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
152025303540

Find the sample standard deviation s of the data: 34, 13, 26, 44, 12, 38, 32.

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

Exam scores are normally distributed with mean 70 and standard deviation 20. Find the z-score z of a score of 50.

μ = 70σ = 20x = 50
z =
3.ProbabilityMedium
P(at least one six) = ??1?2?3P(six) = 1/6 on each roll

A fair die is rolled 3 times. What is the probability P of getting at least one six?

n = 3
P =
4.ProbabilityHard
n = 9, p = 0.2: P(X ≤ 1) = ?k0123456789

9 components are tested, and each one fails with probability 0.2, independently. What is the probability P that at most one fails?

n = 9p = 0.2
P =
5.Normal distributionMedium
120 kg118.7 kgT = X₁ + … + X₁₀?

The masses of boxes are independent and normal with mean 12 kg and standard deviation 0.8 kg. Find the probability P that the total mass of 10 of them is below 118.7 kg. Round z to two decimals.

μ = 12 kgσ = 0.8 kgn = 10< 118.7 kg
P =
6.Normal distributionHard
μ = 100x = ?75%

The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. Find the value x below which 75% of them lie (the 75th percentile).

μ = 100σ = 15P = 75%
x = points
7.Confidence intervalsMedium
n = 6, s = 3.5L = ?Ux̄ = 39E95%

A sample of 6 items has mean 39 and standard deviation s = 3.5. The population is roughly normal. Find the lower limit of the 95% confidence interval for the mean.

n = 6x̄ = 39s = 3.595%t(5) = 2.571
lower limit =
8.Confidence intervalsHard
n = 21σ = ?Us = 5.195%

The diameters of 21 machined shafts have a sample standard deviation of 5.1 µm. Assuming the diameters are normally distributed, find the lower limit σ of the 95% confidence interval for the population standard deviation. With df = 20, 2.5% of the χ² distribution lies below 9.591 and 2.5% lies above 34.17.

n = 21s = 5.1 µmχ² = 9.591, 34.17
σ = µm
9.Hypothesis tests & confidence intervalsMedium
−2.1792.179t = ?df = 12

Two teaching methods are compared. Group 1: 9 students, mean score 32, standard deviation 10.5. Group 2: 5 students, mean score 41.1, standard deviation 7.5. Assume equal population variances and use the pooled two-sample t test. Find the test statistic t for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (group 1 minus group 2).

n₁ = 9x̄₁ = 32s₁ = 10.5n₂ = 5x̄₂ = 41.1s₂ = 7.5
t =
10.Hypothesis tests & confidence intervalsHard
σ = 20, n = ?x̄ − Ex̄ + Ex̄E = 390%

How large must a sample be so that the 90% confidence interval for a mean has a margin of error of at most 3? The standard deviation is 20. Find the smallest sample size n.

σ = 20E = 390%
n =
11.Linear regression & correlationMedium
xy264363472570680782xy2467080

The table gives 6 paired observations of x = hours studied and y = exam score. Find the coefficient of determination r² of the least-squares line.

n = 6
r² =
12.Linear regression & correlationHard
xy128223325420521616xy2462025ŷ = 26

The table gives 6 paired observations of x = car age in years and y = price in k€. Use the least-squares line ŷ = a + b·x. For which value x does the line predict ŷ = 26?

n = 6ŷ = 26
x =

3Solutions

1.Descriptive statistics

Answer: s = 12.19

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 = 199/7 = 28.43

Squared deviations

Σ(x − x̄)² = 5.571² + (−15.43)² + (−2.429)² + 15.57² + (−16.43)² + 9.571² + 3.571² = 891.7

Standard deviation

s = √(891.7/6) = 12.19

Answer

s = 12.19
2.Descriptive statistics

Answer: z = −1

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 = (50 − 70)/20 = −1

Answer

z = −1
3.Probability

Answer: P = 0.4213

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)3 = 0.5787
P = 1 − 0.5787 = 0.4213

Answer

P = 0.4213
4.Probability

Answer: P = 0.4362

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.

Zero and one failure

P(X = 0) = 0.89 = 0.1342
P(X = 1) = 9·0.2·0.88 = 0.302

Add them

P = 0.1342 + 0.302 = 0.4362

Answer

P = 0.4362
5.Normal distribution

Answer: P = 0.305

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 = 120 kg, σT = 0.8·√10 = 2.53 kg

Standardise

z = (118.7 − 120)/2.53 = −0.51

Probability

P = Φ(−0.51) = 0.305

Answer

P = 0.305
6.Normal distribution

Answer: x = 110.1 points

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) = 0.75 → z = 0.6745

Back to x

x = 100 + 0.6745·15 = 110.12 points

Answer

x = 110.12 points
7.Confidence intervals

Answer: lower limit = 35.33

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 = 5, t = 2.571

Margin of error

E = 2.571·3.5/√6 = 3.674

Interval

39 ± 3.674 → [35.33, 42.67]

Answer

lower limit = 35.33
8.Confidence intervals

Answer: σ = 3.902 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 = 20·5.1² = 520.2

Limits for σ²

520.2/34.17 = 15.22
520.2/9.591 = 54.24

Limit for σ

σ = √15.22 = 3.902 µm

Answer

σ = 3.902 µm

95% interval for σ: [3.902, 7.365] µm.

9.Hypothesis tests & confidence intervals

Answer: t = −1.699

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.

Pooled variance

sp² = (8·10.5² + 4·7.5²)/12 = 92.25
sp = 9.605

Standard error

SE = 9.605·√(1/9 + 1/5) = 5.357

Test statistic

t = (32 − 41.1)/5.357 = −1.699

Answer

t = −1.699

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: n = 121

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 = (1.645·20/3)² = 120.3

Round up

n = 121

Answer

n = 121
11.Linear regression & correlation

Answer: r² = 0.8823

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 = 431/6 = 71.83

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = 69.5
Syy = Σ(y − ȳ)² = 312.8

Coefficient of determination

r² = 69.5²/(17.5·312.8) = 0.8823

Answer

r² = 0.8823

About 88.23% of the variation in y is explained by the line.

12.Linear regression & correlation

Answer: x = 1.61

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 = 133/6 = 22.17

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −35.5

Regression line

b = −35.5/17.5 = −2.029
a = 22.17 − (−2.029)·3.5 = 29.27
ŷ = 29.27 − 2.029·x

Solve for x

x = (26 − 29.27)/(−2.029) = 1.61

Answer

x = 1.61

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