Statistics 2: Statistical Inference: cram pack

Formula sheets, the key method steps for every topic and 10 practice problems, with full solutions at the back.

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1Formulas and methods

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

Chi-square tests

Test statistic
χ² = Σ(O − E)²/E
Goodness of fit
E = n·p, df = k − 1 − (estimated parameters)
Independence
E = row total × column total / grand total, df = (r − 1)(c − 1)
Critical values (α = 0.05)
df 1: 3.841, 2: 5.991, 3: 7.815, 4: 9.488
Decision
reject H0 if χ² > χ²crit

2Practice problems

1.Normal distributionMedium
22.438P = ?

The delivery times are normally distributed with mean 32 min and standard deviation 6 min. What fraction lies between 22.4 and 38 min? Give P as a decimal.

μ = 32σ = 6x1 = 22.4x2 = 38
P =
2.Normal distributionHard
4 s0X − Y?

Runner A's lap time X and runner B's lap time Y are independent and normal: X has mean 52 s and standard deviation 4 s, Y has mean 48 s and standard deviation 3 s. Find the probability P that X is greater than Y. Round z to two decimals.

μX = 52 s, σX = 4 sμY = 48 s, σY = 3 s
P =
3.Confidence intervalsMedium
n = 12, s = 2.5L = ?Ux̄ = 36.5E95%

A sample of 12 items has mean 36.5 and standard deviation s = 2.5. The population is roughly normal. Find the lower limit of the 95% confidence interval for the mean.

n = 12x̄ = 36.5s = 2.595%t(11) = 2.201
lower limit =
4.Confidence intervalsHard
σ = 33, n = ?x̄ − Ex̄ + Ex̄E = 0.599%

The population standard deviation is 33. How large a sample n do you need so that a 99% confidence interval for the mean has a margin of error of at most 0.5?

σ = 33E = 0.599%
n =
5.Hypothesis tests & confidence intervalsMedium
−1.961.96z = ?

Two production lines make the same part. Line 1: a sample of 40 parts has mean length 117 mm, and the line's known standard deviation is 14 mm. Line 2: a sample of 50 parts has mean 112.1 mm, known standard deviation 5 mm. Find the test statistic z for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (line 1 minus line 2).

n₁ = 40x̄₁ = 117 mmσ₁ = 14 mmn₂ = 50x̄₂ = 112.1 mmσ₂ = 5 mm
z =
6.Hypothesis tests & confidence intervalsHard
H₀H₁μ₀x̄cμ₀ + 5α = 0.05βσ = 28, power 90%, n = ?

A one-sided z test at α = 0.05 must detect a shift of the mean of 5 units with a power of 90%. The standard deviation is 28. Find the minimum sample size n.

α = 0.05power = 90%δ = 5σ = 28zα = 1.645zβ = 1.282
n =
7.Linear regression & correlationMedium
xy136240343453553xy244050

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 =
8.Linear regression & correlationHard
xy145246350454556662xy2465060x = 1.5

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. Predict ŷ for x = 1.5.

n = 6x = 1.5
ŷ =
9.Chi-square testsMedium
yesnoTotalmen6931100women352560Total10456160

The contingency table shows observed counts. Compute the chi-square statistic χ² for the test of independence between the row and column variables. Do not use a continuity correction.

N = 1602 × 2
χ² =
10.Chi-square testsHard

A chi-square test of independence is carried out on a contingency table with 2 rows and 6 columns (not counting the totals). How many degrees of freedom df does the test have?

2 × 6
df =

3Solutions

1.Normal distribution

Answer: P = 0.7865

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 = (22.4 − 32)/6 = −1.6
z₂ = (38 − 32)/6 = 1

Area between

P = Φ(1) − Φ(−1.6) = 0.8413 − 0.0548 = 0.7865

Answer

P = 0.7865
2.Normal distribution

Answer: P = 0.7881

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 = 52 − 48 = 4 s, σ = 5 s

Standardise

z = (0 − 4)/5 = −0.8

Probability

P = 1 − Φ(−0.8) = 0.7881

Answer

P = 0.7881
3.Confidence intervals

Answer: lower limit = 34.91

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 = 2.201

Margin of error

E = 2.201·2.5/√12 = 1.588

Interval

36.5 ± 1.588 → [34.91, 38.09]

Answer

lower limit = 34.91
4.Confidence intervals

Answer: n = 28906

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.

Solve for n

n = (2.576·33/0.5)² = 28905

Round up

A sample size is a whole number, and rounding down would make the margin too large.

n = 28906

Answer

n = 28906
5.Hypothesis tests & confidence intervals

Answer: z = 2.109

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 = √(14²/40 + 5²/50) = √5.4 = 2.324

Test statistic

z = (117 − 112.1)/2.324 = 2.109

Answer

z = 2.109

|z| > 1.96, so H₀ is rejected at the 5% level (two-sided).

6.Hypothesis tests & confidence intervals

Answer: n = 269

Sample size for a test: to detect a true shift δ with power 1 − β in a one-sided z test at level α, the standard error must satisfy δ = (z_α + z_β)·σ/√n, so n = ((z_α + z_β)·σ/δ)², rounded up.

Solve for n

n = ((1.645 + 1.282)·28/5)² = 268.7

Round up

n = 269

Answer

n = 269
7.Linear regression & correlation

Answer: r = 0.9634

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 = 225/5 = 45

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = 47
Syy = Σ(y − ȳ)² = 238

Correlation coefficient

r = 47/√(10·238) = 0.9634

Answer

r = 0.9634
8.Linear regression & correlation

Answer: ŷ = 45.37

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 = 313/6 = 52.17

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = 59.5

Regression line

b = 59.5/17.5 = 3.4
a = 52.17 − 3.4·3.5 = 40.27
ŷ = 40.27 + 3.4·x

Prediction

ŷ = 40.27 + 3.4·1.5 = 45.37

Answer

ŷ = 45.37
9.Chi-square tests

Answer: χ² = 1.875

Chi-square tests: compare observed counts O with the counts E expected under H₀ using χ² = Σ(O − E)²/E. Goodness of fit: E = n·p for each category, df = k − 1 (minus one more for each parameter estimated from the data). Independence in an r × c table: E = row total × column total / grand total, df = (r − 1)(c − 1). Reject H₀ when χ² exceeds the critical value for the chosen α.

Expected counts

men: 100·104/160 = 65, 100·56/160 = 35
women: 60·104/160 = 39, 60·56/160 = 21

Chi-square statistic

χ² = (69 − 65)²/65 + (31 − 35)²/35 + (35 − 39)²/39 + (25 − 21)²/21
χ² = 1.875

Answer

χ² = 1.875

df = 1. At α = 0.05 the critical value is 3.841, so H₀ (independence) is not rejected.

10.Chi-square tests

Answer: df = 5

Chi-square tests: compare observed counts O with the counts E expected under H₀ using χ² = Σ(O − E)²/E. Goodness of fit: E = n·p for each category, df = k − 1 (minus one more for each parameter estimated from the data). Independence in an r × c table: E = row total × column total / grand total, df = (r − 1)(c − 1). Reject H₀ when χ² exceeds the critical value for the chosen α.

Degrees of freedom

df = (2 − 1)·(6 − 1) = 5

Answer

df = 5

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