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
1015202530354045

Find the sample standard deviation s of the data: 22, 29, 10, 39, 47, 26.

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

A quantity is normally distributed with mean 50 and standard deviation 20. What proportion P of values is below 62?

μ = 50σ = 20x = 62
P =
3.ProbabilityMedium
n = 6k = 4N = ?

In how many ways N can a committee of 4 people be chosen from a group of 6? The order does not matter.

n = 6k = 4
N =
4.ProbabilityHard
n = 5, p = 0.25: P(X = 4) = ?k012345

A test has 5 multiple-choice questions. A student who guesses gets each one right with probability 0.25, independently. What is the probability P of exactly 4 correct answers?

n = 5p = 0.25k = 4
P =
5.Normal distributionMedium
250.25257.5P = ?

The bag weights are normally distributed with mean 250 g and standard deviation 5 g. What fraction lies between 250.25 and 257.5 g? Give P as a decimal.

μ = 250σ = 5x1 = 250.25x2 = 257.5
P =
6.Normal distributionHard
10 g0X − Y?

The mass X of a product from line 1 and Y from line 2 are independent and normal: X has mean 500 g and standard deviation 8 g, Y has mean 490 g and standard deviation 6 g. Find the probability P that X is greater than Y. Round z to two decimals.

μX = 500 g, σX = 8 gμY = 490 g, σY = 6 g
P =
7.Confidence intervalsMedium
n = 15, s = 17.5LU = ?x̄ = 20.5E90%

A sample of 15 items has mean 20.5 and standard deviation s = 17.5. The population is roughly normal. Find the upper limit of the 90% confidence interval for the mean.

n = 15x̄ = 20.5s = 17.590%t(14) = 1.761
upper limit =
8.Confidence intervalsHard
n = 20σ = ?Us = 7.995%

The diameters of 20 machined shafts have a sample standard deviation of 7.9 µm. Assuming the diameters are normally distributed, find the lower 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 = 7.9 µmχ² = 8.907, 32.852
σ = µm
9.Hypothesis tests & confidence intervalsMedium
−1.961.96z = ?

Two production lines make the same part. Line 1: a sample of 50 parts has mean length 98 mm, and the line's known standard deviation is 11 mm. Line 2: a sample of 100 parts has mean 99.2 mm, known standard deviation 8 mm. Find the test statistic z for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (line 1 minus line 2).

n₁ = 50x̄₁ = 98 mmσ₁ = 11 mmn₂ = 100x̄₂ = 99.2 mmσ₂ = 8 mm
z =
10.Hypothesis tests & confidence intervalsHard
−1.961.96z = ?

In a trial, 163 of 250 patients on treatment A recover, and 93 of 150 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.

x₁ = 163n₁ = 250x₂ = 93n₂ = 150
z =
11.Linear regression & correlationMedium
xy294156238271032xy510102030

The table gives 5 paired observations of x = load in kN and y = extension in mm. Their sums are given. Find Pearson's correlation coefficient r from these sums.

n = 5Σx = 30Σy = 106Σx² = 220Σy² = 2588Σxy = 752
r =
12.Linear regression & correlationHard
xy1745204823602672297832101xy2025306080100x = 24.5

The table gives 6 paired observations of x = temperature in °C and y = ice creams sold. Use the least-squares line ŷ = a + b·x. A new observation (24.5, 71) is made. Find its residual e = y − ŷ.

n = 6(24.5, 71)
e =

3Solutions

1.Descriptive statistics

Answer: s = 12.98

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 = 173/6 = 28.83

Squared deviations

Σ(x − x̄)² = (−6.833)² + 0.1667² + (−18.83)² + 10.17² + 18.17² + (−2.833)² = 842.8

Standard deviation

s = √(842.8/5) = 12.98

Answer

s = 12.98
2.Descriptive statistics

Answer: P = 0.7257

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 = 0.6

Standard normal table

Φ(0.6) = 0.7257
P = 0.7257

Answer

P = 0.7257
3.Probability

Answer: N = 15

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.

Combinations

C(6, 4) = (6·5·4·3)/(4·3·2·1) = 15

Answer

N = 15
4.Probability

Answer: P = 0.01465

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 = 4) = 5·0.254·0.751 = 5·0.003906·0.75 = 0.01465

Answer

P = 0.01465
5.Normal distribution

Answer: P = 0.4133

Normal distribution: convert a value x to a z-score, z = (x − μ)/σ, which counts standard deviations from the mean. A z-table (or Φ(z)) gives the area to the left of z, which is P(X < x). The area to the right is 1 − Φ(z), and the area between two values is the difference of their two areas.

z-scores

z₁ = (250.25 − 250)/5 = 0.05
z₂ = (257.5 − 250)/5 = 1.5

Area between

P = Φ(1.5) − Φ(0.05) = 0.9332 − 0.5199 = 0.4133

Answer

P = 0.4133
6.Normal distribution

Answer: P = 0.8413

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 = 500 − 490 = 10 g, σ = 10 g

Standardise

z = (0 − 10)/10 = −1

Probability

P = 1 − Φ(−1) = 0.8413

Answer

P = 0.8413
7.Confidence intervals

Answer: upper limit = 28.46

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 = 14, t = 1.761

Margin of error

E = 1.761·17.5/√15 = 7.957

Interval

20.5 ± 7.957 → [12.54, 28.46]

Answer

upper limit = 28.46
8.Confidence intervals

Answer: σ = 6.008 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 = 19·7.9² = 1186

Limits for σ²

1186/32.852 = 36.09
1186/8.907 = 133.1

Limit for σ

σ = √36.09 = 6.008 µm

Answer

σ = 6.008 µm

95% interval for σ: [6.008, 11.54] µm.

9.Hypothesis tests & confidence intervals

Answer: z = −0.686

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.

Standard error of the difference

SE = √(11²/50 + 8²/100) = √3.06 = 1.749

Test statistic

z = (98 − 99.2)/1.749 = −0.686

Answer

z = −0.686

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

10.Hypothesis tests & confidence intervals

Answer: z = 0.6455

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 = 163/250 = 0.652
p̂₂ = 93/150 = 0.62
p̂ = 256/400 = 0.64

Standard error under H₀

SE = √(0.64·0.36·(1/250 + 1/150)) = 0.04957

Test statistic

z = (0.652 − 0.62)/0.04957 = 0.6455

Answer

z = 0.6455

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

11.Linear regression & correlation

Answer: r = 0.9935

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 = 752 − 30·106/5 = 116
Sxx = 220 − 30²/5 = 40
Syy = 2588 − 106²/5 = 340.8

Correlation coefficient

r = 116/√(40·340.8) = 0.9935

Answer

r = 0.9935
12.Linear regression & correlation

Answer: e = 3.667

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 = 147/6 = 24.5
ȳ = 404/6 = 67.33

Sums of squares and products

Sxx = Σ(x − x̄)² = 157.5
Sxy = Σ(x − x̄)(y − ȳ) = 573

Regression line

b = 573/157.5 = 3.638
a = 67.33 − 3.638·24.5 = −21.8
ŷ = −21.8 + 3.638·x

Predicted value

ŷ = −21.8 + 3.638·24.5 = 67.33

Residual

e = 71 − 67.33 = 3.667

Answer

e = 3.667

The point lies above the line.

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