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
480574P = ?

The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. What fraction lies between 480 and 574 h? Give P as a decimal.

μ = 500σ = 40x1 = 480x2 = 574
P =
2.Normal distributionHard
235240245250255260265gtop 95%

The bag weights are normally distributed with mean 250 g and standard deviation 5 g. Drag the cut-off so the shaded top 95% is to the right of it. Find the cut-off x.

μ = 250 gσ = 5 g95%
x = g
3.Confidence intervalsMedium
μ₁ − μ₂0LU = ?x̄₁ − x̄₂E90%

The battery life of two phone models is compared. Model 1: 80 phones, mean 141 h, known standard deviation 9 h. Model 2: 80 phones, mean 154 h, known standard deviation 7 h. Find the upper limit of the 90% confidence interval for μ₁ − μ₂.

n₁ = 80x̄₁ = 141 hσ₁ = 9 hn₂ = 80x̄₂ = 154 hσ₂ = 7 h90%
upper limit = h
4.Confidence intervalsHard
σ = 3, n = ?x̄ − Ex̄ + Ex̄E = 0.590%

The population standard deviation is 3. 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 0.5?

σ = 3E = 0.590%
n =
5.Hypothesis tests & confidence intervalsMedium
t −2.13102.131

A sample of 16 students has a mean score of 508.7 and a sample standard deviation of 12. Find the test statistic t for H₀: μ = 500.

μ₀ = 500x̄ = 508.7s = 12n = 16
t =
6.Hypothesis tests & confidence intervalsHard
−1.961.96z = ?

In a trial, 135 of 300 patients on treatment A recover, and 37 of 120 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.

x₁ = 135n₁ = 300x₂ = 37n₂ = 120
z =
7.Linear regression & correlationMedium
xy125219318418513xy24152025

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 = 93Σx² = 55Σy² = 1803Σxy = 254
r =
8.Linear regression & correlationHard
xy123222322421517619xy246182022x = 2.5

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 (2.5, 17) is made. Find its residual e = y − ŷ.

n = 6(2.5, 17)
e =
9.Chi-square testsMedium
busbikecarTotalstudents12383080staff285240120Total409070200

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 = 2002 × 3
χ² =
10.Chi-square testsHard
dfα = 0.05α = 0.0113.8416.63525.9919.21037.81511.34549.48813.277511.07015.086612.59216.812714.06718.475815.50720.090916.91921.6661018.30723.209

A chi-square goodness-of-fit test checks whether counts in 7 categories follow a Poisson distribution whose mean is estimated from the same data. Using the table, find the critical value χ²crit at significance level α = 0.01.

α = 0.01
χ²crit =

3Solutions

1.Normal distribution

Answer: P = 0.6593

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 = (480 − 500)/40 = −0.5
z₂ = (574 − 500)/40 = 1.85

Area between

P = Φ(1.85) − Φ(−0.5) = 0.9678 − 0.3085 = 0.6593

Answer

P = 0.6593
2.Normal distribution

Answer: x = 241.8 g

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

Φ(z) = 0.05

z from the table

z ≈ −1.64

Back to the original units

x = 250 − 1.64·5 = 241.8 g

Answer

x = 241.8 g
3.Confidence intervals

Answer: upper limit = −10.9 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 = 141 − 154 = −13

Standard error

SE = √(9²/80 + 7²/80) = 1.275

Margin of error

E = 1.645·1.275 = 2.097

Interval

−13 ± 2.097 → [−15.1, −10.9]

Answer

upper limit = −10.9 h

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

4.Confidence intervals

Answer: n = 98

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 = (1.645·3/0.5)² = 97.42

Round up

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

n = 98

Answer

n = 98
5.Hypothesis tests & confidence intervals

Answer: t = 2.9

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 = 12/√16 = 12/4 = 3

Test statistic

t = (508.7 − 500)/3 = 2.9

Answer

t = 2.9
t0.025; df = 15 = 2.131

|t| > 2.131, so H₀ is rejected at the 5% level (two-sided).

6.Hypothesis tests & confidence intervals

Answer: z = 2.667

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.

Sample proportions

p̂₁ = 135/300 = 0.45
p̂₂ = 37/120 = 0.3083
p̂ = 172/420 = 0.4095

Standard error under H₀

SE = √(0.4095·0.5905·(1/300 + 1/120)) = 0.05311

Test statistic

z = (0.45 − 0.3083)/0.05311 = 2.667

Answer

z = 2.667

|z| > 1.96, so H₀ is rejected at the 5% level (two-sided).

7.Linear regression & correlation

Answer: r = −0.924

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 = 254 − 15·93/5 = −25
Sxx = 55 − 15²/5 = 10
Syy = 1803 − 93²/5 = 73.2

Correlation coefficient

r = −25/√(10·73.2) = −0.924

Answer

r = −0.924
8.Linear regression & correlation

Answer: e = −4.695

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 = 124/6 = 20.67

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −18

Regression line

b = −18/17.5 = −1.029
a = 20.67 − (−1.029)·3.5 = 24.27
ŷ = 24.27 − 1.029·x

Predicted value

ŷ = 24.27 + (−1.029)·2.5 = 21.7

Residual

e = 17 − 21.7 = −4.695

Answer

e = −4.695

The point lies below the line.

9.Chi-square tests

Answer: χ² = 2.09

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: 80·40/200 = 16, 80·90/200 = 36, 80·70/200 = 28
staff: 120·40/200 = 24, 120·90/200 = 54, 120·70/200 = 42

Chi-square statistic

χ² = (12 − 16)²/16 + (38 − 36)²/36 + (30 − 28)²/28 + (28 − 24)²/24 + (52 − 54)²/54 + (40 − 42)²/42
χ² = 2.09

Answer

χ² = 2.09

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

10.Chi-square tests

Answer: χ²crit = 15.09

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 = 7 − 1 − 1 = 5

Table

χ²crit(df = 5, α = 0.01) = 15.086

Answer

χ²crit = 15.086

Reject H₀ if the computed χ² is larger than this value.

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