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
53.583.5P = ?

The exam scores are normally distributed with mean 65 points and standard deviation 10 points. What fraction lies between 53.5 and 83.5 points? Give P as a decimal.

μ = 65σ = 10x1 = 53.5x2 = 83.5
P =
2.Normal distributionHard
μ = 500x = ?80%

The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. Find the value x below which 80% of them lie (the 80th percentile).

μ = 500σ = 40P = 80%
x = h
3.Confidence intervalsMedium
n = 13, sd​ = 5.20LUd̄ = 3E = ?95%

The blood pressure of 13 patients is measured before and after a treatment. The differences d = before − after (mmHg) have mean 3 and standard deviation 5.2. Find the margin of error E of the 95% confidence interval for the mean difference.

n = 13d̄ = 3sd = 5.295%t(12) = 2.179
E = mmHg
4.Confidence intervalsHard
p ≈ 0.1, n = ?p̂ − Ep̂ + Ep̂E = 0.0299%

A pilot study suggests that about 0.1 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 0.02?

p ≈ 0.1E = 0.0299%
n =
5.Hypothesis tests & confidence intervalsMedium
t −2.13102.131

A sample of 16 students has a mean score of 127.3 and a sample standard deviation of 10. Find the test statistic t for H₀: μ = 120.

μ₀ = 120x̄ = 127.3s = 10n = 16
t =
6.Hypothesis tests & confidence intervalsHard
H₀H₁μ₀x̄cμ₀ + 2.5α = 0.01βσ = 20, power 80%, n = ?

A one-sided z test at α = 0.01 must detect a shift of the mean of 2.5 units with a power of 80%. The standard deviation is 20. Find the minimum sample size n.

α = 0.01power = 80%δ = 2.5σ = 20zα = 2.326zβ = 0.842
n =
7.Linear regression & correlationMedium
xy122223319418513xy241520

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

n = 5
r =
8.Linear regression & correlationHard
xy274176198291034xy510102030x = 5

The table gives 5 paired observations of x = load in kN and y = extension in mm. Use the least-squares line ŷ = a + b·x. A new observation (5, 13) is made. Find its residual e = y − ŷ.

n = 5(5, 13)
e =
9.Chi-square testsMedium
lowmediumhighTotalurban272944100rural136126100Total409070200

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 = 2002 × 3
χ² =
10.Chi-square testsHard

A chi-square goodness-of-fit test compares observed counts in 10 categories with fully specified expected proportions (no parameters are estimated from the data). How many degrees of freedom df does the test have?

k = 10
df =

3Solutions

1.Normal distribution

Answer: P = 0.8428

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 = (53.5 − 65)/10 = −1.15
z₂ = (83.5 − 65)/10 = 1.85

Area between

P = Φ(1.85) − Φ(−1.15) = 0.9678 − 0.1251 = 0.8428

Answer

P = 0.8428
2.Normal distribution

Answer: x = 533.7 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) = 0.8 → z = 0.8416

Back to x

x = 500 + 0.8416·40 = 533.66 h

Answer

x = 533.66 h
3.Confidence intervals

Answer: E = 3.143 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 = 12, t = 2.179

Margin of error

E = 2.179·5.2/√13 = 3.143

Answer

E = 3.143 mmHg
4.Confidence intervals

Answer: n = 1494

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 = 2.576²·0.1·0.9/0.02² = 1493

Round up

n = 1494

Answer

n = 1494
5.Hypothesis tests & confidence intervals

Answer: t = 2.92

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/√16 = 10/4 = 2.5

Test statistic

t = (127.3 − 120)/2.5 = 2.92

Answer

t = 2.92
t0.025; df = 15 = 2.131

|t| > 2.131, so H₀ is rejected at the 5% level (two-sided).

6.Hypothesis tests & confidence intervals

Answer: n = 643

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 = ((2.326 + 0.842)·20/2.5)² = 642.3

Round up

n = 643

Answer

n = 643
7.Linear regression & correlation

Answer: r = −0.9237

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 = 95/5 = 19

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = −23
Syy = Σ(y − ȳ)² = 62

Correlation coefficient

r = −23/√(10·62) = −0.9237

Answer

r = −0.9237
8.Linear regression & correlation

Answer: e = −4.9

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 = 30/5 = 6
ȳ = 106/5 = 21.2

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 132

Regression line

b = 132/40 = 3.3
a = 21.2 − 3.3·6 = 1.4
ŷ = 1.4 + 3.3·x

Predicted value

ŷ = 1.4 + 3.3·5 = 17.9

Residual

e = 13 − 17.9 = −4.9

Answer

e = −4.9

The point lies below the line.

9.Chi-square tests

Answer: χ² = 20.91

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

urban: 100·40/200 = 20, 100·90/200 = 45, 100·70/200 = 35
rural: 100·40/200 = 20, 100·90/200 = 45, 100·70/200 = 35

Chi-square statistic

χ² = (27 − 20)²/20 + (29 − 45)²/45 + (44 − 35)²/35 + (13 − 20)²/20 + (61 − 45)²/45 + (26 − 35)²/35
χ² = 20.91

Answer

χ² = 20.91

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

10.Chi-square tests

Answer: df = 9

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 = 10 − 1 = 9

Answer

df = 9

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