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: 36, 31, 25, 27, 20, 35, 59.

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

A quantity is normally distributed with mean 170 and standard deviation 8. What proportion P of values is above 173.2?

μ = 170σ = 8x = 173.2
P =
3.ProbabilityMedium
xp(x)00.210.3520.330.15μ = 1.4

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

x: 0, 1, 2, 3p(x): 0.2, 0.35, 0.3, 0.15
Var(X) =
4.ProbabilityHard
n = 6, p = 0.4: P(X = 1) = ?k0123456

A test has 6 multiple-choice questions. A student who guesses gets each one right with probability 0.4, independently. What is the probability P of exactly 1 correct answers?

n = 6p = 0.4k = 1
P =
5.Normal distributionMedium
200 mm200.3 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 200.3 mm. Round z to two decimals.

μX = 120, σX = 0.3μY = 80, σY = 0.4X + Y > 200.3 mm
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
μ₁ − μ₂0L = ?Ux̄₁ − x̄₂E90%

The battery life of two phone models is compared. Model 1: 30 phones, mean 73.5 h, known standard deviation 16 h. Model 2: 30 phones, mean 71.5 h, known standard deviation 6 h. Find the lower limit of the 90% confidence interval for μ₁ − μ₂.

n₁ = 30x̄₁ = 73.5 hσ₁ = 16 hn₂ = 30x̄₂ = 71.5 hσ₂ = 6 h90%
lower limit = h
8.Confidence intervalsHard
μ₁ − μ₂, t(19) = 2.0930L = ?Ux̄₁ − x̄₂E95%

Two fertilisers are tested. Plot group 1: 13 plots, mean yield 79 kg, standard deviation 9 kg. Plot group 2: 8 plots, mean 81.5 kg, standard deviation 10 kg. Assume equal variances (pooled t, df = n₁ + n₂ − 2). Find the lower limit of the 95% confidence interval for μ₁ − μ₂.

n₁ = 13x̄₁ = 79s₁ = 9n₂ = 8x̄₂ = 81.5s₂ = 1095%t(19) = 2.093
lower limit = kg
9.Hypothesis tests & confidence intervalsMedium
t −2.01102.011

A sample of 49 students has a mean score of 498.7 and a sample standard deviation of 4. Find the test statistic t for H₀: μ = 500.

μ₀ = 500x̄ = 498.7s = 4n = 49
t =
10.Hypothesis tests & confidence intervalsHard
−1.961.96z = ?

In a trial, 72 of 120 patients on treatment A recover, and 90 of 200 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.

x₁ = 72n₁ = 120x₂ = 90n₂ = 200
z =
11.Linear regression & correlationMedium
xy16461952227225872889xy20256080

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

n = 5
r =
12.Linear regression & correlationHard
xy23844164384710541264xy510405060x = 11

The table gives 6 paired observations of x = advertising spend in k€ and y = sales in k€. Use the least-squares line ŷ = a + b·x. A new observation (11, 62) is made. Find its residual e = y − ŷ.

n = 6(11, 62)
e =

3Solutions

1.Descriptive statistics

Answer: s = 12.66

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 = 233/7 = 33.29

Squared deviations

Σ(x − x̄)² = 2.714² + (−2.286)² + (−8.286)² + (−6.286)² + (−13.29)² + 1.714² + 25.71² = 961.4

Standard deviation

s = √(961.4/6) = 12.66

Answer

s = 12.66
2.Descriptive statistics

Answer: P = 0.3446

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 = 0.4

Standard normal table

Φ(0.4) = 0.6554
P = 1 − 0.6554 = 0.3446

Answer

P = 0.3446
3.Probability

Answer: Var(X) = 0.94

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 = 0·0.2 + 1·0.35 + 2·0.3 + 3·0.15 = 1.4

Mean of the squares

E(X²) = 0·0.2 + 1·0.35 + 4·0.3 + 9·0.15 = 2.9

Variance

Var(X) = 2.9 − 1.4² = 0.94

Answer

Var(X) = 0.94
4.Probability

Answer: P = 0.1866

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.

Binomial formula

P(X = 1) = 6·0.41·0.65 = 6·0.4·0.07776 = 0.1866

Answer

P = 0.1866
5.Normal distribution

Answer: P = 0.2743

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 = (200.3 − 200)/0.5 = 0.6

Upper tail

P = 1 − Φ(0.6) = 1 − 0.7257 = 0.2743

Answer

P = 0.2743
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 = −3.132 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 − 71.5 = 2

Standard error

SE = √(16²/30 + 6²/30) = 3.12

Margin of error

E = 1.645·3.12 = 5.132

Interval

2 ± 5.132 → [−3.132, 7.132]

Answer

lower limit = −3.132 h

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

8.Confidence intervals

Answer: lower limit = −11.32 kg

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.

Pooled standard deviation

sp² = (12·9² + 7·10²)/19 = 88
sp = 9.381

Margin of error

E = 2.093·9.381·√(1/13 + 1/8) = 8.823

Interval

−2.5 ± 8.823 → [−11.32, 6.323]

Answer

lower limit = −11.32 kg
9.Hypothesis tests & confidence intervals

Answer: t = −2.275

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 = 4/√49 = 4/7 = 0.5714

Test statistic

t = (498.7 − 500)/0.5714 = −2.275

Answer

t = −2.275
t0.025; df = 48 = 2.011

|t| > 2.011, so H₀ is rejected at the 5% level (two-sided).

10.Hypothesis tests & confidence intervals

Answer: z = 2.598

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 = 72/120 = 0.6
p̂₂ = 90/200 = 0.45
p̂ = 162/320 = 0.5062

Standard error under H₀

SE = √(0.5062·0.4938·(1/120 + 1/200)) = 0.05773

Test statistic

z = (0.6 − 0.45)/0.05773 = 2.598

Answer

z = 2.598

|z| > 1.96, so H₀ is rejected at the 5% level (two-sided).

11.Linear regression & correlation

Answer: r = 0.9716

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 = 110/5 = 22
ȳ = 346/5 = 69.2

Sums of squares and products

Sxx = Σ(x − x̄)² = 90
Sxy = Σ(x − x̄)(y − ȳ) = 363
Syy = Σ(y − ȳ)² = 1551

Correlation coefficient

r = 363/√(90·1551) = 0.9716

Answer

r = 0.9716
12.Linear regression & correlation

Answer: e = 4.281

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 = 42/6 = 7
ȳ = 287/6 = 47.83

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 173

Regression line

b = 173/70 = 2.471
a = 47.83 − 2.471·7 = 30.53
ŷ = 30.53 + 2.471·x

Predicted value

ŷ = 30.53 + 2.471·11 = 57.72

Residual

e = 62 − 57.72 = 4.281

Answer

e = 4.281

The point lies above the line.

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