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.

New problemsBack to the course

In the print window, choose “Save as PDF” to keep a copy on your device.

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
462582P = ?

The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. What fraction lies between 462 and 582 h? Give P as a decimal.

μ = 500σ = 40x1 = 462x2 = 582
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 0.015 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 = 0.015 mm
P =
3.Confidence intervalsMedium
n = 13, s = 10.5L = ?Ux̄ = 65.5E99%

A sample of 13 items has mean 65.5 and standard deviation s = 10.5. The population is roughly normal. Find the lower limit of the 99% confidence interval for the mean.

n = 13x̄ = 65.5s = 10.599%t(12) = 3.055
lower limit =
4.Confidence intervalsHard
σ = 16, n = ?x̄ − Ex̄ + Ex̄E = 595%

The population standard deviation is 16. 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 = 16E = 595%
n =
5.Hypothesis tests & confidence intervalsMedium
−1.22p = ?

The mean of a population is claimed to be 250; the population standard deviation is known to be 12. A random sample of 36 observations has a mean of 247.56. Test H₀: μ = 250 against H₁: μ < 250. Find the p-value p.

μ₀ = 250σ = 12n = 36x̄ = 247.56H₁: μ < μ₀
p =
6.Hypothesis tests & confidence intervalsHard
H₀H₁μ₀x̄cμ₀ + 5α = 0.01βσ = 24, power 90%, n = ?

A one-sided z test at α = 0.01 must detect a shift of the mean of 5 units with a power of 90%. The standard deviation is 24. Find the minimum sample size n.

α = 0.01power = 90%δ = 5σ = 24zα = 2.326zβ = 1.282
n =
7.Linear regression & correlationMedium
xy21041362182910341238xy51010203040

The table gives 6 paired observations of x = load in kN and y = extension in mm. Find Pearson's correlation coefficient r.

n = 6
r =
8.Linear regression & correlationHard
xy141347561762973xy540506070ŷ = 69

The table gives 5 paired observations of x = hours studied and y = exam score. Use the least-squares line ŷ = a + b·x. For which value x does the line predict ŷ = 69?

n = 5ŷ = 69
x =
9.Chi-square testsMedium
busbikecarTotalstudents32202577staff40171875Total723743152

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

row total = 77column total = 37N = 152
E =
10.Chi-square testsHard
passfailTotalage 18–3056460age 31–505347100age 51+451560Total15466220dfα = 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 = 0.8088

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 = (462 − 500)/40 = −0.95
z₂ = (582 − 500)/40 = 2.05

Area between

P = Φ(2.05) − Φ(−0.95) = 0.9798 − 0.1711 = 0.8088

Answer

P = 0.8088
2.Normal distribution

Answer: P = 0.8665

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 = 0.01803 mm

Standardise

z = (0 − 0.02)/0.01803 = −1.11

Probability

P = 1 − Φ(−1.11) = 0.8665

Answer

P = 0.8665
3.Confidence intervals

Answer: lower limit = 56.6

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 = 12, t = 3.055

Margin of error

E = 3.055·10.5/√13 = 8.897

Interval

65.5 ± 8.897 → [56.6, 74.4]

Answer

lower limit = 56.6
4.Confidence intervals

Answer: n = 40

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 = (1.96·16/5)² = 39.34

Round up

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

n = 40

Answer

n = 40
5.Hypothesis tests & confidence intervals

Answer: p = 0.1112

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 = 12/√36 = 2
z = (247.56 − 250)/2 = −1.22

Tail area

Φ(1.22) = 0.8888
1 − Φ(1.22) = 0.1112

p-value

p = 0.1112

Answer

p = 0.1112

p > 0.05, so H₀ is not rejected at the 5% level.

6.Hypothesis tests & confidence intervals

Answer: n = 300

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 + 1.282)·24/5)² = 299.9

Round up

n = 300

Answer

n = 300
7.Linear regression & correlation

Answer: r = 0.9916

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 = 145/6 = 24.17

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 211
Syy = Σ(y − ȳ)² = 646.8

Correlation coefficient

r = 211/√(70·646.8) = 0.9916

Answer

r = 0.9916
8.Linear regression & correlation

Answer: x = 8.089

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 = 25/5 = 5
ȳ = 284/5 = 56.8

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 158

Regression line

b = 158/40 = 3.95
a = 56.8 − 3.95·5 = 37.05
ŷ = 37.05 + 3.95·x

Solve for x

x = (69 − 37.05)/3.95 = 8.089

Answer

x = 8.089
9.Chi-square tests

Answer: E = 18.74

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 = 77·37/152 = 18.74

Answer

E = 18.74
10.Chi-square tests

Answer: χ² = 30.03

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

age 18–30: 42, 18
age 31–50: 70, 30
age 51+: 42, 18

Chi-square statistic

χ² = (56 − 42)²/42 + (4 − 18)²/18 + (53 − 70)²/70 + (47 − 30)²/30 + (45 − 42)²/42 + (15 − 18)²/18
χ² = 30.03

Decision

df = 2, χ²crit = 5.991
30.03 > 5.991

Answer

χ² = 30.03

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

ExamProblems · examproblems.com/courses/statistics-2-inference