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
2943.1P = ?

The delivery times are normally distributed with mean 32 min and standard deviation 6 min. What fraction lies between 29 and 43.1 min? Give P as a decimal.

μ = 32σ = 6x1 = 29x2 = 43.1
P =
2.Normal distributionHard
0.02 mm0D − d?

Hole diameters D are normal with mean 25.02 mm and standard deviation 0.02 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 = 0.02 mmshaft: μ = 25 mm, σ = 0.01 mm
P =
3.Confidence intervalsMedium
n = 10, s = 2LU = ?x̄ = 78.5E95%

A sample of 10 items has mean 78.5 and standard deviation s = 2. The population is roughly normal. Find the upper limit of the 95% confidence interval for the mean.

n = 10x̄ = 78.5s = 295%t(9) = 2.262
upper limit =
4.Confidence intervalsHard
σ = 12, n = ?x̄ − Ex̄ + Ex̄E = 299%

The population standard deviation is 12. 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 2?

σ = 12E = 299%
n =
5.Hypothesis tests & confidence intervalsMedium
−22p = ?

The mean of a population is claimed to be 250; the population standard deviation is known to be 40. A random sample of 100 observations has a mean of 242. Test H₀: μ = 250 against H₁: μ ≠ 250. Find the p-value p.

μ₀ = 250σ = 40n = 100x̄ = 242H₁: μ ≠ μ₀
p =
6.Hypothesis tests & confidence intervalsHard
−1.961.96z = ?

In a trial, 175 of 250 patients on treatment A recover, and 160 of 250 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.

x₁ = 175n₁ = 250x₂ = 160n₂ = 250
z =
7.Linear regression & correlationMedium
xy22843464085310621267xy5104060

The table gives 6 paired observations of x = advertising spend in k€ and y = sales in k€. Find the coefficient of determination r² of the least-squares line.

n = 6
r² =
8.Linear regression & correlationHard
xy147016771885201022211324121xy15202580100120ŷ = 84

The table gives 6 paired observations of x = temperature in °C and y = ice creams sold. Use the least-squares line ŷ = a + b·x. For which value x does the line predict ŷ = 84?

n = 6ŷ = 84
x =
9.Chi-square testsMedium
busbikecarTotalstudents3592064staff27193581Total622855145

The contingency table shows observed counts. For a chi-square test of independence, find the expected count E for the cell “staff, car”.

row total = 81column total = 55N = 145
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 goodness-of-fit test checks whether counts in 9 categories follow a Poisson distribution whose mean is estimated from the same data. Using the table, find the critical value χ²crit at significance level α = 0.05.

α = 0.05
χ²crit =

3Solutions

1.Normal distribution

Answer: P = 0.6593

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 = (29 − 32)/6 = −0.5
z₂ = (43.1 − 32)/6 = 1.85

Area between

P = Φ(1.85) − Φ(−0.5) = 0.9678 − 0.3085 = 0.6593

Answer

P = 0.6593
2.Normal distribution

Answer: P = 0.8133

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 = 0.02236 mm

Standardise

z = (0 − 0.02)/0.02236 = −0.89

Probability

P = 1 − Φ(−0.89) = 0.8133

Answer

P = 0.8133
3.Confidence intervals

Answer: upper limit = 79.93

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.

Critical value

df = n − 1 = 9, t = 2.262

Margin of error

E = 2.262·2/√10 = 1.431

Interval

78.5 ± 1.431 → [77.07, 79.93]

Answer

upper limit = 79.93
4.Confidence intervals

Answer: n = 239

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 = (2.576·12/2)² = 238.9

Round up

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

n = 239

Answer

n = 239
5.Hypothesis tests & confidence intervals

Answer: p = 0.0455

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.

Test statistic

SE = σ/√n = 40/√100 = 4
z = (242 − 250)/4 = −2

Tail area

Φ(2) = 0.9772
1 − Φ(2) = 0.0228

p-value

p = 2·0.0228 = 0.0455

Answer

p = 0.0455

p ≤ 0.05, so H₀ is rejected at the 5% level.

6.Hypothesis tests & confidence intervals

Answer: z = 1.427

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 = 175/250 = 0.7
p̂₂ = 160/250 = 0.64
p̂ = 335/500 = 0.67

Standard error under H₀

SE = √(0.67·0.33·(1/250 + 1/250)) = 0.04206

Test statistic

z = (0.7 − 0.64)/0.04206 = 1.427

Answer

z = 1.427

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

7.Linear regression & correlation

Answer: r² = 0.9828

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 = 284/6 = 47.33

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 292
Syy = Σ(y − ȳ)² = 1239

Coefficient of determination

r² = 292²/(70·1239) = 0.9828

Answer

r² = 0.9828

About 98.28% of the variation in y is explained by the line.

8.Linear regression & correlation

Answer: x = 17.04

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 = 114/6 = 19
ȳ = 568/6 = 94.67

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 380

Regression line

b = 380/70 = 5.429
a = 94.67 − 5.429·19 = −8.476
ŷ = −8.476 + 5.429·x

Solve for x

x = (84 − (−8.476))/5.429 = 17.04

Answer

x = 17.04
9.Chi-square tests

Answer: E = 30.72

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 = 81·55/145 = 30.72

Answer

E = 30.72
10.Chi-square tests

Answer: χ²crit = 14.07

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 = 9 − 1 − 1 = 7

Table

χ²crit(df = 7, α = 0.05) = 14.067

Answer

χ²crit = 14.067

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

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