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
152025303540455055

Find the sample standard deviation s of the data: 28, 12, 55, 44, 18, 13.

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

Exam scores are normally distributed with mean 500 and standard deviation 75. Find the z-score z of a score of 410.

μ = 500σ = 75x = 410
z =
3.ProbabilityMedium
P(at least one six) = ??1?2?3?4?5?6?7P(six) = 1/6 on each roll

A fair die is rolled 7 times. What is the probability P of getting at least one six?

n = 7
P =
4.ProbabilityHard
xp(x)00.320.3540.260.15μ

X has the distribution shown, and Y = 0.5X + 10. Find the standard deviation σY of Y.

x: 0, 2, 4, 6p(x): 0.3, 0.35, 0.2, 0.15Y = 0.5X + 10
σY =
5.Normal distributionMedium
−3−2−10123zshaded area between −z and z = 0.6970

Drag z so that the shaded area between −z and z under the standard normal curve is 0.6970. Find z to two decimals.

A = 0.6970
z =
6.Normal distributionHard
0.04 mm0D − d?

Hole diameters D are normal with mean 25.04 mm and standard deviation 0.015 mm. Shaft diameters d are independent and normal with mean 25 mm and standard deviation 0.015 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 = 0.015 mmshaft: μ = 25 mm, σ = 0.015 mm
P =
7.Confidence intervalsMedium
n = 7, sd​ = 5.70LUd̄ = −1.7E = ?95%

The blood pressure of 7 patients is measured before and after a treatment. The differences d = before − after (mmHg) have mean −1.7 and standard deviation 5.7. Find the margin of error E of the 95% confidence interval for the mean difference.

n = 7d̄ = −1.7sd = 5.795%t(6) = 2.447
E = mmHg
8.Confidence intervalsHard
μ₁ − μ₂, t(19) = 2.8610L = ?Ux̄₁ − x̄₂E99%

Two fertilisers are tested. Plot group 1: 6 plots, mean yield 30 kg, standard deviation 9.5 kg. Plot group 2: 15 plots, mean 42 kg, standard deviation 9.5 kg. Assume equal variances (pooled t, df = n₁ + n₂ − 2). Find the lower limit of the 99% confidence interval for μ₁ − μ₂.

n₁ = 6x̄₁ = 30s₁ = 9.5n₂ = 15x̄₂ = 42s₂ = 9.599%t(19) = 2.861
lower limit = kg
9.Hypothesis tests & confidence intervalsMedium
−0.88p = ?

The mean of a population is claimed to be 250; the population standard deviation is known to be 24. A random sample of 144 observations has a mean of 248.24. Test H₀: μ = 250 against H₁: μ < 250. Find the p-value p.

μ₀ = 250σ = 24n = 144x̄ = 248.24H₁: μ < μ₀
p =
10.Hypothesis tests & confidence intervalsHard
σ = 4, n = ?x̄ − Ex̄ + Ex̄E = 0.590%

How large must a sample be so that the 90% confidence interval for a mean has a margin of error of at most 0.5? The standard deviation is 4. Find the smallest sample size n.

σ = 4E = 0.590%
n =
11.Linear regression & correlationMedium
xy21041361982610311241xy51010203040

The table gives 6 paired observations of x = load in kN and y = extension in mm. Find the coefficient of determination r² of the least-squares line.

n = 6
r² =
12.Linear regression & correlationHard
xy2114196228321036xy510102030x = 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, 11) is made. Find its residual e = y − ŷ.

n = 5(3, 11)
e =

3Solutions

1.Descriptive statistics

Answer: s = 17.69

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 = 170/6 = 28.33

Squared deviations

Σ(x − x̄)² = (−0.3333)² + (−16.33)² + 26.67² + 15.67² + (−10.33)² + (−15.33)² = 1565

Standard deviation

s = √(1565/5) = 17.69

Answer

s = 17.69
2.Descriptive statistics

Answer: z = −1.2

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 = (410 − 500)/75 = −1.2

Answer

z = −1.2
3.Probability

Answer: P = 0.7209

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.

Complement

P(no six) = (5/6)7 = 0.2791
P = 1 − 0.2791 = 0.7209

Answer

P = 0.7209
4.Probability

Answer: σY = 1.03

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 = 2.4, Var(X) = 4.24, σX = 2.059

Linear change

σY = |0.5|·2.059 = 1.03

Answer

σY = 1.03

The mean does move: E(Y) = 0.5·2.4 + 10 = 11.2.

5.Normal distribution

Answer: z = 1.03

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 + 0.6970)/2 = 0.8485

Read the z-table backwards

Find 0.8485 inside the table and read off z.

z = 1.03

Answer

z = 1.03
6.Normal distribution

Answer: P = 0.9706

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 = 0.02121 mm

Standardise

z = (0 − 0.04)/0.02121 = −1.89

Probability

P = 1 − Φ(−1.89) = 0.9706

Answer

P = 0.9706
7.Confidence intervals

Answer: E = 5.272 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 = 6, t = 2.447

Margin of error

E = 2.447·5.7/√7 = 5.272

Answer

E = 5.272 mmHg
8.Confidence intervals

Answer: lower limit = −25.13 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 = (5·9.5² + 14·9.5²)/19 = 90.25
sp = 9.5

Margin of error

E = 2.861·9.5·√(1/6 + 1/15) = 13.13

Interval

−12 ± 13.13 → [−25.13, 1.129]

Answer

lower limit = −25.13 kg
9.Hypothesis tests & confidence intervals

Answer: p = 0.1894

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.

Test statistic

SE = σ/√n = 24/√144 = 2
z = (248.24 − 250)/2 = −0.88

Tail area

Φ(0.88) = 0.8106
1 − Φ(0.88) = 0.1894

p-value

p = 0.1894

Answer

p = 0.1894

p > 0.05, so H₀ is not rejected at the 5% level.

10.Hypothesis tests & confidence intervals

Answer: n = 174

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).

Solve for n

n = (1.645·4/0.5)² = 173.2

Round up

n = 174

Answer

n = 174
11.Linear regression & correlation

Answer: r² = 0.9782

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 = 140/6 = 23.33

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 216
Syy = Σ(y − ȳ)² = 681.3

Coefficient of determination

r² = 216²/(70·681.3) = 0.9782

Answer

r² = 0.9782

About 97.82% of the variation in y is explained by the line.

12.Linear regression & correlation

Answer: e = −3.55

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 = 120/5 = 24

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 126

Regression line

b = 126/40 = 3.15
a = 24 − 3.15·6 = 5.1
ŷ = 5.1 + 3.15·x

Predicted value

ŷ = 5.1 + 3.15·3 = 14.55

Residual

e = 11 − 14.55 = −3.55

Answer

e = −3.55

The point lies below the line.

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