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
240260P = ?

The bag weights are normally distributed with mean 250 g and standard deviation 5 g. What fraction lies between 240 and 260 g? Give P as a decimal.

μ = 250σ = 5x1 = 240x2 = 260
P =
2.Normal distributionHard
μ = 250x = ?80%

The bag weights are normally distributed with mean 250 g and standard deviation 5 g. Find the value x below which 80% of them lie (the 80th percentile).

μ = 250σ = 5P = 80%
x = g
3.Confidence intervalsMedium
μ₁ − μ₂0LUx̄₁ − x̄₂E = ?90%

The battery life of two phone models is compared. Model 1: 60 phones, mean 125.5 h, known standard deviation 5 h. Model 2: 60 phones, mean 137.5 h, known standard deviation 13 h. Find the margin of error E of the 90% confidence interval for μ₁ − μ₂.

n₁ = 60x̄₁ = 125.5 hσ₁ = 5 hn₂ = 60x̄₂ = 137.5 hσ₂ = 13 h90%
E = h
4.Confidence intervalsHard
σ = 12, n = ?x̄ − Ex̄ + Ex̄E = 1.599%

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 1.5?

σ = 12E = 1.599%
n =
5.Hypothesis tests & confidence intervalsMedium
0.92p = ?

The mean of a population is claimed to be 250; the population standard deviation is known to be 6. A random sample of 36 observations has a mean of 250.92. Test H₀: μ = 250 against H₁: μ > 250. Find the p-value p.

μ₀ = 250σ = 6n = 36x̄ = 250.92H₁: μ > μ₀
p =
6.Hypothesis tests & confidence intervalsHard
H₀H₁μ₀x̄cμ₀ + 3α = 0.05βσ = 7, power 90%, n = ?

A one-sided z test at α = 0.05 must detect a shift of the mean of 3 units with a power of 90%. The standard deviation is 7. Find the minimum sample size n.

α = 0.05power = 90%δ = 3σ = 7zα = 1.645zβ = 1.282
n =
7.Linear regression & correlationMedium
xy23043764384910601271xy5104060

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
xy35756777899511104xy5106080100x = 6

The table gives 5 paired observations of x = hours studied and y = exam score. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 6.

n = 5x = 6
ŷ =
9.Chi-square testsMedium
busbikecarTotalstudents304941120staff145036100Total449977220

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 = 2202 × 3
χ² =
10.Chi-square testsHard
yesnoTotalline 1512980line 2641680line 34159100Total156104260dfα = 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 = 2603 × 2α = 0.05
χ² =

3Solutions

1.Normal distribution

Answer: P = 0.9545

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 = (240 − 250)/5 = −2
z₂ = (260 − 250)/5 = 2

Area between

P = Φ(2) − Φ(−2) = 0.9772 − 0.02275 = 0.9545

Answer

P = 0.9545
2.Normal distribution

Answer: x = 254.2 g

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) = 0.8 → z = 0.8416

Back to x

x = 250 + 0.8416·5 = 254.21 g

Answer

x = 254.21 g
3.Confidence intervals

Answer: E = 2.958 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 = 125.5 − 137.5 = −12

Standard error

SE = √(5²/60 + 13²/60) = 1.798

Margin of error

E = 1.645·1.798 = 2.958

Answer

E = 2.958 h

The interval does not contain 0, so the means differ significantly.

4.Confidence intervals

Answer: n = 425

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 = (2.576·12/1.5)² = 424.7

Round up

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

n = 425

Answer

n = 425
5.Hypothesis tests & confidence intervals

Answer: p = 0.1788

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 = 6/√36 = 1
z = (250.92 − 250)/1 = 0.92

Tail area

Φ(0.92) = 0.8212
1 − Φ(0.92) = 0.1788

p-value

p = 0.1788

Answer

p = 0.1788

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

6.Hypothesis tests & confidence intervals

Answer: n = 47

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 = ((1.645 + 1.282)·7/3)² = 46.64

Round up

n = 47

Answer

n = 47
7.Linear regression & correlation

Answer: r² = 0.9796

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 = 290/6 = 48.33

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 280
Syy = Σ(y − ȳ)² = 1143

Coefficient of determination

r² = 280²/(70·1143) = 0.9796

Answer

r² = 0.9796

About 97.96% of the variation in y is explained by the line.

8.Linear regression & correlation

Answer: ŷ = 74.1

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 = 35/5 = 7
ȳ = 401/5 = 80.2

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 244

Regression line

b = 244/40 = 6.1
a = 80.2 − 6.1·7 = 37.5
ŷ = 37.5 + 6.1·x

Prediction

ŷ = 37.5 + 6.1·6 = 74.1

Answer

ŷ = 74.1
9.Chi-square tests

Answer: χ² = 4.371

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

students: 120·44/220 = 24, 120·99/220 = 54, 120·77/220 = 42
staff: 100·44/220 = 20, 100·99/220 = 45, 100·77/220 = 35

Chi-square statistic

χ² = (30 − 24)²/24 + (49 − 54)²/54 + (41 − 42)²/42 + (14 − 20)²/20 + (50 − 45)²/45 + (36 − 35)²/35
χ² = 4.371

Answer

χ² = 4.371

df = 2. At α = 0.05 the critical value is 5.991, so H₀ (independence) is not rejected.

10.Chi-square tests

Answer: χ² = 28.84

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

line 1: 48, 32
line 2: 48, 32
line 3: 60, 40

Chi-square statistic

χ² = (51 − 48)²/48 + (29 − 32)²/32 + (64 − 48)²/48 + (16 − 32)²/32 + (41 − 60)²/60 + (59 − 40)²/40
χ² = 28.84

Decision

df = 2, χ²crit = 5.991
28.84 > 5.991

Answer

χ² = 28.84

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

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