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
76107.5P = ?

The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. What fraction lies between 76 and 107.5 points? Give P as a decimal.

μ = 100σ = 15x1 = 76x2 = 107.5
P =
2.Normal distributionHard
0.03 mm0D − d?

Hole diameters D are normal with mean 25.03 mm and standard deviation 0.02 mm. Shaft diameters d are independent and normal with mean 25 mm and standard deviation 0.02 mm. A randomly chosen shaft fits when D − d > 0. Find the probability P that it fits. Round z to two decimals.

hole: μ = 25.03 mm, σ = 0.02 mmshaft: μ = 25 mm, σ = 0.02 mm
P =
3.Confidence intervalsMedium
n = 10, s = 12.5L = ?Ux̄ = 112E99%

A sample of 10 items has mean 112 and standard deviation s = 12.5. The population is roughly normal. Find the lower limit of the 99% confidence interval for the mean.

n = 10x̄ = 112s = 12.599%t(9) = 3.25
lower limit =
4.Confidence intervalsHard
σ = 9, n = ?x̄ − Ex̄ + Ex̄E = 199%

The population standard deviation is 9. 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 1?

σ = 9E = 199%
n =
5.Hypothesis tests & confidence intervalsMedium
t −1.98401.984

A sample of 100 students has a mean score of 199 and a sample standard deviation of 4. Find the test statistic t for H₀: μ = 200.

μ₀ = 200x̄ = 199s = 4n = 100
t =
6.Hypothesis tests & confidence intervalsHard
H₀H₁μ₀x̄cμ₀ + 2.5α = 0.01βσ = 8, power 80%, n = ?

A one-sided z test at α = 0.01 must detect a shift of the mean of 2.5 units with a power of 80%. The standard deviation is 8. Find the minimum sample size n.

α = 0.01power = 80%δ = 2.5σ = 8zα = 2.326zβ = 0.842
n =
7.Linear regression & correlationMedium
xy126228324424522616xy2462025

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
xy125224324420519619xy246202224x = 5.5

The table gives 6 paired observations of x = car age in years and y = price in k€. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 5.5.

n = 6x = 5.5
ŷ =
9.Chi-square testsMedium
improvednot improvedTotalnew drug172340placebo332760Total5050100

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 = 1002 × 2
χ² =
10.Chi-square testsHard

A chi-square test of independence is carried out on a contingency table with 3 rows and 3 columns (not counting the totals). How many degrees of freedom df does the test have?

3 × 3
df =

3Solutions

1.Normal distribution

Answer: P = 0.6367

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 = (107.5 − 100)/15 = 0.5

Area between

P = Φ(0.5) − Φ(−1.6) = 0.6915 − 0.0548 = 0.6367

Answer

P = 0.6367
2.Normal distribution

Answer: P = 0.8554

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 = 25.03 − 25 = 0.03 mm, σ = 0.02828 mm

Standardise

z = (0 − 0.03)/0.02828 = −1.06

Probability

P = 1 − Φ(−1.06) = 0.8554

Answer

P = 0.8554
3.Confidence intervals

Answer: lower limit = 99.15

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 = 3.25

Margin of error

E = 3.25·12.5/√10 = 12.85

Interval

112 ± 12.85 → [99.15, 124.8]

Answer

lower limit = 99.15
4.Confidence intervals

Answer: n = 538

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 = (2.576·9/1)² = 537.5

Round up

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

n = 538

Answer

n = 538
5.Hypothesis tests & confidence intervals

Answer: t = −2.5

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

SE = 4/√100 = 4/10 = 0.4

Test statistic

t = (199 − 200)/0.4 = −2.5

Answer

t = −2.5
t0.025; df = 99 = 1.984

|t| > 1.984, so H₀ is rejected at the 5% level (two-sided).

6.Hypothesis tests & confidence intervals

Answer: n = 103

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 + 0.842)·8/2.5)² = 102.8

Round up

n = 103

Answer

n = 103
7.Linear regression & correlation

Answer: r² = 0.7741

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 = 140/6 = 23.33

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −34
Syy = Σ(y − ȳ)² = 85.33

Coefficient of determination

r² = −34²/(17.5·85.33) = 0.7741

Answer

r² = 0.7741

About 77.41% of the variation in y is explained by the line.

8.Linear regression & correlation

Answer: ŷ = 19.03

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 = 131/6 = 21.83

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −24.5

Regression line

b = −24.5/17.5 = −1.4
a = 21.83 − (−1.4)·3.5 = 26.73
ŷ = 26.73 − 1.4·x

Prediction

ŷ = 26.73 + (−1.4)·5.5 = 19.03

Answer

ŷ = 19.03
9.Chi-square tests

Answer: χ² = 1.5

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: 40·50/100 = 20, 40·50/100 = 20
placebo: 60·50/100 = 30, 60·50/100 = 30

Chi-square statistic

χ² = (17 − 20)²/20 + (23 − 20)²/20 + (33 − 30)²/30 + (27 − 30)²/30
χ² = 1.5

Answer

χ² = 1.5

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

10.Chi-square tests

Answer: df = 4

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 = (3 − 1)·(3 − 1) = 4

Answer

df = 4

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