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
246256.25P = ?

The bag weights are normally distributed with mean 250 g and standard deviation 5 g. What fraction lies between 246 and 256.25 g? Give P as a decimal.

μ = 250σ = 5x1 = 246x2 = 256.25
P =
2.Normal distributionHard
0.03 mm0D − d?

Hole diameters D are normal with mean 25.03 mm and standard deviation 0.015 mm. Shaft diameters d are independent and normal with mean 25 mm and standard deviation 0.02 mm. A randomly chosen shaft fits when D − d > 0. Find the probability P that it fits. Round z to two decimals.

hole: μ = 25.03 mm, σ = 0.015 mmshaft: μ = 25 mm, σ = 0.02 mm
P =
3.Confidence intervalsMedium
x̄E₁ = 10.999%, n = 100x̄E₂ = ?95%, n = 25

A 99% confidence interval for a mean, based on 100 observations and a known σ, has a margin of error of 10.9. What is the margin of error E₂ of a 95% interval based on 25 observations?

99%n₁ = 100E₁ = 10.995%n₂ = 25
E₂ =
4.Confidence intervalsHard
σ = 14, n = ?x̄ − Ex̄ + Ex̄E = 595%

The population standard deviation is 14. How large a sample n do you need so that a 95% confidence interval for the mean has a margin of error of at most 5?

σ = 14E = 595%
n =
5.Hypothesis tests & confidence intervalsMedium
t −1.98401.984

A sample of 100 students has a mean score of 203.6 and a sample standard deviation of 15. Find the test statistic t for H₀: μ = 200.

μ₀ = 200x̄ = 203.6s = 15n = 100
t =
6.Hypothesis tests & confidence intervalsHard
H₀H₁μ₀x̄cμ₀ + 4α = 0.01βσ = 17, power 95%, n = ?

A one-sided z test at α = 0.01 must detect a shift of the mean of 4 units with a power of 95%. The standard deviation is 17. Find the minimum sample size n.

α = 0.01power = 95%δ = 4σ = 17zα = 2.326zβ = 1.645
n =
7.Linear regression & correlationMedium
xy34255376698111891395xy510406080100

The table gives 6 paired observations of x = advertising spend in k€ and y = sales in k€. Their sums are given. Find Pearson's correlation coefficient r from these sums.

n = 6Σx = 48Σy = 426Σx² = 454Σy² = 32436Σxy = 3796
r =
8.Linear regression & correlationHard
xy23476128181023xy5101020x = 7

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. Predict ŷ for x = 7.

n = 5x = 7
ŷ =
9.Chi-square testsMedium
lowmediumhighTotalurban264846120rural6241040Total327256160

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 × 3
χ² =
10.Chi-square testsHard
yesnoTotalline 1661480line 2211940line 3251540Total11248160dfα = 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 = 0.05 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 = 0.05
χ² =

3Solutions

1.Normal distribution

Answer: P = 0.6825

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 = (246 − 250)/5 = −0.8
z₂ = (256.25 − 250)/5 = 1.25

Area between

P = Φ(1.25) − Φ(−0.8) = 0.8944 − 0.2119 = 0.6825

Answer

P = 0.6825
2.Normal distribution

Answer: P = 0.8849

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 = 25.03 − 25 = 0.03 mm, σ = 0.025 mm

Standardise

z = (0 − 0.03)/0.025 = −1.2

Probability

P = 1 − Φ(−1.2) = 0.8849

Answer

P = 0.8849
3.Confidence intervals

Answer: E₂ = 16.59

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.

Ratio of margins

E₂/E₁ = (1.96/2.576)·√(100/25) = 1.522

New margin

E₂ = 10.9·1.522 = 16.59

Answer

E₂ = 16.59
4.Confidence intervals

Answer: n = 31

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 = (1.96·14/5)² = 30.12

Round up

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

n = 31

Answer

n = 31
5.Hypothesis tests & confidence intervals

Answer: t = 2.4

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 = 15/√100 = 15/10 = 1.5

Test statistic

t = (203.6 − 200)/1.5 = 2.4

Answer

t = 2.4
t0.025; df = 99 = 1.984

|t| > 1.984, so H₀ is rejected at the 5% level (two-sided).

6.Hypothesis tests & confidence intervals

Answer: n = 285

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 + 1.645)·17/4)² = 284.8

Round up

n = 285

Answer

n = 285
7.Linear regression & correlation

Answer: r = 0.991

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 = 3796 − 48·426/6 = 388
Sxx = 454 − 48²/6 = 70
Syy = 32436 − 426²/6 = 2190

Correlation coefficient

r = 388/√(70·2190) = 0.991

Answer

r = 0.991
8.Linear regression & correlation

Answer: ŷ = 15.15

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 = 63/5 = 12.6

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 102

Regression line

b = 102/40 = 2.55
a = 12.6 − 2.55·6 = −2.7
ŷ = −2.7 + 2.55·x

Prediction

ŷ = −2.7 + 2.55·7 = 15.15

Answer

ŷ = 15.15
9.Chi-square tests

Answer: χ² = 4.857

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: 120·32/160 = 24, 120·72/160 = 54, 120·56/160 = 42
rural: 40·32/160 = 8, 40·72/160 = 18, 40·56/160 = 14

Chi-square statistic

χ² = (26 − 24)²/24 + (48 − 54)²/54 + (46 − 42)²/42 + (6 − 8)²/8 + (24 − 18)²/18 + (10 − 14)²/14
χ² = 4.857

Answer

χ² = 4.857

df = 2. At α = 0.05 the critical value is 5.991, so H₀ (independence) is not rejected.

10.Chi-square tests

Answer: χ² = 12.86

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

line 1: 56, 24
line 2: 28, 12
line 3: 28, 12

Chi-square statistic

χ² = (66 − 56)²/56 + (14 − 24)²/24 + (21 − 28)²/28 + (19 − 12)²/12 + (25 − 28)²/28 + (15 − 12)²/12
χ² = 12.86

Decision

df = 2, χ²crit = 5.991
12.86 > 5.991

Answer

χ² = 12.86

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

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