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
5085.5P = ?

The exam scores are normally distributed with mean 65 points and standard deviation 10 points. What fraction lies between 50 and 85.5 points? Give P as a decimal.

μ = 65σ = 10x1 = 50x2 = 85.5
P =
2.Normal distributionHard
μ = 32x = ?20%

The delivery times are normally distributed with mean 32 min and standard deviation 6 min. Find the value x below which 20% of them lie (the 20th percentile).

μ = 32σ = 6P = 20%
x = min
3.Confidence intervalsMedium
n = 6, s = 16L = ?Ux̄ = 70.5E99%

A sample of 6 items has mean 70.5 and standard deviation s = 16. The population is roughly normal. Find the lower limit of the 99% confidence interval for the mean.

n = 6x̄ = 70.5s = 1699%t(5) = 4.032
lower limit =
4.Confidence intervalsHard
p ≈ 0.6, n = ?p̂ − Ep̂ + Ep̂E = 0.0599%

A pilot study suggests that about 0.6 of products are defective. How many products n must be inspected so that a 99% confidence interval for the proportion has a margin of error of at most 0.05?

p ≈ 0.6E = 0.0599%
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 80 mm, and the line's known standard deviation is 4 mm. Line 2: a sample of 40 parts has mean 75.7 mm, known standard deviation 19 mm. Find the test statistic z for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (line 1 minus line 2).

n₁ = 40x̄₁ = 80 mmσ₁ = 4 mmn₂ = 40x̄₂ = 75.7 mmσ₂ = 19 mm
z =
6.Hypothesis tests & confidence intervalsHard
z −2.57602.576

A sample of 36 measurements has mean 196.3. The population standard deviation is 15. Find the lower limit L of the 99% confidence interval for the mean.

x̄ = 196.3σ = 15n = 3699%
L =
7.Linear regression & correlationMedium
xy124222321418518xy2418202224

The table gives 5 paired observations of x = car age in years and y = price in k€. Their sums are given. Find Pearson's correlation coefficient r from these sums.

n = 5Σx = 15Σy = 103Σx² = 55Σy² = 2149Σxy = 293
r =
8.Linear regression & correlationHard
xy1565177019912110023113xy152080100ŷ = 107

The table gives 5 paired observations of x = temperature in °C and y = ice creams sold. Use the least-squares line ŷ = a + b·x. For which value x does the line predict ŷ = 107?

n = 5ŷ = 107
x =
9.Chi-square testsMedium
no defectminormajorTotalday shift21233680night shift331640Total245442120

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 = 1202 × 3
χ² =
10.Chi-square testsHard
yesnoTotalage 18–3051960age 31–50251540age 51+362460Total11248160dfα = 0.05α = 0.0113.8416.63525.9919.21037.81511.34549.48813.277511.07015.086612.59216.812714.06718.475815.50720.090916.91921.6661018.30723.209

The table shows observed counts. Test at α = 0.01 whether the two variables are independent (no continuity correction). Compute the test statistic χ², then compare it with the critical value from the table.

N = 1603 × 2α = 0.01
χ² =

3Solutions

1.Normal distribution

Answer: P = 0.913

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 = (50 − 65)/10 = −1.5
z₂ = (85.5 − 65)/10 = 2.05

Area between

P = Φ(2.05) − Φ(−1.5) = 0.9798 − 0.06681 = 0.913

Answer

P = 0.913
2.Normal distribution

Answer: x = 26.95 min

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) = 0.2 → z = −0.8416

Back to x

x = 32 + −0.8416·6 = 26.95 min

Answer

x = 26.95 min
3.Confidence intervals

Answer: lower limit = 44.16

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 = 4.032

Margin of error

E = 4.032·16/√6 = 26.34

Interval

70.5 ± 26.34 → [44.16, 96.84]

Answer

lower limit = 44.16
4.Confidence intervals

Answer: n = 638

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 = 2.576²·0.6·0.4/0.05² = 637

Round up

n = 638

Answer

n = 638
5.Hypothesis tests & confidence intervals

Answer: z = 1.401

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 = √(4²/40 + 19²/40) = √9.425 = 3.07

Test statistic

z = (80 − 75.7)/3.07 = 1.401

Answer

z = 1.401

|z| < 1.96, so H₀ is not rejected at the 5% level (two-sided).

6.Hypothesis tests & confidence intervals

Answer: L = 189.9

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 = 2.576·15/√36 = 6.44

Interval limit

L = 196.3 − 6.44 = 189.9

Answer

L = 189.9
7.Linear regression & correlation

Answer: r = −0.9701

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.

Centred sums

Sxy = 293 − 15·103/5 = −16
Sxx = 55 − 15²/5 = 10
Syy = 2149 − 103²/5 = 27.2

Correlation coefficient

r = −16/√(10·27.2) = −0.9701

Answer

r = −0.9701
8.Linear regression & correlation

Answer: x = 22.05

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 = 95/5 = 19
ȳ = 439/5 = 87.8

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 252

Regression line

b = 252/40 = 6.3
a = 87.8 − 6.3·19 = −31.9
ŷ = −31.9 + 6.3·x

Solve for x

x = (107 − (−31.9))/6.3 = 22.05

Answer

x = 22.05
9.Chi-square tests

Answer: χ² = 25.63

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

day shift: 80·24/120 = 16, 80·54/120 = 36, 80·42/120 = 28
night shift: 40·24/120 = 8, 40·54/120 = 18, 40·42/120 = 14

Chi-square statistic

χ² = (21 − 16)²/16 + (23 − 36)²/36 + (36 − 28)²/28 + (3 − 8)²/8 + (31 − 18)²/18 + (6 − 14)²/14
χ² = 25.63

Answer

χ² = 25.63

df = 2. At α = 0.05 the critical value is 5.991, so H₀ (independence) is rejected.

10.Chi-square tests

Answer: χ² = 10.36

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

age 18–30: 42, 18
age 31–50: 28, 12
age 51+: 42, 18

Chi-square statistic

χ² = (51 − 42)²/42 + (9 − 18)²/18 + (25 − 28)²/28 + (15 − 12)²/12 + (36 − 42)²/42 + (24 − 18)²/18
χ² = 10.36

Decision

df = 2, χ²crit = 9.210
10.36 > 9.210

Answer

χ² = 10.36

χ² is above the critical value, so H₀ is rejected: the variables appear to be related.

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