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
25303540455055

Find the sample standard deviation s of the data: 41, 40, 47, 44, 59, 22.

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

Exam scores are normally distributed with mean 70 and standard deviation 12. Find the z-score z of a score of 55.6.

μ = 70σ = 12x = 55.6
z =
3.ProbabilityMedium
xp(x)00.320.140.360.3μ = 3.2

For the distribution shown, find the standard deviation σ of X.

x: 0, 2, 4, 6p(x): 0.3, 0.1, 0.3, 0.3
σ =
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
−3−2−10123zshaded area between −z and z = 0.5646

Drag z so that the shaded area between −z and z under the standard normal curve is 0.5646. Find z to two decimals.

A = 0.5646
z =
6.Normal distributionHard
μ = 65x = ?95%

The exam scores are normally distributed with mean 65 points and standard deviation 10 points. Find the value x below which 95% of them lie (the 95th percentile).

μ = 65σ = 10P = 95%
x = points
7.Confidence intervalsMedium
μ₁ − μ₂0LUx̄₁ − x̄₂E = ?95%

The battery life of two phone models is compared. Model 1: 80 phones, mean 129.5 h, known standard deviation 12 h. Model 2: 40 phones, mean 124.5 h, known standard deviation 15 h. Find the margin of error E of the 95% confidence interval for μ₁ − μ₂.

n₁ = 80x̄₁ = 129.5 hσ₁ = 12 hn₂ = 40x̄₂ = 124.5 hσ₂ = 15 h95%
E = h
8.Confidence intervalsHard
p = 0.5, n = ?p̂ − Ep̂ + Ep̂E = 0.0399%

Nothing is known in advance about the proportion of defective products. How many products n must be inspected so that a 99% confidence interval for the proportion has a margin of error of at most 0.03?

p = 0.5E = 0.0399%
n =
9.Hypothesis tests & confidence intervalsMedium
t −1.99801.998

A sample of 64 students has a mean score of 201.1 and a sample standard deviation of 4. Find the test statistic t for H₀: μ = 200.

μ₀ = 200x̄ = 201.1s = 4n = 64
t =
10.Hypothesis tests & confidence intervalsHard
μ₀x̄c

A one-sided z test of H₀: μ = 100 against H₁: μ < 100 is done at α = 0.01 with a sample of 64; the standard deviation is known to be 12. In reality the mean is 97.75. Find the probability of a Type II error β in percent.

μ₀ = 100μ = 97.75σ = 12n = 64α = 0.01
β = %
11.Linear regression & correlationMedium
xy125221324421520xy24202224

The table gives 5 paired observations of x = car age in years and y = price in k€. Find the coefficient of determination r² of the least-squares line.

n = 5
r² =
12.Linear regression & correlationHard
xy234961181410171222xy5101020ŷ = 20

The table gives 6 paired observations of x = load in kN and y = extension in mm. Use the least-squares line ŷ = a + b·x. For which value x does the line predict ŷ = 20?

n = 6ŷ = 20
x =

3Solutions

1.Descriptive statistics

Answer: s = 12.02

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 = 253/6 = 42.17

Squared deviations

Σ(x − x̄)² = (−1.167)² + (−2.167)² + 4.833² + 1.833² + 16.83² + (−20.17)² = 722.8

Standard deviation

s = √(722.8/5) = 12.02

Answer

s = 12.02
2.Descriptive statistics

Answer: z = −1.2

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 = (55.6 − 70)/12 = −1.2

Answer

z = −1.2
3.Probability

Answer: σ = 2.4

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 = 0·0.3 + 2·0.1 + 4·0.3 + 6·0.3 = 3.2

Mean of the squares

E(X²) = 0·0.3 + 4·0.1 + 16·0.3 + 36·0.3 = 16

Variance

Var(X) = 16 − 3.2² = 5.76
σ = √5.76 = 2.4

Answer

σ = 2.4
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: z = 0.78

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 + 0.5646)/2 = 0.7823

Read the z-table backwards

Find 0.7823 inside the table and read off z.

z = 0.78

Answer

z = 0.78
6.Normal distribution

Answer: x = 81.45 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) = 0.95 → z = 1.645

Back to x

x = 65 + 1.645·10 = 81.45 points

Answer

x = 81.45 points
7.Confidence intervals

Answer: E = 5.341 h

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.

Point estimate

x̄₁ − x̄₂ = 129.5 − 124.5 = 5

Standard error

SE = √(12²/80 + 15²/40) = 2.725

Margin of error

E = 1.96·2.725 = 5.341

Answer

E = 5.341 h

The interval contains 0, so the data do not show a significant difference.

8.Confidence intervals

Answer: n = 1844

Confidence interval for a proportion: p̂ ± z·√(p̂(1 − p̂)/n), where p̂ = x/n. To reach a margin E you need n ≥ z²·p(1 − p)/E², rounded up; without a prior estimate use p = 0.5, which gives the largest n.

Solve for n

n = 2.576²·0.5·0.5/0.03² = 1843

Round up

n = 1844

Answer

n = 1844
9.Hypothesis tests & confidence intervals

Answer: t = 2.2

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).

Standard error

SE = 4/√64 = 4/8 = 0.5

Test statistic

t = (201.1 − 200)/0.5 = 2.2

Answer

t = 2.2
t0.025; df = 63 = 1.998

|t| > 1.998, so H₀ is rejected at the 5% level (two-sided).

10.Hypothesis tests & confidence intervals

Answer: β = 79.56 %

Power of a test: power = 1 − β is the probability of rejecting H₀ when it is false. Find the cutoff x̄c from μ₀ and z_α, then compute the probability of landing beyond x̄c using the true mean: z = (x̄c − μ)/(σ/√n).

Critical sample mean

SE = 12/√64 = 1.5
x̄c = 100 − 2.326·1.5 = 96.51

z under the true mean

z = (96.51 − 97.75)/1.5 = −0.826

Type II error

power = P(Z < −0.826) = 20.44%
β = 100% − 20.44% = 79.56%

Answer

β = 79.56%
11.Linear regression & correlation

Answer: r² = 0.5319

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 = 15/5 = 3
ȳ = 111/5 = 22.2

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = −10
Syy = Σ(y − ȳ)² = 18.8

Coefficient of determination

r² = −10²/(10·18.8) = 0.5319

Answer

r² = 0.5319

About 53.19% of the variation in y is explained by the line.

12.Linear regression & correlation

Answer: x = 11.21

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 = 42/6 = 7
ȳ = 76/6 = 12.67

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 122

Regression line

b = 122/70 = 1.743
a = 12.67 − 1.743·7 = 0.4667
ŷ = 0.4667 + 1.743·x

Solve for x

x = (20 − 0.4667)/1.743 = 11.21

Answer

x = 11.21

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