Introductory Statistics: cram pack

Formula sheets, the key method steps for every topic and 12 practice problems, with full solutions at the back.

New problemsBack to the course

In the print window, choose “Save as PDF” to keep a copy on your device.

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
20253035404550

Find the sample standard deviation s of the data: 42, 34, 20, 54, 39, 35, 34.

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

A quantity is normally distributed with mean 60 and standard deviation 12. What proportion P of values is above 50.4?

μ = 60σ = 12x = 50.4
P =
3.ProbabilityMedium
xp(x)00.210.320.1530.35μ = 1.65

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

x: 0, 1, 2, 3p(x): 0.2, 0.3, 0.15, 0.35
σ =
4.ProbabilityHard
n = 7, p = 0.4: P(X = 2) = ?k01234567

A test has 7 multiple-choice questions. A student who guesses gets each one right with probability 0.4, independently. What is the probability P of exactly 2 correct answers?

n = 7p = 0.4k = 2
P =
5.Normal distributionMedium
−3−2−10123zshaded area to the right of z = 0.8830

Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.8830. Find z to two decimals.

A = 0.8830
z =
6.Normal distributionHard
0.04 mm0D − d?

Hole diameters D are normal with mean 25.04 mm and standard deviation 0.01 mm. Shaft diameters d are independent and normal with mean 25 mm and standard deviation 0.02 mm. A randomly chosen shaft fits when D − d > 0. Find the probability P that it fits. Round z to two decimals.

hole: μ = 25.04 mm, σ = 0.01 mmshaft: μ = 25 mm, σ = 0.02 mm
P =
7.Confidence intervalsMedium
n = 6, sd​ = 0.60LU = ?d̄ = 5.7E95%

The blood pressure of 6 patients is measured before and after a treatment. The differences d = before − after (mmHg) have mean 5.7 and standard deviation 0.6. Find the upper limit of the 95% confidence interval for the mean difference.

n = 6d̄ = 5.7sd = 0.695%t(5) = 2.571
upper limit = mmHg
8.Confidence intervalsHard
n = 10σ = ?Us = 2.895%

The diameters of 10 machined shafts have a sample standard deviation of 2.8 µm. Assuming the diameters are normally distributed, find the lower limit σ of the 95% confidence interval for the population standard deviation. With df = 9, 2.5% of the χ² distribution lies below 2.7 and 2.5% lies above 19.023.

n = 10s = 2.8 µmχ² = 2.7, 19.023
σ = µm
9.Hypothesis tests & confidence intervalsMedium
−2.1792.179t = ?df = 12

Two teaching methods are compared. Group 1: 7 students, mean score 119.5, standard deviation 8. Group 2: 7 students, mean score 129.5, standard deviation 7. Assume equal population variances and use the pooled two-sample t test. Find the test statistic t for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (group 1 minus group 2).

n₁ = 7x̄₁ = 119.5s₁ = 8n₂ = 7x̄₂ = 129.5s₂ = 7
t =
10.Hypothesis tests & confidence intervalsHard
−1.961.96z = ?

In a trial, 45 of 150 patients on treatment A recover, and 66 of 300 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.

x₁ = 45n₁ = 150x₂ = 66n₂ = 300
z =
11.Linear regression & correlationMedium
xy2544676748911095xy5106080

The table gives 5 paired observations of x = hours studied and y = exam score. Their sums are given. Find Pearson's correlation coefficient r from these sums.

n = 5Σx = 30Σy = 381Σx² = 220Σy² = 30187Σxy = 2498
r =
12.Linear regression & correlationHard
xy274136148221027xy5101020x = 3

The table gives 5 paired observations of x = load in kN and y = extension in mm. Use the least-squares line ŷ = a + b·x. A new observation (3, 13) is made. Find its residual e = y − ŷ.

