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
105 s98.6 sX + Y?

X is normal with mean 60 s and standard deviation 5 s; Y is independent and normal with mean 45 s and standard deviation 4 s. Find the probability P that X + Y is more than 98.6 s. Round z to two decimals.

μX = 60, σX = 5μY = 45, σY = 4X + Y > 98.6 s
P =
2.Normal distributionHard
μ = 65x = ?75%

The exam scores are normally distributed with mean 65 points and standard deviation 10 points. Find the value x below which 75% of them lie (the 75th percentile).

μ = 65σ = 10P = 75%
x = points
3.Confidence intervalsMedium
x̄E₁ = 1295%, n = 25x̄E₂ = ?99%, n = 25

A 95% confidence interval for a mean, based on 25 observations and a known σ, has a margin of error of 12. What is the margin of error E₂ of a 99% interval from the same data?

95%n₁ = 25E₁ = 1299%
E₂ =
4.Confidence intervalsHard
σ = 33, n = ?x̄ − Ex̄ + Ex̄E = 0.595%

The population standard deviation is 33. 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 0.5?

σ = 33E = 0.595%
n =
5.Hypothesis tests & confidence intervalsMedium
−2.1312.131t = ?df = 15

Two teaching methods are compared. Group 1: 10 students, mean score 31, standard deviation 10. Group 2: 7 students, mean score 38.1, standard deviation 9. Assume equal population variances and use the pooled two-sample t test. Find the test statistic t for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (group 1 minus group 2).

n₁ = 10x̄₁ = 31s₁ = 10n₂ = 7x̄₂ = 38.1s₂ = 9
t =
6.Hypothesis tests & confidence intervalsHard
z −1.9601.96

A sample of 81 measurements has mean 499.5. The population standard deviation is 4. Find the upper limit U of the 95% confidence interval for the mean.

x̄ = 499.5σ = 4n = 8195%
U =
7.Linear regression & correlationMedium
xy137243347448556xy244050

The table gives 5 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 = 5Σx = 15Σy = 231Σx² = 55Σy² = 10867Σxy = 736
r =
8.Linear regression & correlationHard
xy24456781010131217xy51051015x = 3

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. A new observation (3, 6) is made. Find its residual e = y − ŷ.

n = 6(3, 6)
e =
9.Chi-square testsMedium
improvednot improvedTotalnew drug483280placebo154560Total6377140

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 = 1402 × 2
χ² =
10.Chi-square testsHard
passfailTotalplant A433780plant B305080plant C372360Total110110220dfα = 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 = 2203 × 2α = 0.01
χ² =

3Solutions

1.Normal distribution

Answer: P = 0.8413

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 sum

μ = 105 s, σ = √(5² + 4²) = 6.403 s

Standardise

z = (98.6 − 105)/6.403 = −1

Upper tail

P = 1 − Φ(−1) = 1 − 0.1587 = 0.8413

Answer

P = 0.8413
2.Normal distribution

Answer: x = 71.75 points

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) = 0.75 → z = 0.6745

Back to x

x = 65 + 0.6745·10 = 71.75 points

Answer

x = 71.75 points
3.Confidence intervals

Answer: E₂ = 15.77

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 = (2.576/1.96)·√(25/25) = 1.314

New margin

E₂ = 12·1.314 = 15.77

Answer

E₂ = 15.77
4.Confidence intervals

Answer: n = 16735

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 = (1.96·33/0.5)² = 16734

Round up

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

n = 16735

Answer

n = 16735
5.Hypothesis tests & confidence intervals

Answer: t = −1.499

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.

Pooled variance

sp² = (9·10² + 6·9²)/15 = 92.4
sp = 9.612

Standard error

SE = 9.612·√(1/10 + 1/7) = 4.737

Test statistic

t = (31 − 38.1)/4.737 = −1.499

Answer

t = −1.499

With df = 15, the two-sided 5% critical value is 2.131; |t| is smaller, so H₀ is not rejected.

6.Hypothesis tests & confidence intervals

Answer: U = 500.4

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).

Margin of error

E = 1.96·4/√81 = 0.8711

Interval limit

U = 499.5 + 0.8711 = 500.4

Answer

U = 500.4
7.Linear regression & correlation

Answer: r = 0.9743

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 = 736 − 15·231/5 = 43
Sxx = 55 − 15²/5 = 10
Syy = 10867 − 231²/5 = 194.8

Correlation coefficient

r = 43/√(10·194.8) = 0.9743

Answer

r = 0.9743
8.Linear regression & correlation

Answer: e = 1.924

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 = 56/6 = 9.333

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 92

Regression line

b = 92/70 = 1.314
a = 9.333 − 1.314·7 = 0.1333
ŷ = 0.1333 + 1.314·x

Predicted value

ŷ = 0.1333 + 1.314·3 = 4.076

Residual

e = 6 − 4.076 = 1.924

Answer

e = 1.924

The point lies above the line.

9.Chi-square tests

Answer: χ² = 16.97

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

new drug: 80·63/140 = 36, 80·77/140 = 44
placebo: 60·63/140 = 27, 60·77/140 = 33

Chi-square statistic

χ² = (48 − 36)²/36 + (32 − 44)²/44 + (15 − 27)²/27 + (45 − 33)²/33
χ² = 16.97

Answer

χ² = 16.97

df = 1. At α = 0.05 the critical value is 3.841, so H₀ (independence) is rejected.

10.Chi-square tests

Answer: χ² = 8.717

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

plant A: 40, 40
plant B: 40, 40
plant C: 30, 30

Chi-square statistic

χ² = (43 − 40)²/40 + (37 − 40)²/40 + (30 − 40)²/40 + (50 − 40)²/40 + (37 − 30)²/30 + (23 − 30)²/30
χ² = 8.717

Decision

df = 2, χ²crit = 9.210
8.717 < 9.210

Answer

χ² = 8.717

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

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