gn = 4> 158.8 g
P =
6.Normal distributionHard
4 s0X − Y?

Runner A's lap time X and runner B's lap time Y are independent and normal: X has mean 52 s and standard deviation 4 s, Y has mean 48 s and standard deviation 3 s. Find the probability P that X is greater than Y. Round z to two decimals.

μX = 52 s, σX = 4 sμY = 48 s, σY = 3 s
P =
7.Confidence intervalsMedium
n = 10, s = 7L = ?Ux̄ = 54E99%

A sample of 10 items has mean 54 and standard deviation s = 7. The population is roughly normal. Find the lower limit of the 99% confidence interval for the mean.

n = 10x̄ = 54s = 799%t(9) = 3.25
lower limit =
8.Confidence intervalsHard
n = 20Lσ = ?s = 10.995%

The diameters of 20 machined shafts have a sample standard deviation of 10.9 µm. Assuming the diameters are normally distributed, find the upper limit σ of the 95% confidence interval for the population standard deviation. With df = 19, 2.5% of the χ² distribution lies below 8.907 and 2.5% lies above 32.852.

n = 20s = 10.9 µmχ² = 8.907, 32.852
σ = µm
9.Hypothesis tests & confidence intervalsMedium
t −1.99801.998

A sample of 64 students has a mean score of 198.4 and a sample standard deviation of 5. Find the test statistic t for H₀: μ = 200.

μ₀ = 200x̄ = 198.4s = 5n = 64
t =
10.Hypothesis tests & confidence intervalsHard
−1.961.96z = ?

In a trial, 60 of 200 patients on treatment A recover, and 41 of 100 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.

x₁ = 60n₁ = 200x₂ = 41n₂ = 100
z =
11.Linear regression & correlationMedium
xy36257879091071111513132xy510100

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 = 584Σx² = 454Σy² = 60126Σxy = 5150
r =
12.Linear regression & correlationHard
xy15481855216924772790xy1520256080ŷ = 62

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 = 62?

n = 5ŷ = 62
x =

3Solutions

1.Descriptive statistics

Answer: s = 13.89

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 = 268/7 = 38.29

Squared deviations

Σ(x − x̄)² = 11.71² + 0.7143² + (−20.29)² + 3.714² + (−14.29)² + 19.71² + (−1.286)² = 1157

Standard deviation

s = √(1157/6) = 13.89

Answer

s = 13.89
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 = (575 − 500)/50 = 1.5

Answer

z = 1.5
3.Probability

Answer: Var(X) = 2.69

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 = 1·0.3 + 2·0.05 + 3·0.25 + 4·0.05 + 5·0.35 = 3.1

Mean of the squares

E(X²) = 1·0.3 + 4·0.05 + 9·0.25 + 16·0.05 + 25·0.35 = 12.3

Variance

Var(X) = 12.3 − 3.1² = 2.69

Answer

Var(X) = 2.69
4.Probability

Answer: P = 0.244

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.

Zero and one failure

P(X = 0) = 0.7510 = 0.05631
P(X = 1) = 10·0.25·0.759 = 0.1877

Add them

P = 0.05631 + 0.1877 = 0.244

Answer

P = 0.244
5.Normal distribution

Answer: P = 0.6915

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 = 160 g, σT = 1.2·√4 = 2.4 g

Standardise

z = (158.8 − 160)/2.4 = −0.5

Probability

P = 1 − Φ(−0.5) = 0.6915

Answer

P = 0.6915
6.Normal distribution

Answer: P = 0.7881

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 = 52 − 48 = 4 s, σ = 5 s

Standardise

z = (0 − 4)/5 = −0.8

Probability

P = 1 − Φ(−0.8) = 0.7881

Answer

P = 0.7881
7.Confidence intervals

Answer: lower limit = 46.81

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 = 3.25

Margin of error

E = 3.25·7/√10 = 7.194

Interval

54 ± 7.194 → [46.81, 61.19]

Answer

lower limit = 46.81
8.Confidence intervals

Answer: σ = 15.92 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 = 19·10.9² = 2257

Limits for σ²

2257/32.852 = 68.71
2257/8.907 = 253.4

Limit for σ

σ = √253.4 = 15.92 µm

Answer

σ = 15.92 µm

95% interval for σ: [8.289, 15.92] µm.

9.Hypothesis tests & confidence intervals

Answer: t = −2.56

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 = 5/√64 = 5/8 = 0.625

Test statistic

t = (198.4 − 200)/0.625 = −2.56

Answer

t = −2.56
t0.025; df = 63 = 1.998

|t| > 1.998, so H₀ is rejected at the 5% level (two-sided).

10.Hypothesis tests & confidence intervals

Answer: z = −1.901

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 = 60/200 = 0.3
p̂₂ = 41/100 = 0.41
p̂ = 101/300 = 0.3367

Standard error under H₀

SE = √(0.3367·0.6633·(1/200 + 1/100)) = 0.05788

Test statistic

z = (0.3 − 0.41)/0.05788 = −1.901

Answer

z = −1.901

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

11.Linear regression & correlation

Answer: r = 0.9971

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 = 5150 − 48·584/6 = 478
Sxx = 454 − 48²/6 = 70
Syy = 60126 − 584²/6 = 3283

Correlation coefficient

r = 478/√(70·3283) = 0.9971

Answer

r = 0.9971
12.Linear regression & correlation

Answer: x = 19.36

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 = 105/5 = 21
ȳ = 339/5 = 67.8

Sums of squares and products

Sxx = Σ(x − x̄)² = 90
Sxy = Σ(x − x̄)(y − ȳ) = 318

Regression line

b = 318/90 = 3.533
a = 67.8 − 3.533·21 = −6.4
ŷ = −6.4 + 3.533·x

Solve for x

x = (62 − (−6.4))/3.533 = 19.36

Answer

x = 19.36

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