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
30.239.5P = ?

The delivery times are normally distributed with mean 32 min and standard deviation 6 min. What fraction lies between 30.2 and 39.5 min? Give P as a decimal.

μ = 32σ = 6x1 = 30.2x2 = 39.5
P =
2.Normal distributionHard
4 s0X − Y?

Runner A's lap time X and runner B's lap time Y are independent and normal: X has mean 52 s and standard deviation 4 s, Y has mean 48 s and standard deviation 3 s. Find the probability P that X is greater than Y. Round z to two decimals.

μX = 52 s, σX = 4 sμY = 48 s, σY = 3 s
P =
3.Confidence intervalsMedium
n = 13, s = 6LUx̄ = 107.5E = ?90%

A sample of 13 items has mean 107.5 and standard deviation s = 6. The population is roughly normal. Find the margin of error E of a 90% confidence interval for the mean.

n = 13x̄ = 107.5s = 690%t(12) = 1.782
E =
4.Confidence intervalsHard
σ = 7, n = ?x̄ − Ex̄ + Ex̄E = 2.595%

The population standard deviation is 7. 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 2.5?

σ = 7E = 2.595%
n =
5.Hypothesis tests & confidence intervalsMedium
t −1.9901.99

A sample of 81 students has a mean score of 73.84 and a sample standard deviation of 4. Find the test statistic t for H₀: μ = 75.

μ₀ = 75x̄ = 73.84s = 4n = 81
t =
6.Hypothesis tests & confidence intervalsHard
z −1.64501.645

A sample of 36 measurements has mean 206.8. The population standard deviation is 15. Find the lower limit L of the 90% confidence interval for the mean.

x̄ = 206.8σ = 15n = 3690%
L =
7.Linear regression & correlationMedium
xy126223324423521xy24222426

The table gives 5 paired observations of x = car age in years and y = price in k€. Their sums are given. Find Pearson's correlation coefficient r from these sums.

n = 5Σx = 15Σy = 117Σx² = 55Σy² = 2751Σxy = 341
r =
8.Linear regression & correlationHard
xy25476108151019xy5105101520x = 7

The table gives 5 paired observations of x = load in kN and y = extension in mm. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 7.

n = 5x = 7
ŷ =
9.Chi-square testsMedium
busbikecarTotalstudents19272268staff992644Total283648112

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 = 68column total = 36N = 112
E =
10.Chi-square testsHard
yesnoTotalage 18–30192140age 31–50322860age 51+33740Total8456140dfα = 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 = 1403 × 2α = 0.01
χ² =

3Solutions

1.Normal distribution

Answer: P = 0.5123

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 = (30.2 − 32)/6 = −0.3
z₂ = (39.5 − 32)/6 = 1.25

Area between

P = Φ(1.25) − Φ(−0.3) = 0.8944 − 0.3821 = 0.5123

Answer

P = 0.5123
2.Normal distribution

Answer: P = 0.7881

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 = 52 − 48 = 4 s, σ = 5 s

Standardise

z = (0 − 4)/5 = −0.8

Probability

P = 1 − Φ(−0.8) = 0.7881

Answer

P = 0.7881
3.Confidence intervals

Answer: E = 2.965

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 = 1.782

Margin of error

E = 1.782·6/√13 = 2.965

Answer

E = 2.965
4.Confidence intervals

Answer: n = 31

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 = (1.96·7/2.5)² = 30.12

Round up

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

n = 31

Answer

n = 31
5.Hypothesis tests & confidence intervals

Answer: t = −2.61

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/√81 = 4/9 = 0.4444

Test statistic

t = (73.84 − 75)/0.4444 = −2.61

Answer

t = −2.61
t0.025; df = 80 = 1.99

|t| > 1.99, so H₀ is rejected at the 5% level (two-sided).

6.Hypothesis tests & confidence intervals

Answer: L = 202.7

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 = 1.645·15/√36 = 4.112

Interval limit

L = 206.8 − 4.112 = 202.7

Answer

L = 202.7
7.Linear regression & correlation

Answer: r = −0.8704

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 = 341 − 15·117/5 = −10
Sxx = 55 − 15²/5 = 10
Syy = 2751 − 117²/5 = 13.2

Correlation coefficient

r = −10/√(10·13.2) = −0.8704

Answer

r = −0.8704
8.Linear regression & correlation

Answer: ŷ = 13

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 = 30/5 = 6
ȳ = 56/5 = 11.2

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 72

Regression line

b = 72/40 = 1.8
a = 11.2 − 1.8·6 = 0.4
ŷ = 0.4 + 1.8·x

Prediction

ŷ = 0.4 + 1.8·7 = 13

Answer

ŷ = 13
9.Chi-square tests

Answer: E = 21.86

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 = 68·36/112 = 21.86

Answer

E = 21.86
10.Chi-square tests

Answer: χ² = 12.15

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: 24, 16
age 31–50: 36, 24
age 51+: 24, 16

Chi-square statistic

χ² = (19 − 24)²/24 + (21 − 16)²/16 + (32 − 36)²/36 + (28 − 24)²/24 + (33 − 24)²/24 + (7 − 16)²/16
χ² = 12.15

Decision

df = 2, χ²crit = 9.210
12.15 > 9.210

Answer

χ² = 12.15

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

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