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: 55, 10, 29, 52, 11, 38, 48.

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

A quantity is normally distributed with mean 500 and standard deviation 40. What proportion P of values is below 468?

μ = 500σ = 40x = 468
P =
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)00.2510.3520.0530.35μ

X has the distribution shown, and Y = 2X − 3. Find the standard deviation σY of Y.

x: 0, 1, 2, 3p(x): 0.25, 0.35, 0.05, 0.35Y = 2X − 3
σY =
5.Normal distributionMedium
−3−2−10123zshaded area between −z and z = 0.3473

Drag z so that the shaded area between −z and z under the standard normal curve is 0.3473. Find z to two decimals.

A = 0.3473
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 = 16, s = 18LU = ?x̄ = 112E90%

A sample of 16 items has mean 112 and standard deviation s = 18. The population is roughly normal. Find the upper limit of the 90% confidence interval for the mean.

n = 16x̄ = 112s = 1890%t(15) = 1.753
upper limit =
8.Confidence intervalsHard
μ₁ − μ₂, t(12) = 2.1790LU = ?x̄₁ − x̄₂E95%

Two fertilisers are tested. Plot group 1: 9 plots, mean yield 41.5 kg, standard deviation 9.5 kg. Plot group 2: 5 plots, mean 36.5 kg, standard deviation 2 kg. Assume equal variances (pooled t, df = n₁ + n₂ − 2). Find the upper limit of the 95% confidence interval for μ₁ − μ₂.

n₁ = 9x̄₁ = 41.5s₁ = 9.5n₂ = 5x̄₂ = 36.5s₂ = 295%t(12) = 2.179
upper limit = kg
9.Hypothesis tests & confidence intervalsMedium
t −2.13102.131

A sample of 16 students has a mean score of 109 and a sample standard deviation of 15. Find the test statistic t for H₀: μ = 100.

μ₀ = 100x̄ = 109s = 15n = 16
t =
10.Hypothesis tests & confidence intervalsHard
z −1.64501.645

A sample of 64 measurements has mean 101.1. The population standard deviation is 5. Find the lower limit L of the 90% confidence interval for the mean.

x̄ = 101.1σ = 5n = 6490%
L =
11.Linear regression & correlationMedium
xy344450551659757864xy4685060

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 = 33Σy = 325Σx² = 199Σy² = 17863Σxy = 1852
r =
12.Linear regression & correlationHard
xy28486148131020xy510101520x = 5

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 (5, 7) is made. Find its residual e = y − ŷ.

n = 5(5, 7)
e =

3Solutions

1.Descriptive statistics

Answer: s = 18.72

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 = 243/7 = 34.71

Squared deviations

Σ(x − x̄)² = 20.29² + (−24.71)² + (−5.714)² + 17.29² + (−23.71)² + 3.286² + 13.29² = 2103

Standard deviation

s = √(2103/6) = 18.72

Answer

s = 18.72
2.Descriptive statistics

Answer: P = 0.2119

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 = 0.2119

Answer

P = 0.2119
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 = 2.408

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 = 1.5, Var(X) = 1.45, σX = 1.204

Linear change

σY = |2|·1.204 = 2.408

Answer

σY = 2.408

The mean does move: E(Y) = 2·1.5 − 3 = 0.

5.Normal distribution

Answer: z = 0.45

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 + 0.3473)/2 = 0.6736

Read the z-table backwards

Find 0.6736 inside the table and read off z.

z = 0.45

Answer

z = 0.45
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 = 119.9

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 = 15, t = 1.753

Margin of error

E = 1.753·18/√16 = 7.888

Interval

112 ± 7.888 → [104.1, 119.9]

Answer

upper limit = 119.9
8.Confidence intervals

Answer: upper limit = 14.53 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 = (8·9.5² + 4·2²)/12 = 61.5
sp = 7.842

Margin of error

E = 2.179·7.842·√(1/9 + 1/5) = 9.531

Interval

5 ± 9.531 → [−4.531, 14.53]

Answer

upper limit = 14.53 kg
9.Hypothesis tests & confidence intervals

Answer: t = 2.4

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 = 15/√16 = 15/4 = 3.75

Test statistic

t = (109 − 100)/3.75 = 2.4

Answer

t = 2.4
t0.025; df = 15 = 2.131

|t| > 2.131, so H₀ is rejected at the 5% level (two-sided).

10.Hypothesis tests & confidence intervals

Answer: L = 100.1

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).

Margin of error

E = 1.645·5/√64 = 1.028

Interval limit

L = 101.1 − 1.028 = 100.1

Answer

L = 100.1
11.Linear regression & correlation

Answer: r = 0.9584

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 = 1852 − 33·325/6 = 64.5
Sxx = 199 − 33²/6 = 17.5
Syy = 17863 − 325²/6 = 258.8

Correlation coefficient

r = 64.5/√(17.5·258.8) = 0.9584

Answer

r = 0.9584
12.Linear regression & correlation

Answer: e = −4.15

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 = 63/5 = 12.6

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 58

Regression line

b = 58/40 = 1.45
a = 12.6 − 1.45·6 = 3.9
ŷ = 3.9 + 1.45·x

Predicted value

ŷ = 3.9 + 1.45·5 = 11.15

Residual

e = 7 − 11.15 = −4.15

Answer

e = −4.15

The point lies below the line.

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