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: 19, 55, 30, 23, 44, 40.

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

A quantity is normally distributed with mean 70 and standard deviation 12. What proportion P of values is below 94?

μ = 70σ = 12x = 94
P =
3.ProbabilityMedium
xp(x)00.1520.340.260.35μ = 3.5

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

x: 0, 2, 4, 6p(x): 0.15, 0.3, 0.2, 0.35
Var(X) =
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
23.336.8P = ?

The delivery times are normally distributed with mean 32 min and standard deviation 6 min. What fraction lies between 23.3 and 36.8 min? Give P as a decimal.

μ = 32σ = 6x1 = 23.3x2 = 36.8
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
n = 12, s = 20LU = ?x̄ = 92.5E90%

A sample of 12 items has mean 92.5 and standard deviation s = 20. The population is roughly normal. Find the upper limit of the 90% confidence interval for the mean.

n = 12x̄ = 92.5s = 2090%t(11) = 1.796
upper limit =
8.Confidence intervalsHard
n = 10Lσ = ?s = 6.895%

The diameters of 10 machined shafts have a sample standard deviation of 6.8 µm. Assuming the diameters are normally distributed, find the upper 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 = 6.8 µmχ² = 2.7, 19.023
σ = µm
9.Hypothesis tests & confidence intervalsMedium
−2.1792.179t = ?df = 12

13 employees take a typing test before and after a training course. For each person the difference d = after − before (words per minute) is recorded. The mean difference is −1.77 and the standard deviation of the differences is 7.1. Find the paired test statistic t for H₀: μd = 0.

n = 13d̄ = −1.77sd = 7.1
t =
10.Hypothesis tests & confidence intervalsHard
z −1.9601.96

A sample of 64 measurements has mean 497.3. The population standard deviation is 20. Find the upper limit U of the 95% confidence interval for the mean.

x̄ = 497.3σ = 20n = 6495%
U =
11.Linear regression & correlationMedium
xy250362460567664773xy246506070

The table gives 6 paired observations of x = hours studied and y = exam score. Find Pearson's correlation coefficient r.

n = 6
r =
12.Linear regression & correlationHard
xy25246768089510104xy5106080100x = 9

The table gives 5 paired observations of x = hours studied and y = exam score. Use the least-squares line ŷ = a + b·x. A new observation (9, 102) is made. Find its residual e = y − ŷ.

n = 5(9, 102)
e =

3Solutions

1.Descriptive statistics

Answer: s = 13.64

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 = 211/6 = 35.17

Squared deviations

Σ(x − x̄)² = (−16.17)² + 19.83² + (−5.167)² + (−12.17)² + 8.833² + 4.833² = 930.8

Standard deviation

s = √(930.8/5) = 13.64

Answer

s = 13.64
2.Descriptive statistics

Answer: P = 0.9772

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 = 2

Standard normal table

Φ(2) = 0.9772
P = 0.9772

Answer

P = 0.9772
3.Probability

Answer: Var(X) = 4.75

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 = 0·0.15 + 2·0.3 + 4·0.2 + 6·0.35 = 3.5

Mean of the squares

E(X²) = 0·0.15 + 4·0.3 + 16·0.2 + 36·0.35 = 17

Variance

Var(X) = 17 − 3.5² = 4.75

Answer

Var(X) = 4.75
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: P = 0.7146

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 = (23.3 − 32)/6 = −1.45
z₂ = (36.8 − 32)/6 = 0.8

Area between

P = Φ(0.8) − Φ(−1.45) = 0.7881 − 0.07353 = 0.7146

Answer

P = 0.7146
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: upper limit = 102.9

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 = 11, t = 1.796

Margin of error

E = 1.796·20/√12 = 10.37

Interval

92.5 ± 10.37 → [82.13, 102.9]

Answer

upper limit = 102.9
8.Confidence intervals

Answer: σ = 12.42 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 = 9·6.8² = 416.2

Limits for σ²

416.2/19.023 = 21.88
416.2/2.7 = 154.1

Limit for σ

σ = √154.1 = 12.42 µm

Answer

σ = 12.42 µm

95% interval for σ: [4.677, 12.42] µm.

9.Hypothesis tests & confidence intervals

Answer: t = −0.8988

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 of the mean difference

SE = 7.1/√13 = 1.969

Test statistic

t = −1.77/1.969 = −0.8988

Answer

t = −0.8988

With df = 12, the two-sided 5% critical value is 2.179; |t| is smaller, so H₀ is not rejected.

10.Hypothesis tests & confidence intervals

Answer: U = 502.2

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 = 1.96·20/√64 = 4.9

Interval limit

U = 497.3 + 4.9 = 502.2

Answer

U = 502.2
11.Linear regression & correlation

Answer: r = 0.8902

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 = 27/6 = 4.5
ȳ = 376/6 = 62.67

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = 64
Syy = Σ(y − ȳ)² = 295.3

Correlation coefficient

r = 64/√(17.5·295.3) = 0.8902

Answer

r = 0.8902
12.Linear regression & correlation

Answer: e = 2.6

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 = 398/5 = 79.6

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 264

Regression line

b = 264/40 = 6.6
a = 79.6 − 6.6·6 = 40
ŷ = 40 + 6.6·x

Predicted value

ŷ = 40 + 6.6·9 = 99.4

Residual

e = 102 − 99.4 = 2.6

Answer

e = 2.6

The point lies above the line.

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