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
49.567.5P = ?

The exam scores are normally distributed with mean 65 points and standard deviation 10 points. What fraction lies between 49.5 and 67.5 points? Give P as a decimal.

μ = 65σ = 10x1 = 49.5x2 = 67.5
P =
2.Normal distributionHard
10 g0X − Y?

The mass X of a product from line 1 and Y from line 2 are independent and normal: X has mean 500 g and standard deviation 8 g, Y has mean 490 g and standard deviation 6 g. Find the probability P that X is greater than Y. Round z to two decimals.

μX = 500 g, σX = 8 gμY = 490 g, σY = 6 g
P =
3.Confidence intervalsMedium
x̄E₁ = 10.795%, n = 25x̄E₂ = ?90%, n = 225

A 95% confidence interval for a mean, based on 25 observations and a known σ, has a margin of error of 10.7. What is the margin of error E₂ of a 90% interval based on 225 observations?

95%n₁ = 25E₁ = 10.790%n₂ = 225
E₂ =
4.Confidence intervalsHard
σ = 32, n = ?x̄ − Ex̄ + Ex̄E = 599%

The population standard deviation is 32. How large a sample n do you need so that a 99% confidence interval for the mean has a margin of error of at most 5?

σ = 32E = 599%
n =
5.Hypothesis tests & confidence intervalsMedium
−2.2012.201t = ?df = 11

12 employees take a typing test before and after a training course. For each person the difference d = after − before (words per minute) is recorded. The mean difference is 6.14 and the standard deviation of the differences is 5.6. Find the paired test statistic t for H₀: μd = 0.

n = 12d̄ = 6.14sd = 5.6
t =
6.Hypothesis tests & confidence intervalsHard
−1.961.96z = ?

In a trial, 55 of 100 patients on treatment A recover, and 72 of 150 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.

x₁ = 55n₁ = 100x₂ = 72n₂ = 150
z =
7.Linear regression & correlationMedium
xy125226325423523621xy246222426

The table gives 6 paired observations of x = car age in years and y = price in k€. Find the coefficient of determination r² of the least-squares line.

n = 6
r² =
8.Linear regression & correlationHard
xy2941462382910341244xy5102040ŷ = 32

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. For which value x does the line predict ŷ = 32?

n = 6ŷ = 32
x =
9.Chi-square testsMedium
no defectminormajorTotalday shift37341990night shift44172485Total815143175

The contingency table shows observed counts. For a chi-square test of independence, find the expected count E for the cell “day shift, no defect”.

row total = 90column total = 81N = 175
E =
10.Chi-square testsHard
dfα = 0.05α = 0.0113.8416.63525.9919.21037.81511.34549.48813.277511.07015.086612.59216.812714.06718.475815.50720.090916.91921.6661018.30723.209

A chi-square test of independence is carried out on a contingency table with 2 rows and 2 columns (not counting the totals). Using the table, find the critical value χ²crit at significance level α = 0.05.

α = 0.05
χ²crit =

3Solutions

1.Normal distribution

Answer: P = 0.5381

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 = (49.5 − 65)/10 = −1.55
z₂ = (67.5 − 65)/10 = 0.25

Area between

P = Φ(0.25) − Φ(−1.55) = 0.5987 − 0.06057 = 0.5381

Answer

P = 0.5381
2.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 difference

μ = 500 − 490 = 10 g, σ = 10 g

Standardise

z = (0 − 10)/10 = −1

Probability

P = 1 − Φ(−1) = 0.8413

Answer

P = 0.8413
3.Confidence intervals

Answer: E₂ = 2.993

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 = (1.645/1.96)·√(25/225) = 0.2798

New margin

E₂ = 10.7·0.2798 = 2.993

Answer

E₂ = 2.993
4.Confidence intervals

Answer: n = 272

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 = (2.576·32/5)² = 271.8

Round up

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

n = 272

Answer

n = 272
5.Hypothesis tests & confidence intervals

Answer: t = 3.798

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 of the mean difference

SE = 5.6/√12 = 1.617

Test statistic

t = 6.14/1.617 = 3.798

Answer

t = 3.798

With df = 11, the two-sided 5% critical value is 2.201; |t| is larger, so H₀ is rejected.

6.Hypothesis tests & confidence intervals

Answer: z = 1.085

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.

Sample proportions

p̂₁ = 55/100 = 0.55
p̂₂ = 72/150 = 0.48
p̂ = 127/250 = 0.508

Standard error under H₀

SE = √(0.508·0.492·(1/100 + 1/150)) = 0.06454

Test statistic

z = (0.55 − 0.48)/0.06454 = 1.085

Answer

z = 1.085

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

7.Linear regression & correlation

Answer: r² = 0.8156

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 = 21/6 = 3.5
ȳ = 143/6 = 23.83

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −15.5
Syy = Σ(y − ȳ)² = 16.83

Coefficient of determination

r² = −15.5²/(17.5·16.83) = 0.8156

Answer

r² = 0.8156

About 81.56% of the variation in y is explained by the line.

8.Linear regression & correlation

Answer: x = 8.888

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 = 153/6 = 25.5

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 241

Regression line

b = 241/70 = 3.443
a = 25.5 − 3.443·7 = 1.4
ŷ = 1.4 + 3.443·x

Solve for x

x = (32 − 1.4)/3.443 = 8.888

Answer

x = 8.888
9.Chi-square tests

Answer: E = 41.66

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 count

E = 90·81/175 = 41.66

Answer

E = 41.66
10.Chi-square tests

Answer: χ²crit = 3.841

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 = (2 − 1)·(2 − 1) = 1

Table

χ²crit(df = 1, α = 0.05) = 3.841

Answer

χ²crit = 3.841

Reject H₀ if the computed χ² is larger than this value.

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