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
−3−2−10123zshaded area to the right of z = 0.9332

Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.9332. Find z to two decimals.

A = 0.9332
z =
2.Normal distributionHard
μ = 500x = ?10%

The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. Find the value x below which 10% of them lie (the 10th percentile).

μ = 500σ = 40P = 10%
x = h
3.Confidence intervalsMedium
μ₁ − μ₂0LUx̄₁ − x̄₂E = ?90%

The battery life of two phone models is compared. Model 1: 30 phones, mean 50.5 h, known standard deviation 15 h. Model 2: 100 phones, mean 52.5 h, known standard deviation 15 h. Find the margin of error E of the 90% confidence interval for μ₁ − μ₂.

n₁ = 30x̄₁ = 50.5 hσ₁ = 15 hn₂ = 100x̄₂ = 52.5 hσ₂ = 15 h90%
E = h
4.Confidence intervalsHard
σ = 13, n = ?x̄ − Ex̄ + Ex̄E = 1.590%

The population standard deviation is 13. How large a sample n do you need so that a 90% confidence interval for the mean has a margin of error of at most 1.5?

σ = 13E = 1.590%
n =
5.Hypothesis tests & confidence intervalsMedium
t −1.98401.984

A sample of 100 students has a mean score of 121 and a sample standard deviation of 15. Find the test statistic t for H₀: μ = 120.

μ₀ = 120x̄ = 121s = 15n = 100
t =
6.Hypothesis tests & confidence intervalsHard
H₀H₁μ₀x̄cμ₀ + 1.5α = 0.01βσ = 11, power 90%, n = ?

A one-sided z test at α = 0.01 must detect a shift of the mean of 1.5 units with a power of 90%. The standard deviation is 11. Find the minimum sample size n.

α = 0.01power = 90%δ = 1.5σ = 11zα = 2.326zβ = 1.282
n =
7.Linear regression & correlationMedium
xy126223328433531632xy2462530

The table gives 6 paired observations of x = advertising spend in k€ and y = sales in k€. Their sums are given. Find Pearson's correlation coefficient r from these sums.

n = 6Σx = 21Σy = 173Σx² = 91Σy² = 5063Σxy = 635
r =
8.Linear regression & correlationHard
xy1212203164135968xy246101520x = 1.5

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. A new observation (1.5, 23) is made. Find its residual e = y − ŷ.

n = 6(1.5, 23)
e =
9.Chi-square testsMedium
coughno coughTotalsmokers243660non-smokers3268100Total56104160

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 = 1602 × 2
χ² =
10.Chi-square testsHard

A chi-square goodness-of-fit test checks whether counts in 8 categories follow a Poisson distribution whose mean is estimated from the same data. How many degrees of freedom df does the test have?

k = 8
df =

3Solutions

1.Normal distribution

Answer: z = −1.5

From an area back to z: Φ(z) is the area to the left of z. For a left-tail area A, find z with Φ(z) = A in the table. A right-tail area A means Φ(z) = 1 − A. A central area A between −z and z leaves (1 − A)/2 in each tail, so Φ(z) = (1 + A)/2. For a value x, use x = μ + zσ.

Area to the left of z

Φ(z) = 1 − 0.9332 = 0.0668

Read the z-table backwards

Find 0.0668 inside the table and read off z.

z = −1.50

Answer

z = −1.50
2.Normal distribution

Answer: x = 448.7 h

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 for the percentile

Φ(z) = 0.1 → z = −1.282

Back to x

x = 500 + −1.282·40 = 448.74 h

Answer

x = 448.74 h
3.Confidence intervals

Answer: E = 5.137 h

Interval for a difference: estimate ± critical value × standard error. For two independent means with known σ: SE = √(σ₁²/n₁ + σ₂²/n₂) and z. With equal unknown variances: pooled s_p and t with n₁ + n₂ − 2 degrees of freedom. For paired data, work with the differences: d̄ ± t·s_d/√n.

Point estimate

x̄₁ − x̄₂ = 50.5 − 52.5 = −2

Standard error

SE = √(15²/30 + 15²/100) = 3.122

Margin of error

E = 1.645·3.122 = 5.137

Answer

E = 5.137 h

The interval contains 0, so the data do not show a significant difference.

4.Confidence intervals

Answer: n = 204

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 = (1.645·13/1.5)² = 203.3

Round up

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

n = 204

Answer

n = 204
5.Hypothesis tests & confidence intervals

Answer: t = 0.6667

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 = 15/√100 = 15/10 = 1.5

Test statistic

t = (121 − 120)/1.5 = 0.6667

Answer

t = 0.6667
t0.025; df = 99 = 1.984

|t| < 1.984, so H₀ is not rejected at the 5% level (two-sided).

6.Hypothesis tests & confidence intervals

Answer: n = 701

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)·11/1.5)² = 700.1

Round up

n = 701

Answer

n = 701
7.Linear regression & correlation

Answer: r = 0.8152

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 = 635 − 21·173/6 = 29.5
Sxx = 91 − 21²/6 = 17.5
Syy = 5063 − 173²/6 = 74.83

Correlation coefficient

r = 29.5/√(17.5·74.83) = 0.8152

Answer

r = 0.8152
8.Linear regression & correlation

Answer: e = 2.729

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 = 87/6 = 14.5

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −50.5

Regression line

b = −50.5/17.5 = −2.886
a = 14.5 − (−2.886)·3.5 = 24.6
ŷ = 24.6 − 2.886·x

Predicted value

ŷ = 24.6 + (−2.886)·1.5 = 20.27

Residual

e = 23 − 20.27 = 2.729

Answer

e = 2.729

The point lies above the line.

9.Chi-square tests

Answer: χ² = 1.055

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

smokers: 60·56/160 = 21, 60·104/160 = 39
non-smokers: 100·56/160 = 35, 100·104/160 = 65

Chi-square statistic

χ² = (24 − 21)²/21 + (36 − 39)²/39 + (32 − 35)²/35 + (68 − 65)²/65
χ² = 1.055

Answer

χ² = 1.055

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

10.Chi-square tests

Answer: df = 6

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 = 8 − 1 − 1 = 6

Answer

df = 6

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