n = 5(3, 13)
e =

3Solutions

1.Descriptive statistics

Answer: s = 10.24

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 = 258/7 = 36.86

Squared deviations

Σ(x − x̄)² = 5.143² + (−2.857)² + (−16.86)² + 17.14² + 2.143² + (−1.857)² + (−2.857)² = 628.9

Standard deviation

s = √(628.9/6) = 10.24

Answer

s = 10.24
2.Descriptive statistics

Answer: P = 0.7881

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 = −0.8

Standard normal table

Φ(−0.8) = 0.2119
P = 1 − 0.2119 = 0.7881

Answer

P = 0.7881
3.Probability

Answer: σ = 1.152

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 + 1·0.3 + 2·0.15 + 3·0.35 = 1.65

Mean of the squares

E(X²) = 0·0.2 + 1·0.3 + 4·0.15 + 9·0.35 = 4.05

Variance

Var(X) = 4.05 − 1.65² = 1.328
σ = √1.328 = 1.152

Answer

σ = 1.152
4.Probability

Answer: P = 0.2613

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 = 2) = 21·0.42·0.65 = 21·0.16·0.07776 = 0.2613

Answer

P = 0.2613
5.Normal distribution

Answer: z = −1.19

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 of z

Φ(z) = 1 − 0.8830 = 0.1170

Read the z-table backwards

Find 0.1170 inside the table and read off z.

z = −1.19

Answer

z = −1.19
6.Normal distribution

Answer: P = 0.9633

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 = 25.04 − 25 = 0.04 mm, σ = 0.02236 mm

Standardise

z = (0 − 0.04)/0.02236 = −1.79

Probability

P = 1 − Φ(−1.79) = 0.9633

Answer

P = 0.9633
7.Confidence intervals

Answer: upper limit = 6.33 mmHg

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.

Critical value

df = 5, t = 2.571

Margin of error

E = 2.571·0.6/√6 = 0.6298

Interval

5.7 ± 0.6298 → [5.07, 6.33]

Answer

upper limit = 6.33 mmHg
8.Confidence intervals

Answer: σ = 1.926 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 = 9·2.8² = 70.56

Limits for σ²

70.56/19.023 = 3.709
70.56/2.7 = 26.13

Limit for σ

σ = √3.709 = 1.926 µm

Answer

σ = 1.926 µm

95% interval for σ: [1.926, 5.112] µm.

9.Hypothesis tests & confidence intervals

Answer: t = −2.489

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.

Pooled variance

sp² = (6·8² + 6·7²)/12 = 56.5
sp = 7.517

Standard error

SE = 7.517·√(1/7 + 1/7) = 4.018

Test statistic

t = (119.5 − 129.5)/4.018 = −2.489

Answer

t = −2.489

With df = 12, the two-sided 5% critical value is 2.179; |t| is larger, so H₀ is rejected.

10.Hypothesis tests & confidence intervals

Answer: z = 1.856

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 = 45/150 = 0.3
p̂₂ = 66/300 = 0.22
p̂ = 111/450 = 0.2467

Standard error under H₀

SE = √(0.2467·0.7533·(1/150 + 1/300)) = 0.04311

Test statistic

z = (0.3 − 0.22)/0.04311 = 1.856

Answer

z = 1.856

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

11.Linear regression & correlation

Answer: r = 0.9864

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 = 2498 − 30·381/5 = 212
Sxx = 220 − 30²/5 = 40
Syy = 30187 − 381²/5 = 1155

Correlation coefficient

r = 212/√(40·1155) = 0.9864

Answer

r = 0.9864
12.Linear regression & correlation

Answer: e = 3.75

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 = 30/5 = 6
ȳ = 83/5 = 16.6

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 98

Regression line

b = 98/40 = 2.45
a = 16.6 − 2.45·6 = 1.9
ŷ = 1.9 + 2.45·x

Predicted value

ŷ = 1.9 + 2.45·3 = 9.25

Residual

e = 13 − 9.25 = 3.75

Answer

e = 3.75

The point lies above the line.

ExamProblems · examproblems.com/courses/statistics-1