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
76112.75P = ?

The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. What fraction lies between 76 and 112.75 points? Give P as a decimal.

μ = 100σ = 15x1 = 76x2 = 112.75
P =
2.Normal distributionHard
0.02 mm0D − d?

Hole diameters D are normal with mean 25.02 mm and standard deviation 0.01 mm. Shaft diameters d are independent and normal with mean 25 mm and standard deviation 0.01 mm. A randomly chosen shaft fits when D − d > 0. Find the probability P that it fits. Round z to two decimals.

hole: μ = 25.02 mm, σ = 0.01 mmshaft: μ = 25 mm, σ = 0.01 mm
P =
3.Confidence intervalsMedium
n = 7, sd​ = 5.20LU = ?d̄ = −4.3E99%

The blood pressure of 7 patients is measured before and after a treatment. The differences d = before − after (mmHg) have mean −4.3 and standard deviation 5.2. Find the upper limit of the 99% confidence interval for the mean difference.

n = 7d̄ = −4.3sd = 5.299%t(6) = 3.707
upper limit = mmHg
4.Confidence intervalsHard
σ = 17, n = ?x̄ − Ex̄ + Ex̄E = 495%

The population standard deviation is 17. 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 4?

σ = 17E = 495%
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 80.5 mm, and the line's known standard deviation is 3 mm. Line 2: a sample of 30 parts has mean 82.8 mm, known standard deviation 10 mm. Find the test statistic z for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (line 1 minus line 2).

n₁ = 40x̄₁ = 80.5 mmσ₁ = 3 mmn₂ = 30x̄₂ = 82.8 mmσ₂ = 10 mm
z =
6.Hypothesis tests & confidence intervalsHard
z −2.57602.576

A sample of 49 measurements has mean 50.8. The population standard deviation is 5. Find the upper limit U of the 99% confidence interval for the mean.

x̄ = 50.8σ = 5n = 4999%
U =
7.Linear regression & correlationMedium
xy254136208261032xy510102030

The table gives 5 paired observations of x = load in kN and y = extension in mm. Find Pearson's correlation coefficient r.

n = 5
r =
8.Linear regression & correlationHard
xy2741561882510341239xy51010203040x = 7

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. Predict ŷ for x = 7.

n = 6x = 7
ŷ =
9.Chi-square testsMedium
coughno coughTotalsmokers342660non-smokers5664120Total9090180

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 = 1802 × 2
χ² =
10.Chi-square testsHard
passfailTotalline 1421860line 2382260line 37426100Total15466220dfα = 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 = 2203 × 2α = 0.05
χ² =

3Solutions

1.Normal distribution

Answer: P = 0.7475

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 = (76 − 100)/15 = −1.6
z₂ = (112.75 − 100)/15 = 0.85

Area between

P = Φ(0.85) − Φ(−1.6) = 0.8023 − 0.0548 = 0.7475

Answer

P = 0.7475
2.Normal distribution

Answer: P = 0.9207

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 = 25.02 − 25 = 0.02 mm, σ = 0.01414 mm

Standardise

z = (0 − 0.02)/0.01414 = −1.41

Probability

P = 1 − Φ(−1.41) = 0.9207

Answer

P = 0.9207
3.Confidence intervals

Answer: upper limit = 2.986 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 = 6, t = 3.707

Margin of error

E = 3.707·5.2/√7 = 7.286

Interval

−4.3 ± 7.286 → [−11.59, 2.986]

Answer

upper limit = 2.986 mmHg
4.Confidence intervals

Answer: n = 70

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 = (1.96·17/4)² = 69.39

Round up

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

n = 70

Answer

n = 70
5.Hypothesis tests & confidence intervals

Answer: z = −1.219

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 = √(3²/40 + 10²/30) = √3.558 = 1.886

Test statistic

z = (80.5 − 82.8)/1.886 = −1.219

Answer

z = −1.219

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

6.Hypothesis tests & confidence intervals

Answer: U = 52.64

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 = 2.576·5/√49 = 1.84

Interval limit

U = 50.8 + 1.84 = 52.64

Answer

U = 52.64
7.Linear regression & correlation

Answer: r = 0.9979

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 = 96/5 = 19.2

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 134
Syy = Σ(y − ȳ)² = 450.8

Correlation coefficient

r = 134/√(40·450.8) = 0.9979

Answer

r = 0.9979
8.Linear regression & correlation

Answer: ŷ = 23

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 = 138/6 = 23

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 224

Regression line

b = 224/70 = 3.2
a = 23 − 3.2·7 = 0.6
ŷ = 0.6 + 3.2·x

Prediction

ŷ = 0.6 + 3.2·7 = 23

Answer

ŷ = 23
9.Chi-square tests

Answer: χ² = 1.6

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

smokers: 60·90/180 = 30, 60·90/180 = 30
non-smokers: 120·90/180 = 60, 120·90/180 = 60

Chi-square statistic

χ² = (34 − 30)²/30 + (26 − 30)²/30 + (56 − 60)²/60 + (64 − 60)²/60
χ² = 1.6

Answer

χ² = 1.6

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

10.Chi-square tests

Answer: χ² = 2.032

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: 42, 18
line 2: 42, 18
line 3: 70, 30

Chi-square statistic

χ² = (42 − 42)²/42 + (18 − 18)²/18 + (38 − 42)²/42 + (22 − 18)²/18 + (74 − 70)²/70 + (26 − 30)²/30
χ² = 2.032

Decision

df = 2, χ²crit = 5.991
2.032 < 5.991

Answer

χ² = 2.032

χ² is below the critical value, so H₀ is not rejected: there is no significant evidence of a relationship.

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