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: 21, 49, 14, 54, 58, 45, 49.

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

A quantity is normally distributed with mean 500 and standard deviation 50. What proportion P of values is below 425?

μ = 500σ = 50x = 425
P =
3.ProbabilityMedium
red: 8blue: 4

A bag holds 8 red and 4 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?

8 / 4
P =
4.ProbabilityHard
n = 5, p = 0.2: P(X = 2) = ?k012345

A test has 5 multiple-choice questions. A student who guesses gets each one right with probability 0.2, independently. What is the probability P of exactly 2 correct answers?

n = 5p = 0.2k = 2
P =
5.Normal distributionMedium
160 g163.6 gT = X₁ + … + X₄?

The masses of bolts are independent and normal with mean 40 g and standard deviation 1.2 g. Find the probability P that the total mass of 4 of them is below 163.6 g. Round z to two decimals.

μ = 40 gσ = 1.2 gn = 4< 163.6 g
P =
6.Normal distributionHard
μ = 500x = ?97.5%

The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. Find the value x below which 97.5% of them lie (the 97.5th percentile).

μ = 500σ = 40P = 97.5%
x = h
7.Confidence intervalsMedium
n = 8, sd​ = 1.80LUd̄ = −4E = ?99%

The blood pressure of 8 patients is measured before and after a treatment. The differences d = before − after (mmHg) have mean −4 and standard deviation 1.8. Find the margin of error E of the 99% confidence interval for the mean difference.

n = 8d̄ = −4sd = 1.899%t(7) = 3.499
E = mmHg
8.Confidence intervalsHard
n = 15Lσ = ?s = 0.895%

The diameters of 15 machined shafts have a sample standard deviation of 0.8 µm. Assuming the diameters are normally distributed, find the upper limit σ of the 95% confidence interval for the population standard deviation. With df = 14, 2.5% of the χ² distribution lies below 5.629 and 2.5% lies above 26.119.

n = 15s = 0.8 µmχ² = 5.629, 26.119
σ = µm
9.Hypothesis tests & confidence intervalsMedium
t −1.99801.998

A sample of 64 students has a mean score of 50.88 and a sample standard deviation of 10. Find the test statistic t for H₀: μ = 50.

μ₀ = 50x̄ = 50.88s = 10n = 64
t =
10.Hypothesis tests & confidence intervalsHard
σ = 20, n = ?x̄ − Ex̄ + Ex̄E = 0.595%

How large must a sample be so that the 95% confidence interval for a mean has a margin of error of at most 0.5? The standard deviation is 20. Find the smallest sample size n.

σ = 20E = 0.595%
n =
11.Linear regression & correlationMedium
xy342443547654757xy46455055

The table gives 5 paired observations of x = advertising spend in k€ and y = sales in k€. Find Pearson's correlation coefficient r.

n = 5
r =
12.Linear regression & correlationHard
xy126234336441538643xy2463040x = 3.5

The table gives 6 paired observations of x = advertising spend in k€ and y = sales in k€. Use the least-squares line ŷ = a + b·x. A new observation (3.5, 41) is made. Find its residual e = y − ŷ.

n = 6(3.5, 41)
e =

3Solutions

1.Descriptive statistics

Answer: s = 16.98

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 = 290/7 = 41.43

Squared deviations

Σ(x − x̄)² = (−20.43)² + 7.571² + (−27.43)² + 12.57² + 16.57² + 3.571² + 7.571² = 1730

Standard deviation

s = √(1730/6) = 16.98

Answer

s = 16.98
2.Descriptive statistics

Answer: P = 0.06681

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 = −1.5

Standard normal table

Φ(−1.5) = 0.06681
P = 0.06681

Answer

P = 0.06681
3.Probability

Answer: P = 0.4242

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.

First and second draw

P = 8/12 · 7/11 = 56/132 = 0.4242

Answer

P = 0.4242
4.Probability

Answer: P = 0.2048

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) = 10·0.22·0.83 = 10·0.04·0.512 = 0.2048

Answer

P = 0.2048
5.Normal distribution

Answer: P = 0.9332

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 total

μT = 160 g, σT = 1.2·√4 = 2.4 g

Standardise

z = (163.6 − 160)/2.4 = 1.5

Probability

P = Φ(1.5) = 0.9332

Answer

P = 0.9332
6.Normal distribution

Answer: x = 578.4 h

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 for the percentile

Φ(z) = 0.975 → z = 1.96

Back to x

x = 500 + 1.96·40 = 578.4 h

Answer

x = 578.4 h
7.Confidence intervals

Answer: E = 2.227 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 = 7, t = 3.499

Margin of error

E = 3.499·1.8/√8 = 2.227

Answer

E = 2.227 mmHg
8.Confidence intervals

Answer: σ = 1.262 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 = 14·0.8² = 8.96

Limits for σ²

8.96/26.119 = 0.343
8.96/5.629 = 1.592

Limit for σ

σ = √1.592 = 1.262 µm

Answer

σ = 1.262 µm

95% interval for σ: [0.5857, 1.262] µm.

9.Hypothesis tests & confidence intervals

Answer: t = 0.704

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 = 10/√64 = 10/8 = 1.25

Test statistic

t = (50.88 − 50)/1.25 = 0.704

Answer

t = 0.704
t0.025; df = 63 = 1.998

|t| < 1.998, so H₀ is not rejected at the 5% level (two-sided).

10.Hypothesis tests & confidence intervals

Answer: n = 6147

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 = (1.96·20/0.5)² = 6147

Round up

n = 6147

Answer

n = 6147
11.Linear regression & correlation

Answer: r = 0.974

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 = 25/5 = 5
ȳ = 243/5 = 48.6

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = 41
Syy = Σ(y − ȳ)² = 177.2

Correlation coefficient

r = 41/√(10·177.2) = 0.974

Answer

r = 0.974
12.Linear regression & correlation

Answer: e = 4.667

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 = 21/6 = 3.5
ȳ = 218/6 = 36.33

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = 51

Regression line

b = 51/17.5 = 2.914
a = 36.33 − 2.914·3.5 = 26.13
ŷ = 26.13 + 2.914·x

Predicted value

ŷ = 26.13 + 2.914·3.5 = 36.33

Residual

e = 41 − 36.33 = 4.667

Answer

e = 4.667

The point lies above the line.

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